{"id":1070,"date":"2021-11-01T19:17:23","date_gmt":"2021-11-01T19:17:23","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=1070"},"modified":"2022-11-01T04:31:46","modified_gmt":"2022-11-01T04:31:46","slug":"triple-integrals-in-spherical-coordinates","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/triple-integrals-in-spherical-coordinates\/","title":{"raw":"Triple Integrals in Spherical Coordinates","rendered":"Triple Integrals in Spherical Coordinates"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul class=\"os-abstract\">\r\n \t<li><span class=\"os-abstract-content\">Evaluate a triple integral by changing to spherical coordinates.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 data-type=\"title\">Review of Spherical Coordinates<\/h2>\r\nIn three-dimensional space [latex]{\\mathbb{R}^{3}}[\/latex] in the spherical coordinate system, we specify a point [latex]P[\/latex] by its distance [latex]{\\rho}[\/latex] from the origin, the polar angle [latex]{\\theta}[\/latex] from the positive [latex]x[\/latex]-axis (same as in the cylindrical coordinate system), and the angle [latex]{\\varphi}[\/latex] from the positive [latex]z[\/latex]-axis and the line [latex]OP[\/latex] (Figure 1). Note that [latex]{\\rho} \\ {\\geq} \\ {0}[\/latex] and [latex]{0} \\ {\\leq} \\ {\\varphi} \\ {\\leq} \\ {\\pi}[\/latex]. (Refer to\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/introduction-to-cylindrical-and-spherical-coordinates\/\" data-page-slug=\"2-7-cylindrical-and-spherical-coordinates\" data-page-uuid=\"293cc618-b0de-4ec9-9029-6efb4eb1b172\" data-page-fragment=\"page_293cc618-b0de-4ec9-9029-6efb4eb1b172\">Cylindrical and Spherical Coordinates<\/a>\u00a0for a review.) Spherical coordinates are useful for triple integrals over regions that are symmetric with respect to the origin.\r\n\r\n[caption id=\"attachment_1396\" align=\"aligncenter\" width=\"394\"]<img class=\"size-full wp-image-1396\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26173418\/5-5-6.jpeg\" alt=\"A depiction of the spherical coordinate system: a point (x, y, z) is shown, which is equal to (rho, theta, phi) in spherical coordinates. Rho serves as the spherical radius, theta serves as the angle from the x axis in the xy plane, and phi serves as the angle from the z axis.\" width=\"394\" height=\"371\" \/> Figure 1.\u00a0The spherical coordinate system locates points with two angles and a distance from the origin.[\/caption]\r\n<p id=\"fs-id1167794181221\">Recall the relationships that connect rectangular coordinates with spherical coordinates.<\/p>\r\n<p id=\"fs-id1167793852232\">From spherical coordinates to rectangular coordinates:<\/p>\r\n<p style=\"text-align: center;\">[latex]{{x} = {\\rho} \\ {\\sin} \\ {\\varphi} \\ {\\cos} \\ {\\theta}}, {{y} = {\\rho} \\ {\\sin} \\ {\\varphi} \\ {\\sin} \\ {\\theta}}, \\ {\\text{and}} \\ {{z} = {\\rho} \\ {\\cos} \\ {\\varphi}}.[\/latex]<\/p>\r\n<span style=\"font-size: 1rem; text-align: initial;\">From rectangular coordinates to spherical coordinates:<\/span>\r\n<p style=\"text-align: center;\">[latex]{\\rho}^{2} = {x^2} + {y^2} + {z^2}, \\ {\\tan}{\\theta} = {\\frac{y}{x}}, {\\varphi} = {\\arccos}{\\left ( {\\frac{z}{\\sqrt{{x^2} + {y^2} + {z^2}}}} \\right )}.[\/latex]<\/p>\r\nOther relationships that are important to know for conversions are\r\n<ul>\r\n \t<li>[latex]{r} = {\\rho} \\ {\\sin} \\ {\\varphi}[\/latex]<\/li>\r\n \t<li>[latex]{\\theta} = {\\theta}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0These equations are used to convert from spherical coordinates to cylindrical coordinates<\/li>\r\n \t<li>[latex]{z} = {\\rho} \\ {\\cos} \\ {\\varphi}[\/latex]<\/li>\r\n<\/ul>\r\n<div id=\"fs-id1167793475416\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n\r\nand\r\n<ul>\r\n \t<li>[latex]{\\rho} = {\\sqrt{{r^2}+{z^2}}}[\/latex]<\/li>\r\n \t<li>[latex]{\\theta} = {\\theta}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<span id=\"MathJax-Element-178-Frame\" class=\"MathJax\" style=\"box-sizing: border-box; overflow: initial; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: center; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;\" tabindex=\"0\" role=\"presentation\" data-mathml=\"&lt;math xmlns=&quot;http:\/\/www.w3.org\/1998\/Math\/MathML&quot; display=&quot;block&quot;&gt;&lt;semantics&gt;&lt;mrow&gt;&lt;mtable&gt;&lt;mtr \/&gt;&lt;mtr \/&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;&amp;#x2022;&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;mtd \/&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mi&gt;&amp;#x3C1;&lt;\/mi&gt;&lt;mo&gt;=&lt;\/mo&gt;&lt;msqrt&gt;&lt;mrow&gt;&lt;msup&gt;&lt;mi&gt;r&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;mo&gt;+&lt;\/mo&gt;&lt;msup&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;\/mrow&gt;&lt;\/msqrt&gt;&lt;\/mtd&gt;&lt;mtd \/&gt;&lt;mtd \/&gt;&lt;mtd \/&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;&amp;#x2022;&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;mtd \/&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mi&gt;&amp;#x3B8;&lt;\/mi&gt;&lt;mo&gt;=&lt;\/mo&gt;&lt;mi&gt;&amp;#x3B8;&lt;\/mi&gt;&lt;\/mtd&gt;&lt;mtd \/&gt;&lt;mtd \/&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtable&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;These equations are used to convert from&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;cylindrical coordinates to spherical&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;coordinates.&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;\/mtable&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;&amp;#x2022;&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;mtd \/&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mi&gt;&amp;#x3C6;&lt;\/mi&gt;&lt;mo&gt;=&lt;\/mo&gt;&lt;mtext&gt;arccos&lt;\/mtext&gt;&lt;mrow&gt;&lt;mo&gt;(&lt;\/mo&gt;&lt;mrow&gt;&lt;mfrac&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mrow&gt;&lt;msqrt&gt;&lt;mrow&gt;&lt;msup&gt;&lt;mi&gt;r&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;mo&gt;+&lt;\/mo&gt;&lt;msup&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;\/mrow&gt;&lt;\/msqrt&gt;&lt;\/mrow&gt;&lt;\/mfrac&gt;&lt;\/mrow&gt;&lt;mo&gt;)&lt;\/mo&gt;&lt;\/mrow&gt;&lt;\/mtd&gt;&lt;mtd \/&gt;&lt;mtd \/&gt;&lt;mtd \/&gt;&lt;\/mtr&gt;&lt;\/mtable&gt;&lt;\/mrow&gt;&lt;annotation-xml encoding=&quot;MathML-Content&quot;&gt;&lt;mtable&gt;&lt;mtr&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;\u2022&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mi&gt;\u03c1&lt;\/mi&gt;&lt;mo&gt;=&lt;\/mo&gt;&lt;msqrt&gt;&lt;mrow&gt;&lt;msup&gt;&lt;mi&gt;r&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;mo&gt;+&lt;\/mo&gt;&lt;msup&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;\/mrow&gt;&lt;\/msqrt&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;\u2022&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mi&gt;\u03b8&lt;\/mi&gt;&lt;mo&gt;=&lt;\/mo&gt;&lt;mi&gt;\u03b8&lt;\/mi&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtable&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;These equations are used to convert from&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;cylindrical coordinates to spherical&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;coordinates.&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;\/mtable&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;\u2022&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mi&gt;\u03c6&lt;\/mi&gt;&lt;mo&gt;=&lt;\/mo&gt;&lt;mtext&gt;arccos&lt;\/mtext&gt;&lt;mrow&gt;&lt;mo&gt;(&lt;\/mo&gt;&lt;mrow&gt;&lt;mfrac&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mrow&gt;&lt;msqrt&gt;&lt;mrow&gt;&lt;msup&gt;&lt;mi&gt;r&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;mo&gt;+&lt;\/mo&gt;&lt;msup&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;\/mrow&gt;&lt;\/msqrt&gt;&lt;\/mrow&gt;&lt;\/mfrac&gt;&lt;\/mrow&gt;&lt;mo&gt;)&lt;\/mo&gt;&lt;\/mrow&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;\/mtable&gt;&lt;\/annotation-xml&gt;&lt;\/semantics&gt;&lt;\/math&gt;\"><span id=\"MathJax-Span-4025\" class=\"math\"><span id=\"MathJax-Span-4026\" class=\"mrow\"><span id=\"MathJax-Span-4027\" class=\"semantics\"><span id=\"MathJax-Span-4028\" class=\"mrow\"><span id=\"MathJax-Span-4029\" class=\"mtable\"><span id=\"MathJax-Span-4069\" class=\"mtd\"><span id=\"MathJax-Span-4070\" class=\"mrow\"><span id=\"MathJax-Span-4071\" class=\"mtable\"><span id=\"MathJax-Span-4072\" class=\"mtd\"><span id=\"MathJax-Span-4073\" class=\"mrow\"><span id=\"MathJax-Span-4074\" class=\"mtext\">These equations are used to convert from\u00a0<\/span><\/span><\/span><span id=\"MathJax-Span-4075\" class=\"mtd\"><span id=\"MathJax-Span-4076\" class=\"mrow\"><span id=\"MathJax-Span-4077\" class=\"mtext\">cylindrical coordinates to spherical coordinates.<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/li>\r\n \t<li>[latex]{\\varphi} = {\\arccos}{\\left ( {\\frac{z}{\\sqrt{{r^2}+{z^2}}}} \\right )}[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1167794050870\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<div class=\"MathJax_Display\">\r\n\r\n<span id=\"MathJax-Element-178-Frame\" class=\"MathJax\" style=\"box-sizing: border-box; overflow: initial; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: center; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;\" tabindex=\"0\" role=\"presentation\" data-mathml=\"&lt;math xmlns=&quot;http:\/\/www.w3.org\/1998\/Math\/MathML&quot; display=&quot;block&quot;&gt;&lt;semantics&gt;&lt;mrow&gt;&lt;mtable&gt;&lt;mtr \/&gt;&lt;mtr \/&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;&amp;#x2022;&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;mtd \/&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mi&gt;&amp;#x3C1;&lt;\/mi&gt;&lt;mo&gt;=&lt;\/mo&gt;&lt;msqrt&gt;&lt;mrow&gt;&lt;msup&gt;&lt;mi&gt;r&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;mo&gt;+&lt;\/mo&gt;&lt;msup&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;\/mrow&gt;&lt;\/msqrt&gt;&lt;\/mtd&gt;&lt;mtd \/&gt;&lt;mtd \/&gt;&lt;mtd \/&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;&amp;#x2022;&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;mtd \/&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mi&gt;&amp;#x3B8;&lt;\/mi&gt;&lt;mo&gt;=&lt;\/mo&gt;&lt;mi&gt;&amp;#x3B8;&lt;\/mi&gt;&lt;\/mtd&gt;&lt;mtd \/&gt;&lt;mtd \/&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtable&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;These equations are used to convert from&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;cylindrical coordinates to spherical&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;coordinates.&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;\/mtable&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;&amp;#x2022;&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;mtd \/&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mi&gt;&amp;#x3C6;&lt;\/mi&gt;&lt;mo&gt;=&lt;\/mo&gt;&lt;mtext&gt;arccos&lt;\/mtext&gt;&lt;mrow&gt;&lt;mo&gt;(&lt;\/mo&gt;&lt;mrow&gt;&lt;mfrac&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mrow&gt;&lt;msqrt&gt;&lt;mrow&gt;&lt;msup&gt;&lt;mi&gt;r&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;mo&gt;+&lt;\/mo&gt;&lt;msup&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;\/mrow&gt;&lt;\/msqrt&gt;&lt;\/mrow&gt;&lt;\/mfrac&gt;&lt;\/mrow&gt;&lt;mo&gt;)&lt;\/mo&gt;&lt;\/mrow&gt;&lt;\/mtd&gt;&lt;mtd \/&gt;&lt;mtd \/&gt;&lt;mtd \/&gt;&lt;\/mtr&gt;&lt;\/mtable&gt;&lt;\/mrow&gt;&lt;annotation-xml encoding=&quot;MathML-Content&quot;&gt;&lt;mtable&gt;&lt;mtr&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;\u2022&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mi&gt;\u03c1&lt;\/mi&gt;&lt;mo&gt;=&lt;\/mo&gt;&lt;msqrt&gt;&lt;mrow&gt;&lt;msup&gt;&lt;mi&gt;r&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;mo&gt;+&lt;\/mo&gt;&lt;msup&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;\/mrow&gt;&lt;\/msqrt&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;\u2022&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mi&gt;\u03b8&lt;\/mi&gt;&lt;mo&gt;=&lt;\/mo&gt;&lt;mi&gt;\u03b8&lt;\/mi&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtable&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;These equations are used to convert from&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;cylindrical coordinates to spherical&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;coordinates.&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;\/mtable&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;\u2022&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mi&gt;\u03c6&lt;\/mi&gt;&lt;mo&gt;=&lt;\/mo&gt;&lt;mtext&gt;arccos&lt;\/mtext&gt;&lt;mrow&gt;&lt;mo&gt;(&lt;\/mo&gt;&lt;mrow&gt;&lt;mfrac&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mrow&gt;&lt;msqrt&gt;&lt;mrow&gt;&lt;msup&gt;&lt;mi&gt;r&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;mo&gt;+&lt;\/mo&gt;&lt;msup&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;\/mrow&gt;&lt;\/msqrt&gt;&lt;\/mrow&gt;&lt;\/mfrac&gt;&lt;\/mrow&gt;&lt;mo&gt;)&lt;\/mo&gt;&lt;\/mrow&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;\/mtable&gt;&lt;\/annotation-xml&gt;&lt;\/semantics&gt;&lt;\/math&gt;\"><span id=\"MathJax-Span-4025\" class=\"math\"><span id=\"MathJax-Span-4026\" class=\"mrow\"><span id=\"MathJax-Span-4027\" class=\"semantics\"><span id=\"MathJax-Span-4028\" class=\"mrow\"><span id=\"MathJax-Span-4029\" class=\"mtable\"><span id=\"MathJax-Span-4112\" class=\"mtd\"><span id=\"MathJax-Span-4113\" class=\"mrow\"><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><span style=\"font-size: 1rem; text-align: initial;\">The following figure shows a few solid regions that are convenient to express in spherical coordinates.<\/span>\r\n\r\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\" display=\"block\"><\/math><\/div>\r\n<\/div>\r\n[caption id=\"attachment_1397\" align=\"aligncenter\" width=\"923\"]<img class=\"size-full wp-image-1397\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26173500\/5-5-7.jpeg\" alt=\"This figure consists of four figures. In the first, a sphere is shown with the note Sphere rho = c (constant). In the second, a half plane is drawn from the z axis with the note Half plane theta = c (constant). In the last two figures, a half cone is drawn in each with the note Half cone phi = c (constant). In the first of these, the cone opens up and it is marked 0 &lt; c &lt; pi\/2. In the second of these, the cone opens down and it is marked pi\/2 &lt; c &lt; pi.\" width=\"923\" height=\"290\" \/> Figure 2.\u00a0Spherical coordinates are especially convenient for working with solids bounded by these types of surfaces. (The letter [latex]c[\/latex] indicates a constant.)[\/caption]\r\n<h2 data-type=\"title\">Integration in Spherical Coordinates<\/h2>\r\n<p id=\"fs-id1167793522445\">We now establish a triple integral in the spherical coordinate system, as we did before in the cylindrical coordinate system. Let the function [latex]{f}{({\\rho},{\\theta},{\\varphi})}[\/latex] be continuous in a bounded spherical box, [latex]{B} = {\\left \\{ {({\\rho},{\\theta},{\\varphi})}{\\mid}{a} \\ {\\leq} \\ {\\rho} \\ {\\leq} \\ {{b},{\\alpha}} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {{\\beta},{\\gamma}} \\ {\\leq} \\ {\\varphi} \\ {\\leq} \\ {\\psi} \\right \\}}[\/latex]. We then divide each interval into [latex]l[\/latex], [latex]m[\/latex], and [latex]n[\/latex] subdivisions such that [latex]{{\\Delta}{\\rho} = {\\frac{{b}-{a}}{l}}}, \\ {{\\Delta}{\\theta} = {\\frac{{\\beta}-{\\alpha}}{m}}}, \\ {{\\Delta}{\\varphi} = {\\frac{{\\psi}-{\\gamma}}{n}}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793970634\">Now we can illustrate the following theorem for triple integrals in spherical coordinates with [latex]{\\left ( {{\\rho}^{*}_{ijk}}, {{\\theta}^{*}_{ijk}}, {{\\varphi}^{*}_{ijk}} \\right )}[\/latex] being any sample point in the spherical subbox [latex]{B}_{ijk}[\/latex]. For the volume element of the subbox [latex]{\\Delta}{V}[\/latex] in spherical coordinates, we have [latex]{\\Delta}{V} = {({\\Delta}{\\rho})}{({\\rho}{\\Delta}{\\varphi})}{({\\rho} \\ {\\sin} \\ {\\varphi}{\\Delta}{\\theta})}[\/latex], as shown in the following figure.<\/p>\r\n\r\n\r\n[caption id=\"attachment_1398\" align=\"aligncenter\" width=\"756\"]<img class=\"size-full wp-image-1398\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26173534\/5-5-8.jpeg\" alt=\"In the spherical coordinate space, a box is projected onto the polar coordinate plane. On the polar coordinate plane, the projection has area rho sin phi Delta theta. On the z axis, a distance Delta rho is indicated, and from these boundaries, angles are made that project through the edges of the box. There is also a blown up version of the box that shows it has sides Delta rho, rho Delta phi, and rho sin phi Delta theta, with overall volume Delta V = rho squared sin phi Delta rho Delta phi Delta theta.\" width=\"756\" height=\"361\" \/> Figure 3.\u00a0The volume element of a box in spherical coordinates.[\/caption]\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167794327634\">The\u00a0<strong><span id=\"ebd07590-c9ca-4701-95cf-6fa9131d20db_term223\" data-type=\"term\">triple integral in spherical coordinates<\/span><\/strong>\u00a0is the limit of a triple Riemann sum,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{l,m,n\\to\\infty}\\displaystyle\\sum^l_{i=1}\\displaystyle\\sum^m_{j=1}\\displaystyle\\sum^n_{k=1}f(\\rho^*_{ijk},\\theta^*_{ijk},\\varphi^*_{ijk})(\\rho^*_{ijk})^2\\sin\\varphi\\Delta\\rho\\Delta\\theta\\Delta\\varphi[\/latex]<\/p>\r\nprovided the limit exists.\r\n\r\n<\/div>\r\nAs with the other multiple integrals we have examined, all the properties work similarly for a triple integral in the spherical coordinate system, and so do the iterated integrals. Fubini\u2019s theorem takes the following form.\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">theorem: fubini's theorem for spherical coordinates<\/h3>\r\n<p id=\"fs-id1167793433540\">If [latex]{f}{({\\rho},{\\theta},{\\varphi})}[\/latex] is continuous on a spherical solid box [latex]{B} = {[{a},{b}]} \\ {\\times} \\ {[{\\alpha},{\\beta}]} \\ {\\times} {[{\\gamma},{\\psi}]}[\/latex], then<\/p>\r\n<p style=\"text-align: center;\">[latex]\\underset{B}{\\displaystyle\\iiint}{f}{({\\rho},{\\theta},{\\varphi})} \\ {\\rho}^{2}{\\sin} \\ {\\varphi} \\ {{d}{\\rho}} \\ {{d}{\\varphi}} \\ {{d}{\\theta}} = {\\displaystyle\\int^{{\\varphi}={\\psi}}_{{\\varphi}={\\gamma}}} \\ {\\displaystyle\\int^{{\\theta}={\\beta}}_{{\\theta}={\\alpha}}} \\ {\\displaystyle\\int^{{\\rho}={b}}_{{\\rho}={a}}} \\ {f}{({\\rho},{\\theta},{\\varphi})} \\ {\\rho}^{2}{\\sin} \\ {\\varphi} \\ {{d}{\\rho}} \\ {{d}{\\varphi}} \\ {{d}{\\theta}}.[\/latex]<\/p>\r\nThis iterated integral may be replaced by other iterated integrals by integrating with respect to the three variables in other orders.\r\n\r\n<\/div>\r\nAs stated before, spherical coordinate systems work well for solids that are symmetric around a point, such as spheres and cones. Let us look at some examples before we consider triple integrals in spherical coordinates on general spherical regions.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: evaluating a triple integral in spherical coordinates<\/h3>\r\nEvaluate the iterated triple integral\r\n\r\n[latex]{\\displaystyle\\int^{{\\theta}={{2}{\\pi}}}_{{\\theta}={0}}} \\ {\\displaystyle\\int^{{\\varphi}={{\\pi}\/{2}}}_{{\\varphi}={0}}} \\ {\\displaystyle\\int^{{\\rho}={1}}_{{p}={0}}} \\ {\\rho}^{2}{\\sin} \\ {\\varphi} \\ {{d}{\\rho}} \\ {{d}{\\varphi}} \\ {{d}{\\theta}}[\/latex].\r\n\r\n[reveal-answer q=\"426899382\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"426899382\"]\r\n\r\nAs before, in this case the variables in the iterated integral are actually independent of each other and hence we can integrate each piece and multiply:\r\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{{2}{\\pi}}_{0}} \\ {\\displaystyle\\int^{{\\pi}\/2}_{0}} \\ {\\displaystyle\\int^{1}_{0}} \\ {\\rho}^{2}{\\sin} \\ {\\varphi} \\ {{d}{\\rho}} \\ {{d}{\\varphi}} \\ {{d}{\\theta}} = {\\displaystyle\\int^{{2}{\\pi}}_{0}} \\ {{d}{\\theta}} \\ {\\displaystyle\\int^{{\\pi}\/2}_{0}} \\ {\\sin} \\ {\\varphi} \\ {{d}{\\varphi}} \\ {\\displaystyle\\int^{1}_{0}} \\ {\\rho}^{2}{d}{\\rho} = {({2}{\\pi})}{(1)}{\\left ( {\\frac{1}{3}} \\right )} = {\\frac{{2}{\\pi}}{3}}.[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793928675\">The concept of triple integration in spherical coordinates can be extended to integration over a general solid, using the projections onto the coordinate planes. Note that [latex]dV[\/latex] and [latex]dA[\/latex] mean the increments in volume and area, respectively. The variables [latex]V[\/latex] and [latex]A[\/latex] are used as the variables for integration to express the integrals.<\/p>\r\n<p id=\"fs-id1167793421937\">The triple integral of a continuous function [latex]{f}{({\\rho},{\\theta},{\\varphi})}[\/latex] over a general solid region<\/p>\r\n<p style=\"text-align: center;\">[latex]{E} = {\\left \\{ {({\\rho},{\\theta},{\\varphi})}{\\mid}{({\\rho},{\\theta})} \\ {\\in} \\ {{D},{{u_1} \\ {({\\rho},{\\theta}}})} \\ {\\leq} \\ {\\varphi} \\ {\\leq} \\ {{u_2} \\ {({\\rho},{\\theta})}} \\right \\}}[\/latex]<\/p>\r\n<p id=\"fs-id1167793692948\">in [latex]{\\mathbb{R}}^{3}[\/latex], where [latex]D[\/latex] is the projection of [latex]E[\/latex] onto the [latex]{\\rho}{\\theta}[\/latex]-plane, is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{f}{({\\rho},{\\theta},{\\varphi})}{dV} = {\\iint\\limits_{D}}{\\left [ \\ {\\int\\limits^{{u_2}{({\\rho},{\\theta})}}_{{u_1}{({\\rho},{\\theta})}}} \\ {f}{({\\rho},{\\theta},{\\varphi})}{d}{\\varphi} \\right ]}{{d}{A}}.[\/latex]<\/p>\r\nIn particular, [latex]{D} = {\\left \\{ {({\\rho},{\\theta})}{\\mid}{g_1} \\ {(\\theta)} \\ {\\leq} \\ {\\rho} \\ {\\leq} \\ {{g_2} \\ {(\\theta)},{\\alpha}} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\beta} \\right \\}}[\/latex], then we have\r\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{f}{({\\rho},{\\theta},{\\varphi})}{dV} = {\\displaystyle\\int^{\\beta}_{\\alpha}} \\ {\\displaystyle\\int^{{g_2}{({\\theta})}}_{{g_1}{({\\theta})}}} \\ {\\displaystyle\\int^{{u_2}{({\\rho},{\\theta})}}_{{u_1}{({\\rho},{\\theta})}}} \\ {f}{({\\rho},{\\theta},{\\varphi})}{{\\rho}^{2}}{\\sin} \\ {\\varphi} \\ {{d}{\\varphi}} \\ {{d}{\\rho}} \\ {{d}{\\theta}}.[\/latex]<\/p>\r\n<p id=\"fs-id1167793244072\">Similar formulas occur for projections onto the other coordinate planes.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: setting up a triple integral in spherical coordinates<\/h3>\r\nSet up an integral for the volume of the region bounded by the cone [latex]{z} = {\\sqrt{{3}{({x^2}+{y^2})}}}[\/latex] and the hemisphere [latex]{z} = {\\sqrt{{4}-{x^2}-{y^2}}}[\/latex] (see the figure below).\r\n\r\n[caption id=\"attachment_1399\" align=\"aligncenter\" width=\"490\"]<img class=\"size-full wp-image-1399\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26173615\/5-5-9.jpeg\" alt=\"A hemisphere with equation z = the square root of (4 minus x squared minus y squared) in the upper half plane, and within it, a cone with equation z = the square root of (3 times (x squared + y squared)) that is pointing down, with vertex at the origin.\" width=\"490\" height=\"340\" \/> Figure 4.\u00a0A region bounded below by a cone and above by a hemisphere.[\/caption]\r\n\r\n[reveal-answer q=\"972450157\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"972450157\"]\r\n<p id=\"fs-id1167794160034\">Using the conversion formulas from rectangular coordinates to spherical coordinates, we have:<\/p>\r\n<p id=\"fs-id1167793384878\">For the cone: [latex]{z} = {\\sqrt{{3}{({x^2}+{y^2})}}}[\/latex] or [latex]{\\rho} \\ {\\cos} \\ {\\varphi} = {\\sqrt{3}} \\ {\\rho} \\ {\\sin} \\ {\\varphi}[\/latex] or [latex]{\\tan} \\ {\\varphi} = {\\frac{1}{\\sqrt{3}}}[\/latex] or [latex]{\\varphi} = {\\frac{\\pi}{6}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793261178\">For the sphere: [latex]{z} = {\\sqrt{{4}-{x^2}-{y^2}}}[\/latex] or [latex]{z^2} + {x^2} + {y^2} = {4}[\/latex] or [latex]{{\\rho}^{2}} = {4}[\/latex] or [latex]{\\rho} = {2}[\/latex].<\/p>\r\n<p id=\"fs-id1167793433725\">Thus, the triple integral for the volume is [latex]{{V}{(E)}} = {\\displaystyle\\int^{{\\theta} = {{2}{\\pi}}}_{{\\theta} = {0}}} \\ \\ {\\displaystyle\\int^{{\\varphi} = {{\\pi}\/{6}}}_{{\\phi} = {0}}} \\ \\ {\\displaystyle\\int^{{\\rho} = {2}}_{{\\rho} = {0}}} \\ {{\\rho}^{2}}{\\sin} \\ {\\varphi} \\ {{d}{\\rho}} \\ {{d}{\\varphi}} \\ {{d}{\\theta}}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nSet up a triple integral for the volume of the solid region bounded above by the sphere [latex]{\\rho} = {2}[\/latex] and bounded below by the cone [latex]{\\varphi} = {{\\pi}\/{3}}[\/latex].\r\n\r\n[reveal-answer q=\"422572964\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"422572964\"]\r\n\r\n[latex]V(E)=\\displaystyle\\int_{\\theta=0}^{\\theta=2\\pi}\\displaystyle\\int_{\\phi=0}^{\\varphi=\\pi\/3}\\displaystyle\\int_{\\rho=0}^{\\rho=2}\\rho^2\\sin\\varphi \\ d\\rho \\ d\\varphi \\ d\\theta.[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: interchanging order of integration in spherical coordinates<\/h3>\r\n<p id=\"fs-id1167794212109\">Let [latex]E[\/latex]\u00a0<span style=\"font-size: 1rem; text-align: initial;\">be the region bounded below by the cone [latex]{z} = {\\sqrt{{x^2}+{y^2}}}[\/latex] and above by the sphere [latex]{z} = {x^2} + {y^2} + {z^2}[\/latex] (<\/span>Figure 5<span style=\"font-size: 1rem; text-align: initial;\">). Set up a triple integral in spherical coordinates and find the volume of the region using the following orders of integration:<\/span><\/p>\r\n\r\n<ol id=\"fs-id1167793400632\" type=\"a\">\r\n \t<li>[latex]{{d}{\\rho}} \\ {{d}{\\phi}} \\ {{d}{\\theta}}[\/latex],<\/li>\r\n \t<li><span data-type=\"newline\">[latex]{{d}{\\varphi}} \\ {{d}{\\rho}} \\ {{d}{\\theta}}[\/latex].\r\n<\/span><\/li>\r\n<\/ol>\r\n[caption id=\"attachment_1401\" align=\"aligncenter\" width=\"480\"]<img class=\"size-full wp-image-1401\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26173806\/5-5-10.jpeg\" alt=\"A sphere with equation z = x squared + y squared + z squared, and within it, a cone with equation z = the square root of (x squared + y squared) that is pointing down, with vertex at the origin.\" width=\"480\" height=\"341\" \/> Figure 5.\u00a0A region bounded below by a cone and above by a sphere.[\/caption]\r\n\r\n[reveal-answer q=\"178836902\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"178836902\"]\r\n\r\n1. Use the conversion formulas to write the equations of the sphere and cone in spherical coordinates.<span data-type=\"newline\">\r\n<\/span>For the sphere:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned} {x^2} + {y^2} + \\ &amp; {z^2} = {z} \\\\\r\n&amp; {{\\rho}^{2}} = {\\rho} \\ {\\cos} \\ {\\varphi} \\\\\r\n&amp; {\\rho} = {\\cos} \\ {\\varphi}.\r\n\\end{aligned}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">For the cone:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{z} &amp; = {\\sqrt{{x^2}+{y^2}}} \\\\\r\n{\\rho} \\ {\\cos} \\ {\\varphi} &amp; = {\\sqrt{{{\\rho}^{2}}{{\\sin}^{2}}{\\varphi}{{\\cos}^{2}}{\\phi}+{{\\rho}^{2}}{{\\sin}^{2}}{\\varphi}{{\\sin}^{2}}{\\phi}}} \\\\\r\n{\\rho} \\ {\\cos} \\ {\\varphi} &amp; = {\\sqrt{{{\\rho}^{2}}{{\\sin}^{2}}{\\varphi}{({{\\cos}^{2}}{\\phi}+{{\\sin}^{2}}{\\phi})}}} \\\\\r\n{\\rho} \\ {\\cos} \\ {\\varphi} &amp; = {\\rho} \\ {\\sin} \\ {\\varphi} \\\\\r\n{\\cos} \\ {\\varphi} &amp; = {\\sin} \\ {\\varphi} \\\\\r\n{\\varphi} &amp; = {{\\pi}\/{4}}.\r\n\\end{aligned}[\/latex]<\/p>\r\nHence the integral for the volume of the solid region [latex]E[\/latex] becomes\r\n<p style=\"text-align: center;\">[latex]{{V}{(E)}} = {\\displaystyle\\int^{{\\theta} = {{2}{\\pi}}}_{{\\theta} = {0}}} \\ {\\displaystyle\\int^{{\\varphi} = {{\\pi}\/{4}}}_{{\\varphi} = {0}}} \\ {\\displaystyle\\int^{{\\rho} = {{\\cos}{\\varphi}}}_{{\\rho} = {0}}} \\ {{{\\rho}^{2}}{\\sin} \\ {\\varphi} \\ {{d}{\\rho}} \\ {{d}{\\varphi}} \\ {{d}{\\theta}}}.[\/latex]<\/p>\r\n&nbsp;\r\n\r\n2. Consider the [latex]{\\varphi}{\\rho}[\/latex]-plane. Note that the ranges for [latex]{\\varphi}[\/latex] and [latex]{\\rho}[\/latex] (from part a.) are\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{0} &amp; \\ {\\leq} \\ {\\varphi} \\ {\\leq} \\ {{\\pi}\/4} \\\\\r\n{0} &amp; \\ {\\leq} \\ {\\phi} \\ {\\leq} \\ {{\\cos}{\\varphi}}.\r\n\\end{aligned}[\/latex]<\/p>\r\nThe curve [latex]{\\rho} = {\\cos}{\\varphi}[\/latex] meets the line [latex]{\\varphi} = {{\\pi}\/{4}}[\/latex] at the point [latex]({{\\pi}\/{4}},{{\\sqrt{2}}\/{2}})[\/latex]. Thus, to change the order of integration, we need to use two pieces:\r\n\r\n[latex]\\hspace{8cm}\\begin{align}&amp;0\\leq\\rho\\leq\\sqrt2\/2 \\\\&amp;0\\leq\\varphi\\leq\\pi\/4\\end{align}[\/latex] [latex]\\text{ and }[\/latex] [latex]\\begin{align}&amp;\\sqrt2\/2\\leq\\rho\\leq1 \\\\&amp;0\\leq\\varphi\\leq\\cos^{-1}\\rho\\end{align}[\/latex]\r\n\r\nHence the integral for the volume of the solid region [latex]E[\/latex] becomes\r\n<p style=\"text-align: center;\">[latex]{{V}{(E)}} = {\\displaystyle\\int^{{\\theta} = {{2}{\\pi}}}_{{\\theta} = {0}}} \\ {\\displaystyle\\int^{{\\rho} = {{\\sqrt{2}\/{2}}}}_{{\\rho} = {0}}} \\ {\\displaystyle\\int^{{\\varphi} = {{\\pi}\/{4}}}_{{\\varphi} = {0}}} \\ {{{\\rho}^{2}}{\\sin} \\ {\\varphi} \\ {{d}{\\varphi}} \\ {{d}{\\rho}} \\ {{d}{\\theta}}} \\ + {\\displaystyle\\int^{{\\theta} = {{2}{\\pi}}}_{{\\theta} = {0}}} \\ {\\displaystyle\\int^{{\\rho} = {1}}_{{\\rho} = {{\\sqrt{2}\/{2}}}}} \\ {\\displaystyle\\int^{{\\varphi} = {{\\cos}^{-1}}{\\rho}}_{{\\varphi} = {0}}} \\ {{{\\rho}^{2}}{\\sin} \\ {\\varphi} \\ {{d}{\\varphi}} \\ {{d}{\\rho}} \\ {{d}{\\theta}}}.[\/latex]<\/p>\r\nIn each case, the integration results in [latex]{{V}{(E)}} = {\\frac{\\pi}{8}}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167794296748\">Before we end this section, we present a couple of examples that can illustrate the conversion from rectangular coordinates to cylindrical coordinates and from rectangular coordinates to spherical coordinates.<\/p>\r\n\r\n<div id=\"fs-id1167794296753\" class=\"ui-has-child-title\" data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: converting from rectangular coordinates to cylindrical coordinates<\/h3>\r\nConvert the following integral into cylindrical coordinates:\r\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{{y} = {1}}_{{y} = {-1}}}\\displaystyle\\int^{{x} = {\\sqrt{{1} - {{y}^{2}}}}}_{{x} = {0}} \\ {\\displaystyle\\int^{{z} = {\\sqrt{{x^2}+{y^2}}}}_{{z} = {{x}^{2}}+{{y}^{2}}}} \\ {{x}{y}{z}} \\ {{d}{z}} \\ {{d}{x}} \\ {{d}{y}}.[\/latex]<\/p>\r\n[reveal-answer q=\"368178320\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"368178320\"]\r\n<p id=\"fs-id1167793544553\">The ranges of the variables are<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{-1} &amp; \\ \\ \\ {\\leq} \\ \\ \\ {y}{\\leq}{1} \\\\\r\n{0} &amp; \\ \\ \\ {\\leq} \\ \\ \\ {x}{\\leq}{\\sqrt{{1} - {{y}^{2}}}} \\\\\r\n{{x}^{2}} + {{y}^{2}} &amp; \\ \\ \\ {\\leq} \\ \\ \\ {z}{\\leq}{\\sqrt{{{x}^{2}}+{{y}^{2}}}}.\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1167793464597\">The first two inequalities describe the right half of a circle of radius 1. Therefore, the ranges for [latex]{\\theta}[\/latex] and [latex]r[\/latex] are<\/p>\r\n<p style=\"text-align: center;\">[latex]{-{\\frac{\\pi}{2}}} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\frac{\\pi}{2}} \\ {\\text{and}} \\ {0} \\ {\\leq} \\ {r} \\ {\\leq} \\ {1}.[\/latex]<\/p>\r\n<p id=\"fs-id1167793514913\">The limits of [latex]z[\/latex] are [latex]{{r}^{2}}[\/latex], hence<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{{y} = {1}}_{{y} = {-1}}}\\displaystyle\\int^{{x} = {\\sqrt{{1} - {{y}^{2}}}}}_{{x} = {0}}\\displaystyle\\int^{{z} = {\\sqrt{{{x}^{2}} + {{y}^{2}}}}}_{{z} = {{x}^{2}} + {{y}^{2}}} \\ {{x}{y}{z}} \\ {{d}{z}} \\ {{d}{x}} \\ {{d}{y}} = {\\displaystyle\\int^{{\\theta} = {{\\pi}\/{2}}}_{{\\theta} = {-{\\pi}\/{2}}}} \\ {\\displaystyle\\int^{{r} = {1}}_{{r} = {0}}} \\ {\\displaystyle\\int^{{z} = {r}}_{{z} = {{r}^{2}}}} \\ {r}{({r}{\\cos}{\\theta})}{({r}{\\sin}{\\theta})}{z} \\ {{d}{z}} \\ {{d}{r}} \\ {{d}{\\theta}}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: converting from rectangular coordinates to spherical coordinates<\/h3>\r\nConvert the following integral into spherical coordinates:\r\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{{y} = {3}}_{{y} = {0}}}\\displaystyle\\int^{{x} = {\\sqrt{{9} - {{y}^{2}}}}}_{{x} = {0}}\\displaystyle\\int^{{z} = {\\sqrt{{18} - {{x}^{2}} - {{y}^{2}}}}}_{{z} = {\\sqrt{{x}^{2} + {{y}^{2}}}}} \\ {({{x}^{2}} + {{y}^{2}} + {{z}^{2}})} \\ {{d}{z}} \\ {{d}{x}} \\ {{d}{y}}.[\/latex]<\/p>\r\n[reveal-answer q=\"255795354\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"255795354\"]\r\n<p id=\"fs-id1167794052495\">The ranges of the variables are<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{0} \\ \\ \\ &amp; {\\leq} \\ \\ \\ {y} \\ {\\leq} \\ {3} \\\\\r\n{0} \\ \\ \\ &amp; {\\leq} \\ \\ \\ {x} \\ {\\leq} \\ {\\sqrt{{9} - {{y}^{2}}}} \\\\\r\n{\\sqrt{{{x}^{2}} + {{y}^{2}}}} \\ \\ \\ &amp; {\\leq} \\ \\ \\ {z} \\ {\\leq} \\ {\\sqrt{{18} - {{x}^{2}} - {{y}^{2}}}}.\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1167793777086\">The first two ranges of variables describe a quarter disk in the first quadrant of the [latex]xy[\/latex]-plane. Hence the range for [latex]{\\theta}[\/latex] is [latex]{0} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\frac{\\pi}{2}}[\/latex].<\/p>\r\n<p id=\"fs-id1167794067621\">The lower bound [latex]{z} = {\\sqrt{{{x}^{2}} + {{y}^{2}}}}[\/latex] is the upper half of a cone and the upper bound [latex]{z} = {\\sqrt{{18} - {{x}^{2}} - {{y}^{2}}}}[\/latex] is the upper half of a sphere. Therefore, we have [latex]{0} \\ {\\leq} \\ {\\rho} \\ {\\leq} \\ {\\sqrt{18}}[\/latex], which is [latex]{0} \\ {\\leq} \\ {\\rho} \\ {\\leq} \\ {3{\\sqrt{2}}}[\/latex].<\/p>\r\n<p id=\"fs-id1167794168074\">For the ranges of [latex]{\\varphi}[\/latex], we need to find where the cone and the sphere intersect, so solve the equation<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{{{r}^{2}} + {{z}^{2}}} \\ \\ \\ &amp; = \\ \\ \\ {18} \\\\\r\n{({\\sqrt{{{x}^{2}} + {{y}^{2}}}})^2} + {{z}^{2}} \\ \\ \\ &amp; = \\ \\ \\ {18} \\\\\r\n{{z}^{2}} + {{z}^{2}} \\ \\ \\ &amp; = \\ \\ \\ {18} \\\\\r\n{{2}{{z}^{2}}} \\ \\ \\ &amp; = \\ \\ \\ {18} \\\\\r\n{{z}^{2}} \\ \\ \\ &amp; = \\ \\ \\ {9} \\\\\r\n{z} \\ \\ \\ &amp; = \\ \\ \\ {3}.\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1167793610938\">This gives<\/p>\r\n[latex]\\hspace{10cm}\\begin{align}\r\n\r\n3\\sqrt2\\cos\\varphi&amp;=3 \\\\\r\n\r\n\\cos\\varphi&amp;=\\frac1{\\sqrt2} \\\\\r\n\r\n\\varphi&amp;=\\frac{\\pi}4.\r\n\r\n\\end{align}[\/latex]\r\n<p id=\"fs-id1167793924123\">Putting this together, we obtain<\/p>\r\n\r\n<center>[latex]\\displaystyle\\int_{y=0}^{y=3}\\displaystyle\\int_{x=0}^{x=\\sqrt{9-y^2}}\\displaystyle\\int_{z=\\sqrt{x^2+y^2}}^{z=\\sqrt{18-x^2-y^2}}(x^2+y^2+z^2)dz \\ dx \\ dy=\\displaystyle\\int_{\\varphi=0}^{\\varphi=\\pi\/4}\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/2}\\displaystyle\\int_{\\rho=0}^{\\rho=3\\sqrt2}\\rho^4\\sin\\varphi \\ d\\rho \\ d\\theta \\ d\\varphi.[\/latex]<\/center>[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nUse rectangular, cylindrical, and spherical coordinates to set up triple integrals for finding the volume of the region inside the sphere [latex]x^2+y^2+z^2=4[\/latex] but outside the cylinder\u00a0[latex]x^2+y^2=1[\/latex].\r\n\r\n[caption id=\"attachment_1402\" align=\"aligncenter\" width=\"493\"]<img class=\"size-full wp-image-1402\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26173912\/5-5-11.jpeg\" alt=\"A sphere with equation x squared + y squared + z squared = 4, and within it, a cylinder with equation x squared + y squared = 1.\" width=\"493\" height=\"402\" \/> Figure 6.[\/caption]\r\n\r\n[reveal-answer q=\"842692876\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"842692876\"]\r\n\r\nRectangular:\u00a0[latex]\\displaystyle\\int_{x=-2}^{x=2}\\displaystyle\\int_{y=-\\sqrt{4-x^2}}^{y=\\sqrt{4-x^2}}\\displaystyle\\int_{z=-\\sqrt{4-x^2-y^2}}^{z=\\sqrt{4-x^2-y^2}}dz \\ dy \\ dx - \\displaystyle\\int_{x=-1}^{x=1}\\displaystyle\\int_{y=-\\sqrt{1-x^2}}^{y=\\sqrt{1-x^2}}\\displaystyle\\int_{z=-\\sqrt{4-x^2-y^2}}^{z=\\sqrt{4-x^2-y^2}}dz \\ dy \\ dx.[\/latex]\r\n\r\nCylindrical:\u00a0[latex]\\displaystyle\\int_{\\theta=0}^{\\theta=2\\pi}\\displaystyle\\int_{r=1}^{r=2}\\displaystyle\\int_{z=-\\sqrt{4-r^2}}^{z=\\sqrt{4-r^2}}r \\ dz \\ dr \\ d\\theta.[\/latex]\r\n\r\nSpherical:\u00a0[latex]\\displaystyle\\int_{\\varphi=\\pi\/6}^{\\varphi=5\\pi\/6}\\displaystyle\\int_{\\theta=0}^{\\theta=2\\pi}\\displaystyle\\int_{\\rho=\\csc\\varphi}^{\\rho=2} \\rho^2 \\sin{\\varphi} d\\rho \\ d\\theta \\ d\\varphi[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to the above Try It[\/caption]\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8197109&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=7iLbCVweWdI&amp;video_target=tpm-plugin-9elr1s1g-7iLbCVweWdI\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.32_transcript.html\">transcript for \u201cCP 5.32\u201d here (opens in new window).<\/a><\/center>\r\n<p id=\"fs-id1167793442442\">Now that we are familiar with the spherical coordinate system, let\u2019s find the volume of some known geometric figures, such as spheres and ellipsoids.<\/p>\r\n\r\n<div id=\"fs-id1167793442449\" class=\"ui-has-child-title\" data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: chapter opener: finding the volume of L'hemisph\u00c8ric<\/h3>\r\nFind the volume of the spherical planetarium in l\u2019Hemisph\u00e8ric in Valencia, Spain, which is five stories tall and has a radius of approximately [latex]50[\/latex] ft, using the equation [latex]x^2+y^2+z^2=r^2[\/latex].\r\n\r\n[caption id=\"attachment_5569\" align=\"aligncenter\" width=\"325\"]<img class=\"size-full wp-image-5569\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/03022644\/5.60.jpeg\" alt=\"A picture of l\u2019Hemisph\u00e8ric, which is a giant glass structure that is in the shape of an ellipsoid.\" width=\"325\" height=\"219\" \/> Figure 7. (credit: modification of work by Javier Yaya Tur, Wikimedia Commons)[\/caption]\r\n\r\n[reveal-answer q=\"204959485\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"204959485\"]\r\n<p id=\"fs-id1167793416704\">We calculate the volume of the ball in the first octant, where [latex]x\\geq0[\/latex], [latex]y\\geq0[\/latex], and [latex]z\\geq0[\/latex], using spherical coordinates, and then multiply the result by [latex]8[\/latex] for symmetry. Since we consider the region [latex]D[\/latex] as the first octant in the integral, the ranges of the variables are<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{0\\leq\\varphi\\leq\\frac{\\pi}2, \\ 0\\leq\\rho\\leq{r}, \\ 0\\leq\\theta\\leq\\frac{\\pi}2}[\/latex]<\/p>\r\n<p id=\"fs-id1167793500489\">Therefore,<\/p>\r\n[latex]\\hspace{6cm}\\begin{align}\r\n\r\nV&amp;=\\underset{D}{\\displaystyle\\iiint}dx \\ dy \\ dz = 8\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/2}\\displaystyle\\int_{\\rho=0}^{\\rho=\\pi}\\displaystyle\\int_{\\varphi=0}^{\\varphi=\\pi\/2}\\rho^2\\sin\\theta \\ d\\varphi \\ d\\rho \\ d\\theta \\\\\r\n\r\n&amp;=8\\displaystyle\\int_{\\varphi=0}^{\\varphi=\\pi\/2}d\\varphi\\displaystyle\\int_{\\rho=0}^{\\rho=\\pi}\\rho^2 \\ d\\rho\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/2}\\sin\\theta \\ d\\theta \\\\\r\n\r\n&amp;=8\\left(\\frac{\\pi}2\\right)\\left(\\frac{r^3}3\\right)(1) \\\\\r\n\r\n&amp;=\\frac43\\pi{r}^3\r\n\r\n\\end{align}[\/latex]\r\n<p id=\"fs-id1167793522367\">This exactly matches with what we knew. So for a sphere with a radius of approximately [latex]50[\/latex] ft, the volume is [latex]\\frac43\\pi{r}^3\\approx523,000\\text{ ft.}^3[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nFor the next example we find the volume of an ellipsoid.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding the volume of an ellipsoid<\/h3>\r\nFind the volume of the ellipsoid [latex]\\frac{x^2}{a^2}+\\frac{y^2}{b^2}+\\frac{z^2}{c^2}=1[\/latex].\r\n\r\n[reveal-answer q=\"543745872\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"543745872\"]\r\n<p id=\"fs-id1167793251274\">We again use symmetry and evaluate the volume of the ellipsoid using spherical coordinates. As before, we use the first octant [latex]x\\geq0[\/latex], [latex]y\\geq0[\/latex], and [latex]z\\geq0[\/latex] and then multiply the result by 8.<\/p>\r\n<p id=\"fs-id1167793219213\">In this case the ranges of the variables are<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{0\\leq\\varphi\\leq\\frac{\\pi}2, \\ 0\\leq\\rho\\leq\\frac{\\pi}2, \\ 0\\leq\\rho\\leq1, \\ 0\\leq\\theta\\leq\\frac{\\pi}2}[\/latex]<\/p>\r\n<p id=\"fs-id1167793540294\">Also, we need to change the rectangular to spherical coordinates in this way:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{x=a\\rho\\cos\\varphi\\sin\\theta, \\ y=b\\rho\\sin\\varphi\\sin\\theta,\\text{ and }z=c\\rho\\cos\\theta}.[\/latex]<\/p>\r\n<p id=\"fs-id1167794213144\">Then the volume of the ellipsoid becomes<\/p>\r\n[latex]\\hspace{6cm}\\begin{align}\r\n\r\nV&amp;=\\underset{D}{\\displaystyle\\iiint}dx \\ dy \\ dz \\\\\r\n\r\n&amp;=8\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/2}\\displaystyle\\int_{\\rho=0}^{\\rho=1}\\displaystyle\\int_{\\varphi=0}^{\\varphi=\\pi\/2}abc\\rho^2\\sin\\theta \\ d\\varphi \\ d\\rho \\ d\\theta \\\\\r\n\r\n&amp;=8abc\\displaystyle\\int_{\\varphi=0}^{\\varphi=\\pi\/2}d\\varphi\\displaystyle\\int_{\\rho=0}^{\\rho=1}\\rho^2 \\ d\\rho\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/2}\\sin\\theta \\ d\\theta \\\\\r\n\r\n&amp;=8abc\\left(\\frac{\\pi}2\\right)\\left(\\frac13\\right)(1) \\\\\r\n\r\n&amp;=\\frac43\\pi{abc}\r\n\r\n\\end{align}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding the volume of the space inside an ellipsoid and outside a sphere<\/h3>\r\nFind the volume of the space inside the ellipsoid [latex]\\frac{x^2}{75^2}+\\frac{y^2}{80^2}+\\frac{z^2}{90^2}=1[\/latex] and outside the sphere [latex]x^2+y^2+z^2=50^2[\/latex].\r\n\r\n[reveal-answer q=\"234085739\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"234085739\"]\r\n<p id=\"fs-id1167793925860\">This problem is directly related to the l\u2019Hemisph\u00e8ric structure. The volume of space inside the ellipsoid and outside the sphere might be useful to find the expense of heating or cooling that space. We can use the preceding two examples for the volume of the sphere and ellipsoid and then subtract.<\/p>\r\n<p id=\"fs-id1167793732638\">First we find the volume of the ellipsoid using [latex]a=75[\/latex] ft, [latex]b=80[\/latex] ft, and [latex]c=90[\/latex] ft in the result from\u00a0Example \"Finding the Volume of an Ellipsoid\".\u00a0Hence the volume of the ellipsoid is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{V_{\\text{ellipsoid}}=\\frac43\\pi(75)(80)(90)\\approx2,262,000\\text{ ft.}^3}[\/latex]<\/p>\r\n<p id=\"fs-id1167793452781\">From\u00a0Example \"Chapter Opener: Finding the Volume of l\u2019Hemisph\u00e8ric\",\u00a0the volume of the sphere is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{V_{\\text{sphere}}\\approx523,600\\text{ ft.}^3.}[\/latex]<\/p>\r\n<p id=\"fs-id1167793556981\">Therefore, the volume of the space inside the ellipsoid [latex]\\frac{x^2}{75^2}+\\frac{y^2}{80^2}+\\frac{z^2}{90^2}=1[\/latex] and outside the sphere [latex]x^2+y^2+z^2=50^2[\/latex] is approximately<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{V_{\\text{Hemisferic}}=V_{\\text{ellipsoid}}-V_{\\text{sphere}}=1,738,400\\text{ ft.}^3.}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Activity: hot air balloons<\/h3>\r\n<p id=\"fs-id1167793285292\">Hot air ballooning is a relaxing, peaceful pastime that many people enjoy. Many balloonist gatherings take place around the world, such as the Albuquerque International Balloon Fiesta. The Albuquerque event is the largest hot air balloon festival in the world, with over [latex]500[\/latex] balloons participating each year.<\/p>\r\n\r\n\r\n[caption id=\"attachment_1403\" align=\"aligncenter\" width=\"900\"]<img class=\"size-full wp-image-1403\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174048\/5-5-12.jpeg\" alt=\"A picture of many hot air balloons.\" width=\"900\" height=\"600\" \/> Figure 8.\u00a0Balloons lift off at the 2001 Albuquerque International Balloon Fiesta. (credit: David Herrera, Flickr)[\/caption]\r\n<p id=\"fs-id1167793510308\">As the name implies, hot air balloons use hot air to generate lift. (Hot air is less dense than cooler air, so the balloon floats as long as the hot air stays hot.) The heat is generated by a propane burner suspended below the opening of the basket. Once the balloon takes off, the pilot controls the altitude of the balloon, either by using the burner to heat the air and ascend or by using a vent near the top of the balloon to release heated air and descend. The pilot has very little control over where the balloon goes, however\u2014balloons are at the mercy of the winds. The uncertainty over where we will end up is one of the reasons balloonists are attracted to the sport.<\/p>\r\n<p id=\"fs-id1167793510321\">In this project we use triple integrals to learn more about hot air balloons. We model the balloon in two pieces. The top of the balloon is modeled by a half sphere of radius [latex]28[\/latex] feet. The bottom of the balloon is modeled by a frustum of a cone (think of an ice cream cone with the pointy end cut off). The radius of the large end of the frustum is [latex]28[\/latex] feet and the radius of the small end of the frustum is [latex]6[\/latex] feet. A graph of our balloon model and a cross-sectional diagram showing the dimensions are shown in the following figure.<\/p>\r\n\r\n\r\n[caption id=\"attachment_1404\" align=\"aligncenter\" width=\"801\"]<img class=\"size-full wp-image-1404\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174119\/5-5-13.jpeg\" alt=\"This figure consists of two parts, a and b. Figure a shows a representation of a hot air balloon in xyz space as a half sphere on top of a frustrum of a cone. Figure b shows the dimensions, namely, the radius of the half sphere is 28 ft, the distance from the bottom to the top of the frustrum is 44 ft, and the diameter of the circle at the top of the frustrum is 12 ft.\" width=\"801\" height=\"458\" \/> Figure 9.\u00a0(a) Use a half sphere to model the top part of the balloon and a frustum of a cone to model the bottom part of the balloon. (b) A cross section of the balloon showing its dimensions.[\/caption]\r\n<p id=\"fs-id1167794200502\">We first want to find the volume of the balloon. If we look at the top part and the bottom part of the balloon separately, we see that they are geometric solids with known volume formulas. However, it is still worthwhile to set up and evaluate the integrals we would need to find the volume. If we calculate the volume using integration, we can use the known volume formulas to check our answers. This will help ensure that we have the integrals set up correctly for the later, more complicated stages of the project.<\/p>\r\n\r\n<ol id=\"fs-id1167794200510\" type=\"1\">\r\n \t<li>Find the volume of the balloon in two ways.\r\n<ol id=\"fs-id1167794200518\" type=\"a\">\r\n \t<li>Use triple integrals to calculate the volume. Consider each part of the balloon separately. (Consider using spherical coordinates for the top part and cylindrical coordinates for the bottom part.)<\/li>\r\n \t<li>Verify the answer using the formulas for the volume of a sphere, [latex]V=\\frac43\\pi{r}^3[\/latex], and for the volume of a cone, [latex]V=\\frac13\\pi{r}^2h[\/latex].<\/li>\r\n<\/ol>\r\nIn reality, calculating the temperature at a point inside the balloon is a tremendously complicated endeavor. In fact, an entire branch of physics (thermodynamics) is devoted to studying heat and temperature. For the purposes of this project, however, we are going to make some simplifying assumptions about how temperature varies from point to point within the balloon. Assume that just prior to liftoff, the temperature (in degrees Fahrenheit) of the air inside the balloon varies according to the function<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167793937390\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<p style=\"text-align: center;\">[latex]\\large{T_0(r,\\theta,z)=\\frac{z-r}{10}+210}[\/latex]<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li>What is the average temperature of the air in the balloon just prior to liftoff? (Again, look at each part of the balloon separately, and do not forget to convert the function into spherical coordinates when looking at the top part of the balloon.)<span data-type=\"newline\">\r\n<\/span>Now the pilot activates the burner for [latex]10[\/latex] seconds. This action affects the temperature in a [latex]12[\/latex]-foot-wide column [latex]20[\/latex] feet high, directly above the burner. A cross section of the balloon depicting this column in shown in the following figure.<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"CNX_Calc_Figure_15_05_SP3\" class=\"os-figure\">\r\n<figure data-id=\"CNX_Calc_Figure_15_05_SP3\">[caption id=\"attachment_1405\" align=\"aligncenter\" width=\"385\"]<img class=\"size-full wp-image-1405\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174158\/5-5-14.jpeg\" alt=\"This figure shows the dimensions of the balloon and the hot air, namely, the radius of the half sphere is 28 ft, the distance from the bottom to the top of the frustrum is 44 ft, the diameter of the circle at the top of the frustrum is 12 ft, and the inner column of hot air has height 20 ft and diameter 12 ft.\" width=\"385\" height=\"385\" \/> Figure 10.\u00a0<span class=\"os-caption\">Activating the burner heats the air in a [latex]20[\/latex]-foot-high, [latex]12[\/latex]-foot-wide column directly above the burner.<\/span>[\/caption]<\/figure>\r\n<\/div>\r\nAssume that after the pilot activates the burner for [latex]10[\/latex] seconds, the temperature of the air in the column described above\u00a0<em data-effect=\"italics\">increases<\/em>\u00a0according to the formula<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167793376255\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<p style=\"text-align: center;\">[latex]\\large{H(r,\\theta,z)=-2z-48.}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>Then the temperature of the air in the column is given by<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167794067545\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<p style=\"text-align: center;\">[latex]\\large{T_1(r,\\theta,z)=\\frac{z-r}{10}+210+(-2z-48)}[\/latex],<\/p>\r\n\r\n<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>while the temperature in the remainder of the balloon is still given by<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167794292164\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<p style=\"text-align: center;\">[latex]\\large{T_0(r,\\theta,z)=\\frac{z-r}{10}+210}[\/latex].<\/p>\r\n\r\n<\/div><\/li>\r\n \t<li>Find the average temperature of the air in the balloon after the pilot has activated the burner for [latex]10[\/latex] seconds.<\/li>\r\n<\/ol>\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul class=\"os-abstract\">\n<li><span class=\"os-abstract-content\">Evaluate a triple integral by changing to spherical coordinates.<\/span><\/li>\n<\/ul>\n<\/div>\n<h2 data-type=\"title\">Review of Spherical Coordinates<\/h2>\n<p>In three-dimensional space [latex]{\\mathbb{R}^{3}}[\/latex] in the spherical coordinate system, we specify a point [latex]P[\/latex] by its distance [latex]{\\rho}[\/latex] from the origin, the polar angle [latex]{\\theta}[\/latex] from the positive [latex]x[\/latex]-axis (same as in the cylindrical coordinate system), and the angle [latex]{\\varphi}[\/latex] from the positive [latex]z[\/latex]-axis and the line [latex]OP[\/latex] (Figure 1). Note that [latex]{\\rho} \\ {\\geq} \\ {0}[\/latex] and [latex]{0} \\ {\\leq} \\ {\\varphi} \\ {\\leq} \\ {\\pi}[\/latex]. (Refer to\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/introduction-to-cylindrical-and-spherical-coordinates\/\" data-page-slug=\"2-7-cylindrical-and-spherical-coordinates\" data-page-uuid=\"293cc618-b0de-4ec9-9029-6efb4eb1b172\" data-page-fragment=\"page_293cc618-b0de-4ec9-9029-6efb4eb1b172\">Cylindrical and Spherical Coordinates<\/a>\u00a0for a review.) Spherical coordinates are useful for triple integrals over regions that are symmetric with respect to the origin.<\/p>\n<div id=\"attachment_1396\" style=\"width: 404px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1396\" class=\"size-full wp-image-1396\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26173418\/5-5-6.jpeg\" alt=\"A depiction of the spherical coordinate system: a point (x, y, z) is shown, which is equal to (rho, theta, phi) in spherical coordinates. Rho serves as the spherical radius, theta serves as the angle from the x axis in the xy plane, and phi serves as the angle from the z axis.\" width=\"394\" height=\"371\" \/><\/p>\n<p id=\"caption-attachment-1396\" class=\"wp-caption-text\">Figure 1.\u00a0The spherical coordinate system locates points with two angles and a distance from the origin.<\/p>\n<\/div>\n<p id=\"fs-id1167794181221\">Recall the relationships that connect rectangular coordinates with spherical coordinates.<\/p>\n<p id=\"fs-id1167793852232\">From spherical coordinates to rectangular coordinates:<\/p>\n<p style=\"text-align: center;\">[latex]{{x} = {\\rho} \\ {\\sin} \\ {\\varphi} \\ {\\cos} \\ {\\theta}}, {{y} = {\\rho} \\ {\\sin} \\ {\\varphi} \\ {\\sin} \\ {\\theta}}, \\ {\\text{and}} \\ {{z} = {\\rho} \\ {\\cos} \\ {\\varphi}}.[\/latex]<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">From rectangular coordinates to spherical coordinates:<\/span><\/p>\n<p style=\"text-align: center;\">[latex]{\\rho}^{2} = {x^2} + {y^2} + {z^2}, \\ {\\tan}{\\theta} = {\\frac{y}{x}}, {\\varphi} = {\\arccos}{\\left ( {\\frac{z}{\\sqrt{{x^2} + {y^2} + {z^2}}}} \\right )}.[\/latex]<\/p>\n<p>Other relationships that are important to know for conversions are<\/p>\n<ul>\n<li>[latex]{r} = {\\rho} \\ {\\sin} \\ {\\varphi}[\/latex]<\/li>\n<li>[latex]{\\theta} = {\\theta}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0These equations are used to convert from spherical coordinates to cylindrical coordinates<\/li>\n<li>[latex]{z} = {\\rho} \\ {\\cos} \\ {\\varphi}[\/latex]<\/li>\n<\/ul>\n<div id=\"fs-id1167793475416\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<p>and<\/p>\n<ul>\n<li>[latex]{\\rho} = {\\sqrt{{r^2}+{z^2}}}[\/latex]<\/li>\n<li>[latex]{\\theta} = {\\theta}[\/latex]\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0<span id=\"MathJax-Element-178-Frame\" class=\"MathJax\" style=\"box-sizing: border-box; overflow: initial; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: center; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;\" tabindex=\"0\" role=\"presentation\" data-mathml=\"&lt;math xmlns=&quot;http:\/\/www.w3.org\/1998\/Math\/MathML&quot; display=&quot;block&quot;&gt;&lt;semantics&gt;&lt;mrow&gt;&lt;mtable&gt;&lt;mtr \/&gt;&lt;mtr \/&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;&amp;#x2022;&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;mtd \/&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mi&gt;&amp;#x3C1;&lt;\/mi&gt;&lt;mo&gt;=&lt;\/mo&gt;&lt;msqrt&gt;&lt;mrow&gt;&lt;msup&gt;&lt;mi&gt;r&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;mo&gt;+&lt;\/mo&gt;&lt;msup&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;\/mrow&gt;&lt;\/msqrt&gt;&lt;\/mtd&gt;&lt;mtd \/&gt;&lt;mtd \/&gt;&lt;mtd \/&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;&amp;#x2022;&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;mtd \/&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mi&gt;&amp;#x3B8;&lt;\/mi&gt;&lt;mo&gt;=&lt;\/mo&gt;&lt;mi&gt;&amp;#x3B8;&lt;\/mi&gt;&lt;\/mtd&gt;&lt;mtd \/&gt;&lt;mtd \/&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtable&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;These equations are used to convert from&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;cylindrical coordinates to spherical&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;coordinates.&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;\/mtable&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;&amp;#x2022;&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;mtd \/&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mi&gt;&amp;#x3C6;&lt;\/mi&gt;&lt;mo&gt;=&lt;\/mo&gt;&lt;mtext&gt;arccos&lt;\/mtext&gt;&lt;mrow&gt;&lt;mo&gt;(&lt;\/mo&gt;&lt;mrow&gt;&lt;mfrac&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mrow&gt;&lt;msqrt&gt;&lt;mrow&gt;&lt;msup&gt;&lt;mi&gt;r&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;mo&gt;+&lt;\/mo&gt;&lt;msup&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;\/mrow&gt;&lt;\/msqrt&gt;&lt;\/mrow&gt;&lt;\/mfrac&gt;&lt;\/mrow&gt;&lt;mo&gt;)&lt;\/mo&gt;&lt;\/mrow&gt;&lt;\/mtd&gt;&lt;mtd \/&gt;&lt;mtd \/&gt;&lt;mtd \/&gt;&lt;\/mtr&gt;&lt;\/mtable&gt;&lt;\/mrow&gt;&lt;annotation-xml encoding=&quot;MathML-Content&quot;&gt;&lt;mtable&gt;&lt;mtr&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;\u2022&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mi&gt;\u03c1&lt;\/mi&gt;&lt;mo&gt;=&lt;\/mo&gt;&lt;msqrt&gt;&lt;mrow&gt;&lt;msup&gt;&lt;mi&gt;r&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;mo&gt;+&lt;\/mo&gt;&lt;msup&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;\/mrow&gt;&lt;\/msqrt&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;\u2022&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mi&gt;\u03b8&lt;\/mi&gt;&lt;mo&gt;=&lt;\/mo&gt;&lt;mi&gt;\u03b8&lt;\/mi&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtable&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;These equations are used to convert from&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;cylindrical coordinates to spherical&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;coordinates.&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;\/mtable&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;\u2022&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mi&gt;\u03c6&lt;\/mi&gt;&lt;mo&gt;=&lt;\/mo&gt;&lt;mtext&gt;arccos&lt;\/mtext&gt;&lt;mrow&gt;&lt;mo&gt;(&lt;\/mo&gt;&lt;mrow&gt;&lt;mfrac&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mrow&gt;&lt;msqrt&gt;&lt;mrow&gt;&lt;msup&gt;&lt;mi&gt;r&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;mo&gt;+&lt;\/mo&gt;&lt;msup&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;\/mrow&gt;&lt;\/msqrt&gt;&lt;\/mrow&gt;&lt;\/mfrac&gt;&lt;\/mrow&gt;&lt;mo&gt;)&lt;\/mo&gt;&lt;\/mrow&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;\/mtable&gt;&lt;\/annotation-xml&gt;&lt;\/semantics&gt;&lt;\/math&gt;\"><span id=\"MathJax-Span-4025\" class=\"math\"><span id=\"MathJax-Span-4026\" class=\"mrow\"><span id=\"MathJax-Span-4027\" class=\"semantics\"><span id=\"MathJax-Span-4028\" class=\"mrow\"><span id=\"MathJax-Span-4029\" class=\"mtable\"><span id=\"MathJax-Span-4069\" class=\"mtd\"><span id=\"MathJax-Span-4070\" class=\"mrow\"><span id=\"MathJax-Span-4071\" class=\"mtable\"><span id=\"MathJax-Span-4072\" class=\"mtd\"><span id=\"MathJax-Span-4073\" class=\"mrow\"><span id=\"MathJax-Span-4074\" class=\"mtext\">These equations are used to convert from\u00a0<\/span><\/span><\/span><span id=\"MathJax-Span-4075\" class=\"mtd\"><span id=\"MathJax-Span-4076\" class=\"mrow\"><span id=\"MathJax-Span-4077\" class=\"mtext\">cylindrical coordinates to spherical coordinates.<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/li>\n<li>[latex]{\\varphi} = {\\arccos}{\\left ( {\\frac{z}{\\sqrt{{r^2}+{z^2}}}} \\right )}[\/latex]<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1167794050870\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<div class=\"MathJax_Display\">\n<p><span id=\"MathJax-Element-178-Frame\" class=\"MathJax\" style=\"box-sizing: border-box; overflow: initial; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 16px; text-indent: 0px; text-align: center; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;\" tabindex=\"0\" role=\"presentation\" data-mathml=\"&lt;math xmlns=&quot;http:\/\/www.w3.org\/1998\/Math\/MathML&quot; display=&quot;block&quot;&gt;&lt;semantics&gt;&lt;mrow&gt;&lt;mtable&gt;&lt;mtr \/&gt;&lt;mtr \/&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;&amp;#x2022;&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;mtd \/&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mi&gt;&amp;#x3C1;&lt;\/mi&gt;&lt;mo&gt;=&lt;\/mo&gt;&lt;msqrt&gt;&lt;mrow&gt;&lt;msup&gt;&lt;mi&gt;r&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;mo&gt;+&lt;\/mo&gt;&lt;msup&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;\/mrow&gt;&lt;\/msqrt&gt;&lt;\/mtd&gt;&lt;mtd \/&gt;&lt;mtd \/&gt;&lt;mtd \/&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;&amp;#x2022;&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;mtd \/&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mi&gt;&amp;#x3B8;&lt;\/mi&gt;&lt;mo&gt;=&lt;\/mo&gt;&lt;mi&gt;&amp;#x3B8;&lt;\/mi&gt;&lt;\/mtd&gt;&lt;mtd \/&gt;&lt;mtd \/&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtable&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;These equations are used to convert from&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;cylindrical coordinates to spherical&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;coordinates.&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;\/mtable&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;&amp;#x2022;&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;mtd \/&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mi&gt;&amp;#x3C6;&lt;\/mi&gt;&lt;mo&gt;=&lt;\/mo&gt;&lt;mtext&gt;arccos&lt;\/mtext&gt;&lt;mrow&gt;&lt;mo&gt;(&lt;\/mo&gt;&lt;mrow&gt;&lt;mfrac&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mrow&gt;&lt;msqrt&gt;&lt;mrow&gt;&lt;msup&gt;&lt;mi&gt;r&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;mo&gt;+&lt;\/mo&gt;&lt;msup&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;\/mrow&gt;&lt;\/msqrt&gt;&lt;\/mrow&gt;&lt;\/mfrac&gt;&lt;\/mrow&gt;&lt;mo&gt;)&lt;\/mo&gt;&lt;\/mrow&gt;&lt;\/mtd&gt;&lt;mtd \/&gt;&lt;mtd \/&gt;&lt;mtd \/&gt;&lt;\/mtr&gt;&lt;\/mtable&gt;&lt;\/mrow&gt;&lt;annotation-xml encoding=&quot;MathML-Content&quot;&gt;&lt;mtable&gt;&lt;mtr&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;\u2022&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mi&gt;\u03c1&lt;\/mi&gt;&lt;mo&gt;=&lt;\/mo&gt;&lt;msqrt&gt;&lt;mrow&gt;&lt;msup&gt;&lt;mi&gt;r&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;mo&gt;+&lt;\/mo&gt;&lt;msup&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;\/mrow&gt;&lt;\/msqrt&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;\u2022&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mi&gt;\u03b8&lt;\/mi&gt;&lt;mo&gt;=&lt;\/mo&gt;&lt;mi&gt;\u03b8&lt;\/mi&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtable&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;These equations are used to convert from&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;cylindrical coordinates to spherical&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;coordinates.&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;\/mtable&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;mtr&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mtext&gt;\u2022&lt;\/mtext&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd columnalign=&quot;left&quot;&gt;&lt;mi&gt;\u03c6&lt;\/mi&gt;&lt;mo&gt;=&lt;\/mo&gt;&lt;mtext&gt;arccos&lt;\/mtext&gt;&lt;mrow&gt;&lt;mo&gt;(&lt;\/mo&gt;&lt;mrow&gt;&lt;mfrac&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mrow&gt;&lt;msqrt&gt;&lt;mrow&gt;&lt;msup&gt;&lt;mi&gt;r&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;mo&gt;+&lt;\/mo&gt;&lt;msup&gt;&lt;mi&gt;z&lt;\/mi&gt;&lt;mn&gt;2&lt;\/mn&gt;&lt;\/msup&gt;&lt;\/mrow&gt;&lt;\/msqrt&gt;&lt;\/mrow&gt;&lt;\/mfrac&gt;&lt;\/mrow&gt;&lt;mo&gt;)&lt;\/mo&gt;&lt;\/mrow&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;mtd&gt;&lt;\/mtd&gt;&lt;\/mtr&gt;&lt;\/mtable&gt;&lt;\/annotation-xml&gt;&lt;\/semantics&gt;&lt;\/math&gt;\"><span id=\"MathJax-Span-4025\" class=\"math\"><span id=\"MathJax-Span-4026\" class=\"mrow\"><span id=\"MathJax-Span-4027\" class=\"semantics\"><span id=\"MathJax-Span-4028\" class=\"mrow\"><span id=\"MathJax-Span-4029\" class=\"mtable\"><span id=\"MathJax-Span-4112\" class=\"mtd\"><span id=\"MathJax-Span-4113\" class=\"mrow\"><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><span style=\"font-size: 1rem; text-align: initial;\">The following figure shows a few solid regions that are convenient to express in spherical coordinates.<\/span><\/p>\n<p><math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\" display=\"block\"><\/math><\/div>\n<\/div>\n<div id=\"attachment_1397\" style=\"width: 933px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1397\" class=\"size-full wp-image-1397\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26173500\/5-5-7.jpeg\" alt=\"This figure consists of four figures. In the first, a sphere is shown with the note Sphere rho = c (constant). In the second, a half plane is drawn from the z axis with the note Half plane theta = c (constant). In the last two figures, a half cone is drawn in each with the note Half cone phi = c (constant). In the first of these, the cone opens up and it is marked 0 &lt; c &lt; pi\/2. In the second of these, the cone opens down and it is marked pi\/2 &lt; c &lt; pi.\" width=\"923\" height=\"290\" \/><\/p>\n<p id=\"caption-attachment-1397\" class=\"wp-caption-text\">Figure 2.\u00a0Spherical coordinates are especially convenient for working with solids bounded by these types of surfaces. (The letter [latex]c[\/latex] indicates a constant.)<\/p>\n<\/div>\n<h2 data-type=\"title\">Integration in Spherical Coordinates<\/h2>\n<p id=\"fs-id1167793522445\">We now establish a triple integral in the spherical coordinate system, as we did before in the cylindrical coordinate system. Let the function [latex]{f}{({\\rho},{\\theta},{\\varphi})}[\/latex] be continuous in a bounded spherical box, [latex]{B} = {\\left \\{ {({\\rho},{\\theta},{\\varphi})}{\\mid}{a} \\ {\\leq} \\ {\\rho} \\ {\\leq} \\ {{b},{\\alpha}} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {{\\beta},{\\gamma}} \\ {\\leq} \\ {\\varphi} \\ {\\leq} \\ {\\psi} \\right \\}}[\/latex]. We then divide each interval into [latex]l[\/latex], [latex]m[\/latex], and [latex]n[\/latex] subdivisions such that [latex]{{\\Delta}{\\rho} = {\\frac{{b}-{a}}{l}}}, \\ {{\\Delta}{\\theta} = {\\frac{{\\beta}-{\\alpha}}{m}}}, \\ {{\\Delta}{\\varphi} = {\\frac{{\\psi}-{\\gamma}}{n}}}[\/latex].<\/p>\n<p id=\"fs-id1167793970634\">Now we can illustrate the following theorem for triple integrals in spherical coordinates with [latex]{\\left ( {{\\rho}^{*}_{ijk}}, {{\\theta}^{*}_{ijk}}, {{\\varphi}^{*}_{ijk}} \\right )}[\/latex] being any sample point in the spherical subbox [latex]{B}_{ijk}[\/latex]. For the volume element of the subbox [latex]{\\Delta}{V}[\/latex] in spherical coordinates, we have [latex]{\\Delta}{V} = {({\\Delta}{\\rho})}{({\\rho}{\\Delta}{\\varphi})}{({\\rho} \\ {\\sin} \\ {\\varphi}{\\Delta}{\\theta})}[\/latex], as shown in the following figure.<\/p>\n<div id=\"attachment_1398\" style=\"width: 766px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1398\" class=\"size-full wp-image-1398\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26173534\/5-5-8.jpeg\" alt=\"In the spherical coordinate space, a box is projected onto the polar coordinate plane. On the polar coordinate plane, the projection has area rho sin phi Delta theta. On the z axis, a distance Delta rho is indicated, and from these boundaries, angles are made that project through the edges of the box. There is also a blown up version of the box that shows it has sides Delta rho, rho Delta phi, and rho sin phi Delta theta, with overall volume Delta V = rho squared sin phi Delta rho Delta phi Delta theta.\" width=\"756\" height=\"361\" \/><\/p>\n<p id=\"caption-attachment-1398\" class=\"wp-caption-text\">Figure 3.\u00a0The volume element of a box in spherical coordinates.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">definition<\/h3>\n<hr \/>\n<p id=\"fs-id1167794327634\">The\u00a0<strong><span id=\"ebd07590-c9ca-4701-95cf-6fa9131d20db_term223\" data-type=\"term\">triple integral in spherical coordinates<\/span><\/strong>\u00a0is the limit of a triple Riemann sum,<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\lim_{l,m,n\\to\\infty}\\displaystyle\\sum^l_{i=1}\\displaystyle\\sum^m_{j=1}\\displaystyle\\sum^n_{k=1}f(\\rho^*_{ijk},\\theta^*_{ijk},\\varphi^*_{ijk})(\\rho^*_{ijk})^2\\sin\\varphi\\Delta\\rho\\Delta\\theta\\Delta\\varphi[\/latex]<\/p>\n<p>provided the limit exists.<\/p>\n<\/div>\n<p>As with the other multiple integrals we have examined, all the properties work similarly for a triple integral in the spherical coordinate system, and so do the iterated integrals. Fubini\u2019s theorem takes the following form.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">theorem: fubini&#8217;s theorem for spherical coordinates<\/h3>\n<p id=\"fs-id1167793433540\">If [latex]{f}{({\\rho},{\\theta},{\\varphi})}[\/latex] is continuous on a spherical solid box [latex]{B} = {[{a},{b}]} \\ {\\times} \\ {[{\\alpha},{\\beta}]} \\ {\\times} {[{\\gamma},{\\psi}]}[\/latex], then<\/p>\n<p style=\"text-align: center;\">[latex]\\underset{B}{\\displaystyle\\iiint}{f}{({\\rho},{\\theta},{\\varphi})} \\ {\\rho}^{2}{\\sin} \\ {\\varphi} \\ {{d}{\\rho}} \\ {{d}{\\varphi}} \\ {{d}{\\theta}} = {\\displaystyle\\int^{{\\varphi}={\\psi}}_{{\\varphi}={\\gamma}}} \\ {\\displaystyle\\int^{{\\theta}={\\beta}}_{{\\theta}={\\alpha}}} \\ {\\displaystyle\\int^{{\\rho}={b}}_{{\\rho}={a}}} \\ {f}{({\\rho},{\\theta},{\\varphi})} \\ {\\rho}^{2}{\\sin} \\ {\\varphi} \\ {{d}{\\rho}} \\ {{d}{\\varphi}} \\ {{d}{\\theta}}.[\/latex]<\/p>\n<p>This iterated integral may be replaced by other iterated integrals by integrating with respect to the three variables in other orders.<\/p>\n<\/div>\n<p>As stated before, spherical coordinate systems work well for solids that are symmetric around a point, such as spheres and cones. Let us look at some examples before we consider triple integrals in spherical coordinates on general spherical regions.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: evaluating a triple integral in spherical coordinates<\/h3>\n<p>Evaluate the iterated triple integral<\/p>\n<p>[latex]{\\displaystyle\\int^{{\\theta}={{2}{\\pi}}}_{{\\theta}={0}}} \\ {\\displaystyle\\int^{{\\varphi}={{\\pi}\/{2}}}_{{\\varphi}={0}}} \\ {\\displaystyle\\int^{{\\rho}={1}}_{{p}={0}}} \\ {\\rho}^{2}{\\sin} \\ {\\varphi} \\ {{d}{\\rho}} \\ {{d}{\\varphi}} \\ {{d}{\\theta}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q426899382\">Show Solution<\/span><\/p>\n<div id=\"q426899382\" class=\"hidden-answer\" style=\"display: none\">\n<p>As before, in this case the variables in the iterated integral are actually independent of each other and hence we can integrate each piece and multiply:<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{{2}{\\pi}}_{0}} \\ {\\displaystyle\\int^{{\\pi}\/2}_{0}} \\ {\\displaystyle\\int^{1}_{0}} \\ {\\rho}^{2}{\\sin} \\ {\\varphi} \\ {{d}{\\rho}} \\ {{d}{\\varphi}} \\ {{d}{\\theta}} = {\\displaystyle\\int^{{2}{\\pi}}_{0}} \\ {{d}{\\theta}} \\ {\\displaystyle\\int^{{\\pi}\/2}_{0}} \\ {\\sin} \\ {\\varphi} \\ {{d}{\\varphi}} \\ {\\displaystyle\\int^{1}_{0}} \\ {\\rho}^{2}{d}{\\rho} = {({2}{\\pi})}{(1)}{\\left ( {\\frac{1}{3}} \\right )} = {\\frac{{2}{\\pi}}{3}}.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167793928675\">The concept of triple integration in spherical coordinates can be extended to integration over a general solid, using the projections onto the coordinate planes. Note that [latex]dV[\/latex] and [latex]dA[\/latex] mean the increments in volume and area, respectively. The variables [latex]V[\/latex] and [latex]A[\/latex] are used as the variables for integration to express the integrals.<\/p>\n<p id=\"fs-id1167793421937\">The triple integral of a continuous function [latex]{f}{({\\rho},{\\theta},{\\varphi})}[\/latex] over a general solid region<\/p>\n<p style=\"text-align: center;\">[latex]{E} = {\\left \\{ {({\\rho},{\\theta},{\\varphi})}{\\mid}{({\\rho},{\\theta})} \\ {\\in} \\ {{D},{{u_1} \\ {({\\rho},{\\theta}}})} \\ {\\leq} \\ {\\varphi} \\ {\\leq} \\ {{u_2} \\ {({\\rho},{\\theta})}} \\right \\}}[\/latex]<\/p>\n<p id=\"fs-id1167793692948\">in [latex]{\\mathbb{R}}^{3}[\/latex], where [latex]D[\/latex] is the projection of [latex]E[\/latex] onto the [latex]{\\rho}{\\theta}[\/latex]-plane, is<\/p>\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{f}{({\\rho},{\\theta},{\\varphi})}{dV} = {\\iint\\limits_{D}}{\\left [ \\ {\\int\\limits^{{u_2}{({\\rho},{\\theta})}}_{{u_1}{({\\rho},{\\theta})}}} \\ {f}{({\\rho},{\\theta},{\\varphi})}{d}{\\varphi} \\right ]}{{d}{A}}.[\/latex]<\/p>\n<p>In particular, [latex]{D} = {\\left \\{ {({\\rho},{\\theta})}{\\mid}{g_1} \\ {(\\theta)} \\ {\\leq} \\ {\\rho} \\ {\\leq} \\ {{g_2} \\ {(\\theta)},{\\alpha}} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\beta} \\right \\}}[\/latex], then we have<\/p>\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{f}{({\\rho},{\\theta},{\\varphi})}{dV} = {\\displaystyle\\int^{\\beta}_{\\alpha}} \\ {\\displaystyle\\int^{{g_2}{({\\theta})}}_{{g_1}{({\\theta})}}} \\ {\\displaystyle\\int^{{u_2}{({\\rho},{\\theta})}}_{{u_1}{({\\rho},{\\theta})}}} \\ {f}{({\\rho},{\\theta},{\\varphi})}{{\\rho}^{2}}{\\sin} \\ {\\varphi} \\ {{d}{\\varphi}} \\ {{d}{\\rho}} \\ {{d}{\\theta}}.[\/latex]<\/p>\n<p id=\"fs-id1167793244072\">Similar formulas occur for projections onto the other coordinate planes.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: setting up a triple integral in spherical coordinates<\/h3>\n<p>Set up an integral for the volume of the region bounded by the cone [latex]{z} = {\\sqrt{{3}{({x^2}+{y^2})}}}[\/latex] and the hemisphere [latex]{z} = {\\sqrt{{4}-{x^2}-{y^2}}}[\/latex] (see the figure below).<\/p>\n<div id=\"attachment_1399\" style=\"width: 500px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1399\" class=\"size-full wp-image-1399\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26173615\/5-5-9.jpeg\" alt=\"A hemisphere with equation z = the square root of (4 minus x squared minus y squared) in the upper half plane, and within it, a cone with equation z = the square root of (3 times (x squared + y squared)) that is pointing down, with vertex at the origin.\" width=\"490\" height=\"340\" \/><\/p>\n<p id=\"caption-attachment-1399\" class=\"wp-caption-text\">Figure 4.\u00a0A region bounded below by a cone and above by a hemisphere.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q972450157\">Show Solution<\/span><\/p>\n<div id=\"q972450157\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794160034\">Using the conversion formulas from rectangular coordinates to spherical coordinates, we have:<\/p>\n<p id=\"fs-id1167793384878\">For the cone: [latex]{z} = {\\sqrt{{3}{({x^2}+{y^2})}}}[\/latex] or [latex]{\\rho} \\ {\\cos} \\ {\\varphi} = {\\sqrt{3}} \\ {\\rho} \\ {\\sin} \\ {\\varphi}[\/latex] or [latex]{\\tan} \\ {\\varphi} = {\\frac{1}{\\sqrt{3}}}[\/latex] or [latex]{\\varphi} = {\\frac{\\pi}{6}}[\/latex].<\/p>\n<p id=\"fs-id1167793261178\">For the sphere: [latex]{z} = {\\sqrt{{4}-{x^2}-{y^2}}}[\/latex] or [latex]{z^2} + {x^2} + {y^2} = {4}[\/latex] or [latex]{{\\rho}^{2}} = {4}[\/latex] or [latex]{\\rho} = {2}[\/latex].<\/p>\n<p id=\"fs-id1167793433725\">Thus, the triple integral for the volume is [latex]{{V}{(E)}} = {\\displaystyle\\int^{{\\theta} = {{2}{\\pi}}}_{{\\theta} = {0}}} \\ \\ {\\displaystyle\\int^{{\\varphi} = {{\\pi}\/{6}}}_{{\\phi} = {0}}} \\ \\ {\\displaystyle\\int^{{\\rho} = {2}}_{{\\rho} = {0}}} \\ {{\\rho}^{2}}{\\sin} \\ {\\varphi} \\ {{d}{\\rho}} \\ {{d}{\\varphi}} \\ {{d}{\\theta}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Set up a triple integral for the volume of the solid region bounded above by the sphere [latex]{\\rho} = {2}[\/latex] and bounded below by the cone [latex]{\\varphi} = {{\\pi}\/{3}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q422572964\">Show Solution<\/span><\/p>\n<div id=\"q422572964\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]V(E)=\\displaystyle\\int_{\\theta=0}^{\\theta=2\\pi}\\displaystyle\\int_{\\phi=0}^{\\varphi=\\pi\/3}\\displaystyle\\int_{\\rho=0}^{\\rho=2}\\rho^2\\sin\\varphi \\ d\\rho \\ d\\varphi \\ d\\theta.[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: interchanging order of integration in spherical coordinates<\/h3>\n<p id=\"fs-id1167794212109\">Let [latex]E[\/latex]\u00a0<span style=\"font-size: 1rem; text-align: initial;\">be the region bounded below by the cone [latex]{z} = {\\sqrt{{x^2}+{y^2}}}[\/latex] and above by the sphere [latex]{z} = {x^2} + {y^2} + {z^2}[\/latex] (<\/span>Figure 5<span style=\"font-size: 1rem; text-align: initial;\">). Set up a triple integral in spherical coordinates and find the volume of the region using the following orders of integration:<\/span><\/p>\n<ol id=\"fs-id1167793400632\" type=\"a\">\n<li>[latex]{{d}{\\rho}} \\ {{d}{\\phi}} \\ {{d}{\\theta}}[\/latex],<\/li>\n<li><span data-type=\"newline\">[latex]{{d}{\\varphi}} \\ {{d}{\\rho}} \\ {{d}{\\theta}}[\/latex].<br \/>\n<\/span><\/li>\n<\/ol>\n<div id=\"attachment_1401\" style=\"width: 490px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1401\" class=\"size-full wp-image-1401\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26173806\/5-5-10.jpeg\" alt=\"A sphere with equation z = x squared + y squared + z squared, and within it, a cone with equation z = the square root of (x squared + y squared) that is pointing down, with vertex at the origin.\" width=\"480\" height=\"341\" \/><\/p>\n<p id=\"caption-attachment-1401\" class=\"wp-caption-text\">Figure 5.\u00a0A region bounded below by a cone and above by a sphere.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q178836902\">Show Solution<\/span><\/p>\n<div id=\"q178836902\" class=\"hidden-answer\" style=\"display: none\">\n<p>1. Use the conversion formulas to write the equations of the sphere and cone in spherical coordinates.<span data-type=\"newline\"><br \/>\n<\/span>For the sphere:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned} {x^2} + {y^2} + \\ & {z^2} = {z} \\\\  & {{\\rho}^{2}} = {\\rho} \\ {\\cos} \\ {\\varphi} \\\\  & {\\rho} = {\\cos} \\ {\\varphi}.  \\end{aligned}[\/latex]<\/p>\n<p style=\"text-align: left;\">For the cone:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {z} & = {\\sqrt{{x^2}+{y^2}}} \\\\  {\\rho} \\ {\\cos} \\ {\\varphi} & = {\\sqrt{{{\\rho}^{2}}{{\\sin}^{2}}{\\varphi}{{\\cos}^{2}}{\\phi}+{{\\rho}^{2}}{{\\sin}^{2}}{\\varphi}{{\\sin}^{2}}{\\phi}}} \\\\  {\\rho} \\ {\\cos} \\ {\\varphi} & = {\\sqrt{{{\\rho}^{2}}{{\\sin}^{2}}{\\varphi}{({{\\cos}^{2}}{\\phi}+{{\\sin}^{2}}{\\phi})}}} \\\\  {\\rho} \\ {\\cos} \\ {\\varphi} & = {\\rho} \\ {\\sin} \\ {\\varphi} \\\\  {\\cos} \\ {\\varphi} & = {\\sin} \\ {\\varphi} \\\\  {\\varphi} & = {{\\pi}\/{4}}.  \\end{aligned}[\/latex]<\/p>\n<p>Hence the integral for the volume of the solid region [latex]E[\/latex] becomes<\/p>\n<p style=\"text-align: center;\">[latex]{{V}{(E)}} = {\\displaystyle\\int^{{\\theta} = {{2}{\\pi}}}_{{\\theta} = {0}}} \\ {\\displaystyle\\int^{{\\varphi} = {{\\pi}\/{4}}}_{{\\varphi} = {0}}} \\ {\\displaystyle\\int^{{\\rho} = {{\\cos}{\\varphi}}}_{{\\rho} = {0}}} \\ {{{\\rho}^{2}}{\\sin} \\ {\\varphi} \\ {{d}{\\rho}} \\ {{d}{\\varphi}} \\ {{d}{\\theta}}}.[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>2. Consider the [latex]{\\varphi}{\\rho}[\/latex]-plane. Note that the ranges for [latex]{\\varphi}[\/latex] and [latex]{\\rho}[\/latex] (from part a.) are<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {0} & \\ {\\leq} \\ {\\varphi} \\ {\\leq} \\ {{\\pi}\/4} \\\\  {0} & \\ {\\leq} \\ {\\phi} \\ {\\leq} \\ {{\\cos}{\\varphi}}.  \\end{aligned}[\/latex]<\/p>\n<p>The curve [latex]{\\rho} = {\\cos}{\\varphi}[\/latex] meets the line [latex]{\\varphi} = {{\\pi}\/{4}}[\/latex] at the point [latex]({{\\pi}\/{4}},{{\\sqrt{2}}\/{2}})[\/latex]. Thus, to change the order of integration, we need to use two pieces:<\/p>\n<p>[latex]\\hspace{8cm}\\begin{align}&0\\leq\\rho\\leq\\sqrt2\/2 \\\\&0\\leq\\varphi\\leq\\pi\/4\\end{align}[\/latex] [latex]\\text{ and }[\/latex] [latex]\\begin{align}&\\sqrt2\/2\\leq\\rho\\leq1 \\\\&0\\leq\\varphi\\leq\\cos^{-1}\\rho\\end{align}[\/latex]<\/p>\n<p>Hence the integral for the volume of the solid region [latex]E[\/latex] becomes<\/p>\n<p style=\"text-align: center;\">[latex]{{V}{(E)}} = {\\displaystyle\\int^{{\\theta} = {{2}{\\pi}}}_{{\\theta} = {0}}} \\ {\\displaystyle\\int^{{\\rho} = {{\\sqrt{2}\/{2}}}}_{{\\rho} = {0}}} \\ {\\displaystyle\\int^{{\\varphi} = {{\\pi}\/{4}}}_{{\\varphi} = {0}}} \\ {{{\\rho}^{2}}{\\sin} \\ {\\varphi} \\ {{d}{\\varphi}} \\ {{d}{\\rho}} \\ {{d}{\\theta}}} \\ + {\\displaystyle\\int^{{\\theta} = {{2}{\\pi}}}_{{\\theta} = {0}}} \\ {\\displaystyle\\int^{{\\rho} = {1}}_{{\\rho} = {{\\sqrt{2}\/{2}}}}} \\ {\\displaystyle\\int^{{\\varphi} = {{\\cos}^{-1}}{\\rho}}_{{\\varphi} = {0}}} \\ {{{\\rho}^{2}}{\\sin} \\ {\\varphi} \\ {{d}{\\varphi}} \\ {{d}{\\rho}} \\ {{d}{\\theta}}}.[\/latex]<\/p>\n<p>In each case, the integration results in [latex]{{V}{(E)}} = {\\frac{\\pi}{8}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1167794296748\">Before we end this section, we present a couple of examples that can illustrate the conversion from rectangular coordinates to cylindrical coordinates and from rectangular coordinates to spherical coordinates.<\/p>\n<div id=\"fs-id1167794296753\" class=\"ui-has-child-title\" data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: converting from rectangular coordinates to cylindrical coordinates<\/h3>\n<p>Convert the following integral into cylindrical coordinates:<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{{y} = {1}}_{{y} = {-1}}}\\displaystyle\\int^{{x} = {\\sqrt{{1} - {{y}^{2}}}}}_{{x} = {0}} \\ {\\displaystyle\\int^{{z} = {\\sqrt{{x^2}+{y^2}}}}_{{z} = {{x}^{2}}+{{y}^{2}}}} \\ {{x}{y}{z}} \\ {{d}{z}} \\ {{d}{x}} \\ {{d}{y}}.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q368178320\">Show Solution<\/span><\/p>\n<div id=\"q368178320\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793544553\">The ranges of the variables are<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {-1} & \\ \\ \\ {\\leq} \\ \\ \\ {y}{\\leq}{1} \\\\  {0} & \\ \\ \\ {\\leq} \\ \\ \\ {x}{\\leq}{\\sqrt{{1} - {{y}^{2}}}} \\\\  {{x}^{2}} + {{y}^{2}} & \\ \\ \\ {\\leq} \\ \\ \\ {z}{\\leq}{\\sqrt{{{x}^{2}}+{{y}^{2}}}}.  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1167793464597\">The first two inequalities describe the right half of a circle of radius 1. Therefore, the ranges for [latex]{\\theta}[\/latex] and [latex]r[\/latex] are<\/p>\n<p style=\"text-align: center;\">[latex]{-{\\frac{\\pi}{2}}} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\frac{\\pi}{2}} \\ {\\text{and}} \\ {0} \\ {\\leq} \\ {r} \\ {\\leq} \\ {1}.[\/latex]<\/p>\n<p id=\"fs-id1167793514913\">The limits of [latex]z[\/latex] are [latex]{{r}^{2}}[\/latex], hence<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{{y} = {1}}_{{y} = {-1}}}\\displaystyle\\int^{{x} = {\\sqrt{{1} - {{y}^{2}}}}}_{{x} = {0}}\\displaystyle\\int^{{z} = {\\sqrt{{{x}^{2}} + {{y}^{2}}}}}_{{z} = {{x}^{2}} + {{y}^{2}}} \\ {{x}{y}{z}} \\ {{d}{z}} \\ {{d}{x}} \\ {{d}{y}} = {\\displaystyle\\int^{{\\theta} = {{\\pi}\/{2}}}_{{\\theta} = {-{\\pi}\/{2}}}} \\ {\\displaystyle\\int^{{r} = {1}}_{{r} = {0}}} \\ {\\displaystyle\\int^{{z} = {r}}_{{z} = {{r}^{2}}}} \\ {r}{({r}{\\cos}{\\theta})}{({r}{\\sin}{\\theta})}{z} \\ {{d}{z}} \\ {{d}{r}} \\ {{d}{\\theta}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: converting from rectangular coordinates to spherical coordinates<\/h3>\n<p>Convert the following integral into spherical coordinates:<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{{y} = {3}}_{{y} = {0}}}\\displaystyle\\int^{{x} = {\\sqrt{{9} - {{y}^{2}}}}}_{{x} = {0}}\\displaystyle\\int^{{z} = {\\sqrt{{18} - {{x}^{2}} - {{y}^{2}}}}}_{{z} = {\\sqrt{{x}^{2} + {{y}^{2}}}}} \\ {({{x}^{2}} + {{y}^{2}} + {{z}^{2}})} \\ {{d}{z}} \\ {{d}{x}} \\ {{d}{y}}.[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q255795354\">Show Solution<\/span><\/p>\n<div id=\"q255795354\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794052495\">The ranges of the variables are<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {0} \\ \\ \\ & {\\leq} \\ \\ \\ {y} \\ {\\leq} \\ {3} \\\\  {0} \\ \\ \\ & {\\leq} \\ \\ \\ {x} \\ {\\leq} \\ {\\sqrt{{9} - {{y}^{2}}}} \\\\  {\\sqrt{{{x}^{2}} + {{y}^{2}}}} \\ \\ \\ & {\\leq} \\ \\ \\ {z} \\ {\\leq} \\ {\\sqrt{{18} - {{x}^{2}} - {{y}^{2}}}}.  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1167793777086\">The first two ranges of variables describe a quarter disk in the first quadrant of the [latex]xy[\/latex]-plane. Hence the range for [latex]{\\theta}[\/latex] is [latex]{0} \\ {\\leq} \\ {\\theta} \\ {\\leq} \\ {\\frac{\\pi}{2}}[\/latex].<\/p>\n<p id=\"fs-id1167794067621\">The lower bound [latex]{z} = {\\sqrt{{{x}^{2}} + {{y}^{2}}}}[\/latex] is the upper half of a cone and the upper bound [latex]{z} = {\\sqrt{{18} - {{x}^{2}} - {{y}^{2}}}}[\/latex] is the upper half of a sphere. Therefore, we have [latex]{0} \\ {\\leq} \\ {\\rho} \\ {\\leq} \\ {\\sqrt{18}}[\/latex], which is [latex]{0} \\ {\\leq} \\ {\\rho} \\ {\\leq} \\ {3{\\sqrt{2}}}[\/latex].<\/p>\n<p id=\"fs-id1167794168074\">For the ranges of [latex]{\\varphi}[\/latex], we need to find where the cone and the sphere intersect, so solve the equation<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {{{r}^{2}} + {{z}^{2}}} \\ \\ \\ & = \\ \\ \\ {18} \\\\  {({\\sqrt{{{x}^{2}} + {{y}^{2}}}})^2} + {{z}^{2}} \\ \\ \\ & = \\ \\ \\ {18} \\\\  {{z}^{2}} + {{z}^{2}} \\ \\ \\ & = \\ \\ \\ {18} \\\\  {{2}{{z}^{2}}} \\ \\ \\ & = \\ \\ \\ {18} \\\\  {{z}^{2}} \\ \\ \\ & = \\ \\ \\ {9} \\\\  {z} \\ \\ \\ & = \\ \\ \\ {3}.  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1167793610938\">This gives<\/p>\n<p>[latex]\\hspace{10cm}\\begin{align}    3\\sqrt2\\cos\\varphi&=3 \\\\    \\cos\\varphi&=\\frac1{\\sqrt2} \\\\    \\varphi&=\\frac{\\pi}4.    \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1167793924123\">Putting this together, we obtain<\/p>\n<div style=\"text-align: center;\">[latex]\\displaystyle\\int_{y=0}^{y=3}\\displaystyle\\int_{x=0}^{x=\\sqrt{9-y^2}}\\displaystyle\\int_{z=\\sqrt{x^2+y^2}}^{z=\\sqrt{18-x^2-y^2}}(x^2+y^2+z^2)dz \\ dx \\ dy=\\displaystyle\\int_{\\varphi=0}^{\\varphi=\\pi\/4}\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/2}\\displaystyle\\int_{\\rho=0}^{\\rho=3\\sqrt2}\\rho^4\\sin\\varphi \\ d\\rho \\ d\\theta \\ d\\varphi.[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Use rectangular, cylindrical, and spherical coordinates to set up triple integrals for finding the volume of the region inside the sphere [latex]x^2+y^2+z^2=4[\/latex] but outside the cylinder\u00a0[latex]x^2+y^2=1[\/latex].<\/p>\n<div id=\"attachment_1402\" style=\"width: 503px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1402\" class=\"size-full wp-image-1402\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26173912\/5-5-11.jpeg\" alt=\"A sphere with equation x squared + y squared + z squared = 4, and within it, a cylinder with equation x squared + y squared = 1.\" width=\"493\" height=\"402\" \/><\/p>\n<p id=\"caption-attachment-1402\" class=\"wp-caption-text\">Figure 6.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q842692876\">Show Solution<\/span><\/p>\n<div id=\"q842692876\" class=\"hidden-answer\" style=\"display: none\">\n<p>Rectangular:\u00a0[latex]\\displaystyle\\int_{x=-2}^{x=2}\\displaystyle\\int_{y=-\\sqrt{4-x^2}}^{y=\\sqrt{4-x^2}}\\displaystyle\\int_{z=-\\sqrt{4-x^2-y^2}}^{z=\\sqrt{4-x^2-y^2}}dz \\ dy \\ dx - \\displaystyle\\int_{x=-1}^{x=1}\\displaystyle\\int_{y=-\\sqrt{1-x^2}}^{y=\\sqrt{1-x^2}}\\displaystyle\\int_{z=-\\sqrt{4-x^2-y^2}}^{z=\\sqrt{4-x^2-y^2}}dz \\ dy \\ dx.[\/latex]<\/p>\n<p>Cylindrical:\u00a0[latex]\\displaystyle\\int_{\\theta=0}^{\\theta=2\\pi}\\displaystyle\\int_{r=1}^{r=2}\\displaystyle\\int_{z=-\\sqrt{4-r^2}}^{z=\\sqrt{4-r^2}}r \\ dz \\ dr \\ d\\theta.[\/latex]<\/p>\n<p>Spherical:\u00a0[latex]\\displaystyle\\int_{\\varphi=\\pi\/6}^{\\varphi=5\\pi\/6}\\displaystyle\\int_{\\theta=0}^{\\theta=2\\pi}\\displaystyle\\int_{\\rho=\\csc\\varphi}^{\\rho=2} \\rho^2 \\sin{\\varphi} d\\rho \\ d\\theta \\ d\\varphi[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try It<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8197109&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=7iLbCVweWdI&amp;video_target=tpm-plugin-9elr1s1g-7iLbCVweWdI\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.32_transcript.html\">transcript for \u201cCP 5.32\u201d here (opens in new window).<\/a><\/div>\n<p id=\"fs-id1167793442442\">Now that we are familiar with the spherical coordinate system, let\u2019s find the volume of some known geometric figures, such as spheres and ellipsoids.<\/p>\n<div id=\"fs-id1167793442449\" class=\"ui-has-child-title\" data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: chapter opener: finding the volume of L&#8217;hemisph\u00c8ric<\/h3>\n<p>Find the volume of the spherical planetarium in l\u2019Hemisph\u00e8ric in Valencia, Spain, which is five stories tall and has a radius of approximately [latex]50[\/latex] ft, using the equation [latex]x^2+y^2+z^2=r^2[\/latex].<\/p>\n<div id=\"attachment_5569\" style=\"width: 335px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5569\" class=\"size-full wp-image-5569\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/03022644\/5.60.jpeg\" alt=\"A picture of l\u2019Hemisph\u00e8ric, which is a giant glass structure that is in the shape of an ellipsoid.\" width=\"325\" height=\"219\" \/><\/p>\n<p id=\"caption-attachment-5569\" class=\"wp-caption-text\">Figure 7. (credit: modification of work by Javier Yaya Tur, Wikimedia Commons)<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q204959485\">Show Solution<\/span><\/p>\n<div id=\"q204959485\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793416704\">We calculate the volume of the ball in the first octant, where [latex]x\\geq0[\/latex], [latex]y\\geq0[\/latex], and [latex]z\\geq0[\/latex], using spherical coordinates, and then multiply the result by [latex]8[\/latex] for symmetry. Since we consider the region [latex]D[\/latex] as the first octant in the integral, the ranges of the variables are<\/p>\n<p style=\"text-align: center;\">[latex]\\large{0\\leq\\varphi\\leq\\frac{\\pi}2, \\ 0\\leq\\rho\\leq{r}, \\ 0\\leq\\theta\\leq\\frac{\\pi}2}[\/latex]<\/p>\n<p id=\"fs-id1167793500489\">Therefore,<\/p>\n<p>[latex]\\hspace{6cm}\\begin{align}    V&=\\underset{D}{\\displaystyle\\iiint}dx \\ dy \\ dz = 8\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/2}\\displaystyle\\int_{\\rho=0}^{\\rho=\\pi}\\displaystyle\\int_{\\varphi=0}^{\\varphi=\\pi\/2}\\rho^2\\sin\\theta \\ d\\varphi \\ d\\rho \\ d\\theta \\\\    &=8\\displaystyle\\int_{\\varphi=0}^{\\varphi=\\pi\/2}d\\varphi\\displaystyle\\int_{\\rho=0}^{\\rho=\\pi}\\rho^2 \\ d\\rho\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/2}\\sin\\theta \\ d\\theta \\\\    &=8\\left(\\frac{\\pi}2\\right)\\left(\\frac{r^3}3\\right)(1) \\\\    &=\\frac43\\pi{r}^3    \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1167793522367\">This exactly matches with what we knew. So for a sphere with a radius of approximately [latex]50[\/latex] ft, the volume is [latex]\\frac43\\pi{r}^3\\approx523,000\\text{ ft.}^3[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>For the next example we find the volume of an ellipsoid.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: finding the volume of an ellipsoid<\/h3>\n<p>Find the volume of the ellipsoid [latex]\\frac{x^2}{a^2}+\\frac{y^2}{b^2}+\\frac{z^2}{c^2}=1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q543745872\">Show Solution<\/span><\/p>\n<div id=\"q543745872\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793251274\">We again use symmetry and evaluate the volume of the ellipsoid using spherical coordinates. As before, we use the first octant [latex]x\\geq0[\/latex], [latex]y\\geq0[\/latex], and [latex]z\\geq0[\/latex] and then multiply the result by 8.<\/p>\n<p id=\"fs-id1167793219213\">In this case the ranges of the variables are<\/p>\n<p style=\"text-align: center;\">[latex]\\large{0\\leq\\varphi\\leq\\frac{\\pi}2, \\ 0\\leq\\rho\\leq\\frac{\\pi}2, \\ 0\\leq\\rho\\leq1, \\ 0\\leq\\theta\\leq\\frac{\\pi}2}[\/latex]<\/p>\n<p id=\"fs-id1167793540294\">Also, we need to change the rectangular to spherical coordinates in this way:<\/p>\n<p style=\"text-align: center;\">[latex]\\large{x=a\\rho\\cos\\varphi\\sin\\theta, \\ y=b\\rho\\sin\\varphi\\sin\\theta,\\text{ and }z=c\\rho\\cos\\theta}.[\/latex]<\/p>\n<p id=\"fs-id1167794213144\">Then the volume of the ellipsoid becomes<\/p>\n<p>[latex]\\hspace{6cm}\\begin{align}    V&=\\underset{D}{\\displaystyle\\iiint}dx \\ dy \\ dz \\\\    &=8\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/2}\\displaystyle\\int_{\\rho=0}^{\\rho=1}\\displaystyle\\int_{\\varphi=0}^{\\varphi=\\pi\/2}abc\\rho^2\\sin\\theta \\ d\\varphi \\ d\\rho \\ d\\theta \\\\    &=8abc\\displaystyle\\int_{\\varphi=0}^{\\varphi=\\pi\/2}d\\varphi\\displaystyle\\int_{\\rho=0}^{\\rho=1}\\rho^2 \\ d\\rho\\displaystyle\\int_{\\theta=0}^{\\theta=\\pi\/2}\\sin\\theta \\ d\\theta \\\\    &=8abc\\left(\\frac{\\pi}2\\right)\\left(\\frac13\\right)(1) \\\\    &=\\frac43\\pi{abc}    \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: finding the volume of the space inside an ellipsoid and outside a sphere<\/h3>\n<p>Find the volume of the space inside the ellipsoid [latex]\\frac{x^2}{75^2}+\\frac{y^2}{80^2}+\\frac{z^2}{90^2}=1[\/latex] and outside the sphere [latex]x^2+y^2+z^2=50^2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q234085739\">Show Solution<\/span><\/p>\n<div id=\"q234085739\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793925860\">This problem is directly related to the l\u2019Hemisph\u00e8ric structure. The volume of space inside the ellipsoid and outside the sphere might be useful to find the expense of heating or cooling that space. We can use the preceding two examples for the volume of the sphere and ellipsoid and then subtract.<\/p>\n<p id=\"fs-id1167793732638\">First we find the volume of the ellipsoid using [latex]a=75[\/latex] ft, [latex]b=80[\/latex] ft, and [latex]c=90[\/latex] ft in the result from\u00a0Example &#8220;Finding the Volume of an Ellipsoid&#8221;.\u00a0Hence the volume of the ellipsoid is<\/p>\n<p style=\"text-align: center;\">[latex]\\large{V_{\\text{ellipsoid}}=\\frac43\\pi(75)(80)(90)\\approx2,262,000\\text{ ft.}^3}[\/latex]<\/p>\n<p id=\"fs-id1167793452781\">From\u00a0Example &#8220;Chapter Opener: Finding the Volume of l\u2019Hemisph\u00e8ric&#8221;,\u00a0the volume of the sphere is<\/p>\n<p style=\"text-align: center;\">[latex]\\large{V_{\\text{sphere}}\\approx523,600\\text{ ft.}^3.}[\/latex]<\/p>\n<p id=\"fs-id1167793556981\">Therefore, the volume of the space inside the ellipsoid [latex]\\frac{x^2}{75^2}+\\frac{y^2}{80^2}+\\frac{z^2}{90^2}=1[\/latex] and outside the sphere [latex]x^2+y^2+z^2=50^2[\/latex] is approximately<\/p>\n<p style=\"text-align: center;\">[latex]\\large{V_{\\text{Hemisferic}}=V_{\\text{ellipsoid}}-V_{\\text{sphere}}=1,738,400\\text{ ft.}^3.}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Activity: hot air balloons<\/h3>\n<p id=\"fs-id1167793285292\">Hot air ballooning is a relaxing, peaceful pastime that many people enjoy. Many balloonist gatherings take place around the world, such as the Albuquerque International Balloon Fiesta. The Albuquerque event is the largest hot air balloon festival in the world, with over [latex]500[\/latex] balloons participating each year.<\/p>\n<div id=\"attachment_1403\" style=\"width: 910px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1403\" class=\"size-full wp-image-1403\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174048\/5-5-12.jpeg\" alt=\"A picture of many hot air balloons.\" width=\"900\" height=\"600\" \/><\/p>\n<p id=\"caption-attachment-1403\" class=\"wp-caption-text\">Figure 8.\u00a0Balloons lift off at the 2001 Albuquerque International Balloon Fiesta. (credit: David Herrera, Flickr)<\/p>\n<\/div>\n<p id=\"fs-id1167793510308\">As the name implies, hot air balloons use hot air to generate lift. (Hot air is less dense than cooler air, so the balloon floats as long as the hot air stays hot.) The heat is generated by a propane burner suspended below the opening of the basket. Once the balloon takes off, the pilot controls the altitude of the balloon, either by using the burner to heat the air and ascend or by using a vent near the top of the balloon to release heated air and descend. The pilot has very little control over where the balloon goes, however\u2014balloons are at the mercy of the winds. The uncertainty over where we will end up is one of the reasons balloonists are attracted to the sport.<\/p>\n<p id=\"fs-id1167793510321\">In this project we use triple integrals to learn more about hot air balloons. We model the balloon in two pieces. The top of the balloon is modeled by a half sphere of radius [latex]28[\/latex] feet. The bottom of the balloon is modeled by a frustum of a cone (think of an ice cream cone with the pointy end cut off). The radius of the large end of the frustum is [latex]28[\/latex] feet and the radius of the small end of the frustum is [latex]6[\/latex] feet. A graph of our balloon model and a cross-sectional diagram showing the dimensions are shown in the following figure.<\/p>\n<div id=\"attachment_1404\" style=\"width: 811px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1404\" class=\"size-full wp-image-1404\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174119\/5-5-13.jpeg\" alt=\"This figure consists of two parts, a and b. Figure a shows a representation of a hot air balloon in xyz space as a half sphere on top of a frustrum of a cone. Figure b shows the dimensions, namely, the radius of the half sphere is 28 ft, the distance from the bottom to the top of the frustrum is 44 ft, and the diameter of the circle at the top of the frustrum is 12 ft.\" width=\"801\" height=\"458\" \/><\/p>\n<p id=\"caption-attachment-1404\" class=\"wp-caption-text\">Figure 9.\u00a0(a) Use a half sphere to model the top part of the balloon and a frustum of a cone to model the bottom part of the balloon. (b) A cross section of the balloon showing its dimensions.<\/p>\n<\/div>\n<p id=\"fs-id1167794200502\">We first want to find the volume of the balloon. If we look at the top part and the bottom part of the balloon separately, we see that they are geometric solids with known volume formulas. However, it is still worthwhile to set up and evaluate the integrals we would need to find the volume. If we calculate the volume using integration, we can use the known volume formulas to check our answers. This will help ensure that we have the integrals set up correctly for the later, more complicated stages of the project.<\/p>\n<ol id=\"fs-id1167794200510\" type=\"1\">\n<li>Find the volume of the balloon in two ways.\n<ol id=\"fs-id1167794200518\" type=\"a\">\n<li>Use triple integrals to calculate the volume. Consider each part of the balloon separately. (Consider using spherical coordinates for the top part and cylindrical coordinates for the bottom part.)<\/li>\n<li>Verify the answer using the formulas for the volume of a sphere, [latex]V=\\frac43\\pi{r}^3[\/latex], and for the volume of a cone, [latex]V=\\frac13\\pi{r}^2h[\/latex].<\/li>\n<\/ol>\n<p>In reality, calculating the temperature at a point inside the balloon is a tremendously complicated endeavor. In fact, an entire branch of physics (thermodynamics) is devoted to studying heat and temperature. For the purposes of this project, however, we are going to make some simplifying assumptions about how temperature varies from point to point within the balloon. Assume that just prior to liftoff, the temperature (in degrees Fahrenheit) of the air inside the balloon varies according to the function<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167793937390\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<p style=\"text-align: center;\">[latex]\\large{T_0(r,\\theta,z)=\\frac{z-r}{10}+210}[\/latex]<\/p>\n<\/div>\n<\/li>\n<li>What is the average temperature of the air in the balloon just prior to liftoff? (Again, look at each part of the balloon separately, and do not forget to convert the function into spherical coordinates when looking at the top part of the balloon.)<span data-type=\"newline\"><br \/>\n<\/span>Now the pilot activates the burner for [latex]10[\/latex] seconds. This action affects the temperature in a [latex]12[\/latex]-foot-wide column [latex]20[\/latex] feet high, directly above the burner. A cross section of the balloon depicting this column in shown in the following figure.<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"CNX_Calc_Figure_15_05_SP3\" class=\"os-figure\">\n<figure data-id=\"CNX_Calc_Figure_15_05_SP3\">\n<div id=\"attachment_1405\" style=\"width: 395px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1405\" class=\"size-full wp-image-1405\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/26174158\/5-5-14.jpeg\" alt=\"This figure shows the dimensions of the balloon and the hot air, namely, the radius of the half sphere is 28 ft, the distance from the bottom to the top of the frustrum is 44 ft, the diameter of the circle at the top of the frustrum is 12 ft, and the inner column of hot air has height 20 ft and diameter 12 ft.\" width=\"385\" height=\"385\" \/><\/p>\n<p id=\"caption-attachment-1405\" class=\"wp-caption-text\">Figure 10.\u00a0<span class=\"os-caption\">Activating the burner heats the air in a [latex]20[\/latex]-foot-high, [latex]12[\/latex]-foot-wide column directly above the burner.<\/span><\/p>\n<\/div>\n<\/figure>\n<\/div>\n<p>Assume that after the pilot activates the burner for [latex]10[\/latex] seconds, the temperature of the air in the column described above\u00a0<em data-effect=\"italics\">increases<\/em>\u00a0according to the formula<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167793376255\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<p style=\"text-align: center;\">[latex]\\large{H(r,\\theta,z)=-2z-48.}[\/latex]<\/p>\n<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Then the temperature of the air in the column is given by<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167794067545\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<p style=\"text-align: center;\">[latex]\\large{T_1(r,\\theta,z)=\\frac{z-r}{10}+210+(-2z-48)}[\/latex],<\/p>\n<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span>while the temperature in the remainder of the balloon is still given by<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167794292164\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<p style=\"text-align: center;\">[latex]\\large{T_0(r,\\theta,z)=\\frac{z-r}{10}+210}[\/latex].<\/p>\n<\/div>\n<\/li>\n<li>Find the average temperature of the air in the balloon after the pilot has activated the burner for [latex]10[\/latex] seconds.<\/li>\n<\/ol>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1070\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>CP 5.32. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":428269,"menu_order":22,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"CP 5.32\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1070","chapter","type-chapter","status-publish","hentry"],"part":23,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1070","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/428269"}],"version-history":[{"count":152,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1070\/revisions"}],"predecessor-version":[{"id":6456,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1070\/revisions\/6456"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/23"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1070\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=1070"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=1070"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=1070"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=1070"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}