{"id":817,"date":"2021-08-27T19:56:56","date_gmt":"2021-08-27T19:56:56","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=817"},"modified":"2022-10-26T03:03:04","modified_gmt":"2022-10-26T03:03:04","slug":"spherical-coordinates","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/spherical-coordinates\/","title":{"raw":"Spherical Coordinates","rendered":"Spherical Coordinates"},"content":{"raw":"<div data-type=\"note\">\r\n<div id=\"fs-id1163724036465\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<div id=\"fs-id1163724074875\" class=\"theorem ui-has-child-title\" data-type=\"note\">\r\n<div id=\"fs-id1163724080388\" class=\"ui-has-child-title\" data-type=\"example\">\r\n<div data-type=\"example\">\r\n<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\">Convert from spherical to rectangular coordinates.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Convert from rectangular to spherical coordinates.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1163723669991\">In the Cartesian coordinate system, the location of a point in space is described using an ordered triple in which each coordinate represents a distance. In the cylindrical coordinate system, location of a point in space is described using two distances ([latex]r[\/latex] and[latex]z[\/latex]) and an angle measure ([latex]\\theta[\/latex]). In the spherical coordinate system, we again use an ordered triple to describe the location of a point in space. In this case, the triple describes one distance and two angles. Spherical coordinates make it simple to describe a sphere, just as cylindrical coordinates make it easy to describe a cylinder. Grid lines for spherical coordinates are based on angle measures, like those for polar coordinates.<\/p>\r\n\r\n<\/div>\r\n<div data-type=\"example\">\r\n<div class=\"textbox shaded\"><header>\r\n<h3 class=\"os-title\" style=\"text-align: center;\" data-type=\"title\"><span id=\"2\" class=\"os-title-label\" data-type=\"\">DEFINITION<\/span><\/h3>\r\n\r\n<hr \/>\r\n\r\n<\/header><section>\r\n<div class=\"os-note-body\">\r\n<p id=\"fs-id1163723830824\">In the\u00a0<strong><span id=\"293cc618-b0de-4ec9-9029-6efb4eb1b172_term107\" data-type=\"term\">spherical coordinate system<\/span><\/strong>, a point [latex]P[\/latex] in space (Figure 1) is represented by the ordered triple [latex](\\rho,\\theta,\\varphi)[\/latex] where<\/p>\r\n\r\n<ul id=\"fs-id1163723954970\" data-bullet-style=\"bullet\">\r\n \t<li>[latex]\\rho[\/latex] (the Greek letter rho) is the distance between [latex]P[\/latex] and the origin [latex](\\rho\\ne0)[\/latex];<\/li>\r\n \t<li>[latex]\\theta[\/latex] is the same angle used to describe the location in cylindrical coordinates;<\/li>\r\n \t<li>[latex]\\varphi[\/latex] (the Greek letter phi) is the angle formed by the positive [latex]z[\/latex]-axis and line segment [latex]\\overline{OP}[\/latex], where [latex]O[\/latex] is the origin and [latex]0\\leq\\varphi\\leq\\pi[\/latex].<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/section><\/div>\r\n\r\n[caption id=\"attachment_5293\" align=\"aligncenter\" width=\"437\"]<img class=\"size-full wp-image-5293\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31230733\/2.97.jpg\" alt=\"This figure is the first quadrant of the 3-dimensional coordinate system. It has a point labeled \u201c(x, y, z) = (rho, theta, phi).\u201d There is a line segment from the origin to the point. It is labeled \u201crho.\u201d The angle between this line segment and the z-axis is phi. There is a line segment in the x y-plane from the origin to the shadow of the point. This segment is labeled \u201cr.\u201d The angle between the x-axis and r is theta.\" width=\"437\" height=\"457\" \/> Figure 1. The relationship among spherical, rectangular, and cylindrical coordinates.[\/caption]\r\n\r\nBy convention, the origin is represented as [latex](0, 0, 0)[\/latex] in spherical coordinates.\r\n<div class=\"textbox shaded\"><header><header>\r\n<h3 class=\"os-title\" style=\"text-align: center;\"><span class=\"os-title-label\">THEOREM: converting among Spherical, cylindrical, and rectangular coordinates<\/span><\/h3>\r\n\r\n<hr \/>\r\n\r\n<\/header><section>\r\n<div class=\"os-note-body\">\r\n<p id=\"fs-id1163723772032\">Rectangular coordinates [latex](x, y, z)[\/latex] and spherical coordinates [latex](\\rho,\\theta,\\varphi)[\/latex] of a point are related as follows:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nx&amp;=\\rho\\sin\\varphi\\cos\\theta \\quad \\text{These equations are used to convert from spherical coordinates to rectangular coordinates.} \\\\\r\ny&amp;=\\rho\\sin\\varphi\\sin\\theta \\\\\r\nz&amp;=\\rho\\cos\\varphi \\\\\r\n&amp;\\text{and} \\\\\r\n\\rho^2&amp;=x^2+y^2+z^2 \\quad \\text{These equations are used to convert from rectangular coordinates to spherical coordinates.} \\\\\r\n\\tan\\theta&amp;=\\frac{y}x \\\\\r\n\\varphi&amp;=\\arccos\\left(\\frac{z}{\\sqrt{x^2+y^2+z^2}}\\right)\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1163723968448\">If a point has cylindrical coordinates [latex](r,\\theta,z)[\/latex], then these equations define the relationship between cylindrical and spherical coordinates.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nr&amp;=\\rho\\sin\\varphi \\quad \\text{These equations are used to convert from spherical coordinates to cylindrical coordinates} \\\\\r\n\\theta&amp;=\\theta \\\\\r\nz&amp;=\\rho\\cos\\varphi \\\\\r\n&amp;\\text{and} \\\\\r\n\\rho&amp;=\\sqrt{r^2+z^2} \\quad \\text{These equations are used to convert from cylindrical coordinates to spherical coordinates} \\\\\r\n\\theta&amp;=\\theta \\\\\r\n\\varphi&amp;=\\arccos\\left(\\frac{z}{\\sqrt{r^2+z^2}}\\right)\r\n\\end{aligned}[\/latex]<\/p>\r\n\r\n<div id=\"fs-id1163723563904\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\r\n<div class=\"MathJax_Display\"><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><\/header><\/div>\r\n<p id=\"fs-id1163723656797\"><span style=\"font-size: 1rem; text-align: initial;\">The formulas to convert from spherical coordinates to rectangular coordinates may seem complex, but they are straightforward applications of trigonometry. Looking at\u00a0Figure 2, it is easy to see that [latex]r=\\rho\\cos\\varphi[\/latex]. Then, looking at the triangle in the [latex]xy[\/latex]-plane with [latex]r[\/latex] as its hypotenuse, we have [latex]x=r\\cos\\theta=\\rho\\sin\\varphi\\cos\\theta[\/latex]. The derivation of the formula for [latex]y[\/latex] is similar.\u00a0Figure 9 in <a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/cylindrical-coordinates\/\" target=\"_blank\" rel=\"noopener\">Cylindrical Coordinates<\/a>\u00a0also shows that [latex]\\rho^2=r^2+z^2=x^2+y^2+z^2[\/latex] and [latex]z=\\rho\\cos\\varphi[\/latex]. Solving this last equation for [latex]\\varphi[\/latex] and then substituting [latex]\\rho=\\sqrt{r^2+z^2}[\/latex] (from the first equation) yields [latex]\\varphi=\\arccos\\left(\\frac{z}{\\sqrt{r^2+z^2}}\\right)[\/latex]. Also, note that, as before, we must be careful when using the formula [latex]\\tan\\theta=\\frac{y}x[\/latex] to choose the correct value of [latex]\\theta[\/latex].\u00a0<\/span><\/p>\r\n\r\n\r\n[caption id=\"attachment_5294\" align=\"aligncenter\" width=\"437\"]<img class=\"size-full wp-image-5294\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31230837\/2.98.jpg\" alt=\"This figure is the first quadrant of the 3-dimensional coordinate system. It has a point labeled \u201c(x, y, z) = (r, theta, z) = (rho, theta, phi).\u201d There is a line segment from the origin to the point. It is labeled \u201crho.\u201d The angle between this line segment and the z-axis is phi. There is a line segment in the x y-plane from the origin to the shadow of the point. This segment is labeled \u201cr.\u201d The angle between the x-axis and r is theta.The distance from r to the point is labeled \u201cz.\u201d\" width=\"437\" height=\"457\" \/> Figure 2. The equations that convert from one system to another are derived from right-triangle relationships.[\/caption]\r\n\r\nAs we did with cylindrical coordinates, let\u2019s consider the surfaces that are generated when each of the coordinates is held constant. Let [latex]c[\/latex] be a constant, and consider surfaces of the form [latex]\\rho=c[\/latex]. Points on these surfaces are at a fixed distance from the origin and form a sphere. The coordinate [latex]\\theta[\/latex] in the spherical coordinate system is the same as in the cylindrical coordinate system, so surfaces of the form [latex]\\theta=c[\/latex] are half-planes, as before. Last, consider surfaces of the form [latex]\\varphi=c[\/latex]. The points on these surfaces are at a fixed angle from the [latex]z[\/latex]-axis and form a half-cone (Figure 3).\r\n\r\n[caption id=\"attachment_5296\" align=\"aligncenter\" width=\"623\"]<img class=\"size-full wp-image-5296\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31230916\/2.99.jpg\" alt=\"This figure has three images. The first image is a sphere centered in the 3-dimensional coordinate system. The second figure is a vertical plane with an edge on the z-axis in the 3-dimensional coordinate system. The third image is an elliptical cone with the center at the origin of the 3-dimensional coordinate system.\" width=\"623\" height=\"262\" \/> Figure 3. In spherical coordinates, surfaces of the form [latex]\\rho=c[\/latex] are spheres of radius [latex]\\rho[\/latex] (a), surfaces of the form [latex]\\theta=c[\/latex] are half-planes at an angle [latex]\\theta[\/latex] from the [latex]x[\/latex]-axis (b), and surfaces of the form [latex]\\varphi=c[\/latex] are half-cones at an angle [latex]\\varphi[\/latex] from the [latex]z[\/latex]-axis (c).[\/caption]\r\n<div class=\"textbox exercises\">\r\n<h3>Example: converting from spherical coordinates<\/h3>\r\nPlot the point with spherical coordinates [latex]\\left(8,\\frac{\\pi}3,\\frac{\\pi}6\\right)[\/latex] and express its location in both rectangular and cylindrical coordinates.\r\n\r\n[reveal-answer q=\"874652471\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"874652471\"]\r\n<p id=\"fs-id1163723236542\">Use the equations in\u00a0Theorem: Converting among Spherical, Cylindrical, and Rectangular Coordinates\u00a0to translate between spherical and cylindrical coordinates (Figure 4):<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nx&amp;=\\rho\\sin\\varphi\\cos\\theta=8\\sin\\left(\\frac{\\pi}6\\right)\\cos\\left(\\frac{\\pi}3\\right)=8\\left(\\frac12\\right)\\frac12=2 \\\\\r\ny&amp;=\\rho\\sin\\varphi\\sin\\theta=8\\sin\\left(\\frac{\\pi}6\\right)\\sin\\left(\\frac{\\pi}3\\right)=8\\left(\\frac12\\right)\\frac{\\sqrt{3}}2=2\\sqrt3 \\\\\r\nz&amp;=\\rho\\cos\\varphi=8\\cos\\left(\\frac{\\pi}6\\right)=8\\left(\\frac{\\sqrt{3}}2\\right)=4\\sqrt3.\r\n\\end{aligned}[\/latex]<\/p>\r\n\r\n[caption id=\"attachment_5304\" align=\"aligncenter\" width=\"676\"]<img class=\"size-full wp-image-5304\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31231600\/2.100.jpg\" alt=\"This figure is the first quadrant of the 3-dimensional coordinate system. It has a point labeled \u201c(8, pi\/3, pi\/6).\u201d There is a line segment from the origin to the point. It is labeled \u201crho = 8.\u201d The angle between this line segment and the z-axis is labeled \u201cphi = pi\/6.\u201d There is a line segment in the x y-plane from the origin to the shadow of the point. The angle between the x-axis and r is labeled \u201ctheta = pi\/3.\u201d\" width=\"676\" height=\"538\" \/> Figure 4. The projection of the point in the [latex]xy[\/latex]-plane is [latex]4[\/latex] units from the origin. The line from the origin to the point\u2019s projection forms an angle of [latex]\\frac{\\pi}{3}[\/latex] with the positive [latex]x[\/latex]-axis. The point lies [latex]4\\sqrt3[\/latex] units above the [latex]xy[\/latex]-plane.[\/caption]\r\n<p id=\"fs-id1163724035807\">The point with spherical coordinates [latex]\\left(8,\\frac{\\pi}3,\\frac{\\pi}6\\right)[\/latex] has rectangular coordinates [latex](2,2\\sqrt3,4\\sqrt3)[\/latex].<\/p>\r\n<p id=\"fs-id1163723242019\">Finding the values in cylindrical coordinates is equally straightforward:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nr&amp;=\\rho\\sin\\varphi=8\\sin\\frac{\\pi}6=4 \\\\\r\n\\theta&amp;=\\theta \\\\\r\nz&amp;=\\rho\\cos\\varphi=8\\cos\\frac{\\pi}6=4\\sqrt3\r\n\\end{aligned}[\/latex].<\/p>\r\n<p id=\"fs-id1163723994378\">Thus, cylindrical coordinates for the point are [latex]\\left(4,\\frac{\\pi}3,4\\sqrt3\\right)[\/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\nPlot the point with spherical coordinates [latex]\\left(2,-\\frac{5\\pi}6,\\frac{\\pi}6\\right)[\/latex] and describe its location in both rectangular and cylindrical coordinates.\r\n\r\n[reveal-answer q=\"387456286\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"387456286\"]\r\n\r\n[caption id=\"attachment_5307\" align=\"aligncenter\" width=\"657\"]<img class=\"size-full wp-image-5307\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31231825\/2.58.jpg\" alt=\"This figure is of the 3-dimensional coordinate system. It has a point. There is a line segment from the origin to the point. The angle between this line segment and the z-axis is phi. There is a line segment in the x y-plane from the origin to the shadow of the point.The angle between the x-axis and rho is theta.\" width=\"657\" height=\"392\" \/> Figure 5. The point with spherical coordinates [latex]\\left(2,-\\frac{5\\pi}6,\\frac{\\pi}6\\right)[\/latex][\/caption]Cartesian: [latex]\\left(-\\frac{\\sqrt{3}}2,-\\frac12,\\sqrt3\\right)[\/latex], cylindrical: [latex]\\left(1,-\\frac{5\\pi}6,\\sqrt3\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try IT.\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7809469&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=yrKGmi_JFbM&amp;video_target=tpm-plugin-f16mh34p-yrKGmi_JFbM\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the transcript for <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP2.58_transcript.html\">\u201cCP 2.58\u201d here (opens in new window).<\/a><\/center>\r\n<div id=\"fs-id1163723335773\" class=\"ui-has-child-title\" data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: converting from rectangular coordinates<\/h3>\r\nConvert the rectangular coordinates [latex](-1,1,\\sqrt6)[\/latex] to both spherical and cylindrical coordinates.\r\n\r\n[reveal-answer q=\"989282734\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"989282734\"]\r\n<p id=\"fs-id1163723719347\">Start by converting from rectangular to spherical coordinates:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\rho^2=x^2+y^2+z^2=(-1)^2+1^2+\\left(\\sqrt6\\right)^2=8 \\quad \\tan\\theta&amp;=\\frac{1}{-1} \\\\\r\n\\rho&amp;=2\\sqrt2 \\quad \\theta&amp;=\\arctan(-1)=\\frac{3\\pi}4\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1163724120398\">Because [latex](x,y)=(-1,1)[\/latex], then the correct choice for [latex]\\theta[\/latex] is\u00a0[latex]\\frac{3\\pi}4[\/latex].<\/p>\r\n<p id=\"fs-id1163723727982\">There are actually two ways to identify [latex]\\varphi[\/latex]. We can use the equation [latex]\\varphi=\\arccos\\left(\\frac{z}{\\sqrt{x^2+y^2+z^2}}\\right)[\/latex]. A more simple approach, however, is to use equation [latex]z=\\rho\\cos\\varphi[\/latex]. We know that [latex]z=\\sqrt6[\/latex] and [latex]\\rho=2\\sqrt2[\/latex], so<\/p>\r\n<p style=\"text-align: center;\">[latex]\\sqrt6=2\\sqrt2\\cos\\varphi\\text{, so}\\cos\\varphi=\\frac{\\sqrt6}{2\\sqrt2}=\\frac{\\sqrt3}2[\/latex]<\/p>\r\n<p id=\"fs-id1163724023218\">and therefore [latex]\\varphi=\\frac{\\pi}6[\/latex]. The spherical coordinates of the point are [latex](2\\sqrt2,\\frac{3\\pi}4,\\frac{\\pi}6)[\/latex].<\/p>\r\n<p id=\"fs-id1163724081108\">To find the cylindrical coordinates for the point, we need only find\u00a0[latex]r[\/latex]:<\/p>\r\n<p style=\"text-align: center;\">[latex]r=\\rho\\sin\\varphi=2\\sqrt2\\sin\\left(\\frac{\\pi}6\\right)=\\sqrt2[\/latex].<\/p>\r\n<p id=\"fs-id1163723095252\">The cylindrical coordinates for the point are [latex](\\sqrt2,\\frac{3\\pi}4,\\sqrt6)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: identifying surfaces in the spherical coordinate system<\/h3>\r\n<p id=\"fs-id1163724128608\">Describe the surfaces with the given spherical equations.<\/p>\r\n\r\n<ol id=\"fs-id1163724128611\" type=\"a\">\r\n \t<li>[latex]\\theta=\\frac{\\pi}3[\/latex]<\/li>\r\n \t<li>[latex]\\varphi=\\frac{5\\pi}6[\/latex]<\/li>\r\n \t<li>[latex]\\rho=6[\/latex]<\/li>\r\n \t<li>[latex]\\rho=\\sin\\theta\\sin\\varphi[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"746526133\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"746526133\"]\r\n<ol id=\"fs-id1163724069760\" type=\"a\">\r\n \t<li>The variable [latex]\\theta[\/latex] represents the measure of the same angle in both the cylindrical and spherical coordinate systems. Points with coordinates [latex](\\rho,\\frac{\\pi}3,\\varphi)[\/latex] lie on the plane that forms angle [latex]\\theta=\\frac{\\pi}3[\/latex] with the positive [latex]x[\/latex]-axis. Because [latex]\\rho&gt;0[\/latex], the surface described by equation [latex]\\theta=\\frac{\\pi}3[\/latex] is the half-plane shown in\u00a0Figure 6.<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"CNX_Calc_Figure_12_07_016\" class=\"os-figure\">\r\n<div class=\"os-caption-container\">[caption id=\"attachment_5311\" align=\"aligncenter\" width=\"279\"]<img class=\"size-full wp-image-5311\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31232103\/2.101.jpg\" alt=\"This figure is the first quadrant of the 3-dimensional coordinate system. There is a plane attached to the z-axis, dividing the x y-plane with a diagonal line. The angle between the x-axis and this plane is theta = pi\/3.\" width=\"279\" height=\"431\" \/> Figure 6. The surface described by equation [latex]\\theta=\\frac{\\pi}3[\/latex] is a half-plane.[\/caption]<\/div>\r\n<\/div><\/li>\r\n \t<li>Equation [latex]\\varphi=\\frac{5\\pi}6[\/latex] describes all points in the spherical coordinate system that lie on a line from the origin forming an angle measuring [latex]\\frac{5\\pi}6[\/latex] rad with the positive [latex]z[\/latex]-axis. These points form a half-cone (Figure 7). Because there is only one value for [latex]\\varphi[\/latex] that is measured from the positive [latex]z[\/latex]-axis, we do not get the full cone (with two pieces).<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"CNX_Calc_Figure_12_07_017\" class=\"os-figure\">\r\n<div class=\"os-caption-container\">[caption id=\"attachment_5312\" align=\"aligncenter\" width=\"445\"]<img class=\"size-full wp-image-5312\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31232128\/2.102.jpg\" alt=\"This figure is the upper part of an elliptical cone. The bottom point of the cone is at the origin of the 3-dimensional coordinate system.\" width=\"445\" height=\"360\" \/> Figure 7. The equation [latex]\\varphi=\\frac{5\\pi}6[\/latex] describes a cone.[\/caption]<\/div>\r\n<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>To find the equation in rectangular coordinates, use equation [latex]\\varphi=\\arccos\\left(\\frac{z}{\\sqrt{x^2+y^2+z^2}}\\right)[\/latex].\u00a0<span data-type=\"newline\">\r\n<\/span>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\frac{5\\pi}6&amp;=\\arccos\\left(\\frac{z}{\\sqrt{x^2+y^2+z^2}}\\right) \\\\\r\n\\cos\\frac{5\\pi}6&amp;=\\frac{z}{\\sqrt{x^2+y^2+z^2}} \\\\\r\n-\\frac{\\sqrt3}2&amp;=\\frac{z}{\\sqrt{x^2+y^2+z^2}} \\\\\r\n\\frac34&amp;=\\frac{z}{\\sqrt{x^2+y^2+z^2}} \\\\\r\n\\frac{3x^2}4+\\frac{3y^2}4+\\frac{3z^2}4&amp;=z^2 \\\\\r\n\\frac{3x^2}4+\\frac{3y^2}4-\\frac{z^2}4&amp;=0.\r\n\\end{aligned}[\/latex]<\/p>\r\nThis is the equation of a cone centered on the [latex]z[\/latex]-axis.<\/li>\r\n \t<li>Equation [latex]\\rho=6[\/latex] describes the set of all points [latex]6[\/latex] units away from the origin\u2014a sphere with radius [latex]6[\/latex] (Figure 8).\r\n[caption id=\"attachment_5313\" align=\"aligncenter\" width=\"461\"]<img class=\"size-full wp-image-5313\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31232157\/2.103.jpg\" alt=\"This figure is a sphere. The z-axis is vertically through the center and intersects the sphere at (0, 0, 6). The y-axis is horizontally through the center and intersects the sphere at (0, 6, 0).\" width=\"461\" height=\"437\" \/> Figure 8. Equation [latex]\\rho=6[\/latex] describes a sphere with radius [latex]6[\/latex].[\/caption]<\/li>\r\n \t<li>To identify this surface, convert the equation from spherical to rectangular coordinates, using equations [latex]y=\\rho\\sin\\varphi\\sin\\theta[\/latex] and\u00a0[latex]\\rho^2=x^2+y^2+z^2[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\rho&amp;=\\sin\\theta\\sin\\varphi \\\\\r\n\\rho^2&amp;=\\rho\\sin\\theta\\sin\\varphi&amp;\\text{Multiply both sides of the equation by }\\varphi \\\\\r\nx^2+y^2+z^2&amp;=y&amp;\\text{Substitute rectangular variables using the equations above.} \\\\\r\nx^2+y^2-y+z^2&amp;=0&amp;\\text{Subtract }y\\text{ from both sides of the equation.} \\\\\r\nx^2+y^2-y+\\frac14+z^2&amp;=\\frac14&amp;\\text{Complete the square.} \\\\\r\nx^2+\\left(y-\\frac12\\right)^2+z^2&amp;=\\frac14&amp;\\text{Rewrite the middle terms as a perfect square.}\r\n\\end{aligned}[\/latex]<\/p>\r\nThe equation describes a sphere centered at point [latex] (0,\\frac{1}{2},0)[\/latex] with radius [latex]\\frac{1}{2}[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n<p id=\"fs-id1163723523037\">Describe the surfaces defined by the following equations.<\/p>\r\n\r\n<ol id=\"fs-id1163723523040\" type=\"a\">\r\n \t<li>[latex]\\rho=13[\/latex]<\/li>\r\n \t<li>[latex]\\theta=\\frac{2\\pi}3[\/latex]<\/li>\r\n \t<li>[latex]\\varphi=\\frac{\\pi}4[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"874652711\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"874652711\"]\r\n\r\na. This is the set of all points [latex]13[\/latex] units from the origin. This set forms a sphere with radius [latex]13[\/latex].\r\n\r\nb. This set of points forms a half plane. The angle between the half plane and the positive [latex]x[\/latex]-axis is [latex]\\theta=\\frac{2\\pi}3[\/latex].\r\n\r\nc. Let [latex]P[\/latex] be a point on this surface. The position vector of this point forms an angle of [latex]\\varphi=\\frac{\\pi}4[\/latex] with the positive [latex]z[\/latex]-axis, which means that points closer to the origin are closer to the axis. These points form a half-cone.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1163724023236\">Spherical coordinates are useful in analyzing systems that have some degree of symmetry about a point, such as the volume of the space inside a domed stadium or wind speeds in a planet\u2019s atmosphere. A sphere that has Cartesian equation [latex]x^{2}+y^{2}+z^{2}=c^{2}[\/latex] has the simple equation [latex]\\rho=c[\/latex] in spherical coordinates.<\/p>\r\n<p id=\"fs-id1163724023526\">In geography, latitude and longitude are used to describe locations on Earth\u2019s surface, as shown in\u00a0Figure 9. Although the shape of Earth is not a perfect sphere, we use spherical coordinates to communicate the locations of points on Earth. Let\u2019s assume Earth has the shape of a sphere with radius 4000 mi. We express angle measures in degrees rather than radians because latitude and longitude are measured in degrees.<\/p>\r\n\r\n\r\n[caption id=\"attachment_5317\" align=\"aligncenter\" width=\"483\"]<img class=\"size-full wp-image-5317\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31232343\/2.104.jpg\" alt=\"This figure is an image of the Earth. It has the prime meridian labeled, which is a circle on the surface circumnavigating the Earth vertically through the poles. The equator is also labeled which is a horizontal circle circumnavigating the Earth. Three vectors extend out from the center of Earth. Two of them extend to the equator and indicate a measurement of longitude. Two of them extend to a vertical polar circle and indicate a measurement of latitude.\" width=\"483\" height=\"436\" \/> Figure 9. In the latitude\u2013longitude system, angles describe the location of a point on Earth relative to the equator and the prime meridian.[\/caption]\r\n\r\nLet the center of Earth be the center of the sphere, with the ray from the center through the North Pole representing the positive [latex]z[\/latex]-axis. The prime meridian represents the trace of the surface as it intersects the [latex]xz[\/latex]-plane. The equator is the trace of the sphere intersecting the [latex]xy[\/latex]-plane.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: converting latitude and longitude to spherical coordinates<\/h3>\r\nThe latitude of Columbus, Ohio, is [latex]40^{\\small\\circ}[\/latex] N and the longitude is [latex]83^{\\small\\circ}[\/latex] W, which means that Columbus is [latex]40^{\\small\\circ}[\/latex] north of the equator. Imagine a ray from the center of Earth through Columbus and a ray from the center of Earth through the equator directly south of Columbus. The measure of the angle formed by the rays is [latex]40^{\\small\\circ}[\/latex]. In the same way, measuring from the prime meridian, Columbus lies [latex]83^{\\small\\circ}[\/latex] to the west. Express the location of Columbus in spherical coordinates.\r\n\r\n[reveal-answer q=\"233355186\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"233355186\"]\r\n\r\nThe radius of Earth is [latex]4,000[\/latex] mi, so [latex]\\rho=4,000[\/latex]. The intersection of the prime meridian and the equator lies on the positive [latex]x[\/latex]-axis. Movement to the west is then described with negative angle measures, which shows that [latex]\\theta=-83^{\\small\\circ}[\/latex], Because Columbus lies [latex]40^{\\small\\circ}[\/latex] north of the equator, it lies [latex]50^{\\small\\circ}[\/latex] south of the North Pole, so [latex]\\varphi-50^{\\small\\circ}[\/latex]. In spherical coordinates, Columbus lies at point [latex](4,000,-83^{\\small\\circ},50^{\\small\\circ})[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nSydney, Australia is at [latex]34^{\\small\\circ}[\/latex]S and [latex]151^{\\small\\circ}[\/latex]E. Express Sydney\u2019s location in spherical coordinates.\r\n\r\n[reveal-answer q=\"467253716\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"467253716\"]\r\n\r\n[latex](4,000,151^{\\small\\circ},124^{\\small\\circ})[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nCylindrical and spherical coordinates give us the flexibility to select a coordinate system appropriate to the problem at hand. A thoughtful choice of coordinate system can make a problem much easier to solve, whereas a poor choice can lead to unnecessarily complex calculations. In the following example, we examine several different problems and discuss how to select the best coordinate system for each one.\r\n\r\n<\/div>\r\n<div data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: choosing the best coordinate system<\/h3>\r\n<p id=\"fs-id1163723217907\">In each of the following situations, we determine which coordinate system is most appropriate and describe how we would orient the coordinate axes. There could be more than one right answer for how the axes should be oriented, but we select an orientation that makes sense in the context of the problem.\u00a0<em data-effect=\"italics\">Note<\/em>: There is not enough information to set up or solve these problems; we simply select the coordinate system (Figure 10).<\/p>\r\n\r\n<ol id=\"fs-id1163723217922\" type=\"a\">\r\n \t<li>Find the center of gravity of a bowling ball.<\/li>\r\n \t<li>Determine the velocity of a submarine subjected to an ocean current.<\/li>\r\n \t<li>Calculate the pressure in a conical water tank.<\/li>\r\n \t<li>Find the volume of oil flowing through a pipeline.<\/li>\r\n \t<li>Determine the amount of leather required to make a football.<\/li>\r\n<\/ol>\r\n[caption id=\"attachment_5319\" align=\"aligncenter\" width=\"731\"]<img class=\"size-full wp-image-5319\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31232437\/2.105.jpg\" alt=\"This figure has 5 images. The first image shows bowling balls. The second image is a submarine traveling on an ocean surface. The third image is a traffic cone. The fourth image is a pipeline across some barren land. The fifth image is a football.\" width=\"731\" height=\"492\" \/> Figure 10. (credit: (a) modification of work by scl hua, Wikimedia, (b) modification of work by DVIDSHUB, Flickr, (c) modification of work by Michael Malak, Wikimedia, (d) modification of work by Sean Mack, Wikimedia, (e) modification of work by Elvert Barnes, Flickr)[\/caption]\r\n\r\n[reveal-answer q=\"757372734\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"757372734\"]\r\n<ol id=\"fs-id1163723567687\" type=\"a\">\r\n \t<li>Clearly, a bowling ball is a sphere, so spherical coordinates would probably work best here. The origin should be located at the physical center of the ball. There is no obvious choice for how the [latex]x[\/latex]-, [latex]y[\/latex]- and [latex]z[\/latex]-axes should be oriented. Bowling balls normally have a weight block in the center. One possible choice is to align the [latex]z[\/latex]-axis with the axis of symmetry of the weight block.<\/li>\r\n \t<li>A submarine generally moves in a straight line. There is no rotational or spherical symmetry that applies in this situation, so rectangular coordinates are a good choice. The [latex]z[\/latex]-axis should probably point upward. The [latex]x[\/latex]- and [latex]y[\/latex]-axes could be aligned to point east and north, respectively. The origin should be some convenient physical location, such as the starting position of the submarine or the location of a particular port.<\/li>\r\n \t<li>A cone has several kinds of symmetry. In cylindrical coordinates, a cone can be represented by equation [latex]z=kr[\/latex], where [latex]k[\/latex] is a constant. In spherical coordinates, we have seen that surfaces of the form [latex]\\varphi=c[\/latex] are half-cones. Last, in rectangular coordinates, elliptic cones are quadric surfaces and can be represented by equations of the form [latex]z^2=\\frac{x^2}{a^2}+\\frac{y^2}{b^2}[\/latex]. In this case, we could choose any of the three. However, the equation for the surface is more complicated in rectangular coordinates than in the other two systems, so we might want to avoid that choice. In addition, we are talking about a water tank, and the depth of the water might come into play at some point in our calculations, so it might be nice to have a component that represents height and depth directly. Based on this reasoning, cylindrical coordinates might be the best choice. Choose the [latex]z[\/latex]-axis to align with the axis of the cone. The orientation of the other two axes is arbitrary. The origin should be the bottom point of the cone.<\/li>\r\n \t<li>A pipeline is a cylinder, so cylindrical coordinates would be best the best choice. In this case, however, we would likely choose to orient our [latex]z[\/latex]-axis with the center axis of the pipeline. The [latex]x[\/latex]-axis could be chosen to point straight downward or to some other logical direction. The origin should be chosen based on the problem statement. Note that this puts the [latex]z[\/latex]-axis in a horizontal orientation, which is a little different from what we usually do. It may make sense to choose an unusual orientation for the axes if it makes sense for the problem.<\/li>\r\n \t<li>A football has rotational symmetry about a central axis, so cylindrical coordinates would work best. The [latex]z[\/latex]-axis should align with the axis of the ball. The origin could be the center of the ball or perhaps one of the ends. The position of the [latex]x[\/latex]-axis is arbitrary.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Choosing the Best Coordinate System.\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7809470&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=1WaXMBJU_j4&amp;video_target=tpm-plugin-9k5unai6-1WaXMBJU_j4\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the transcript for <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/Ex2.67_transcript.html\">\u201cEx 2.67\u201d here (opens in new window).<\/a><\/center>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nWhich coordinate system is most appropriate for creating a star map, as viewed from Earth (see the following figure)?\r\n\r\n[caption id=\"attachment_5322\" align=\"aligncenter\" width=\"731\"]<img class=\"size-full wp-image-5322\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31232545\/2.61.jpg\" alt=\"This figure is a circle with a star chart in the middle.\" width=\"731\" height=\"731\" \/> Figure 11. Star map as viewed from Earth.[\/caption]\r\n\r\nHow should we orient the coordinate axes?\r\n\r\n[reveal-answer q=\"883476235\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"883476235\"]\r\n\r\nSpherical coordinates with the origin located at the center of the earth, the [latex]z[\/latex]-axis aligned with the North Pole, and the [latex]x[\/latex]-axis aligned with the prime meridian.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<\/div>\r\n<div data-type=\"example\"><\/div>","rendered":"<div data-type=\"note\">\n<div id=\"fs-id1163724036465\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<div id=\"fs-id1163724074875\" class=\"theorem ui-has-child-title\" data-type=\"note\">\n<div id=\"fs-id1163724080388\" class=\"ui-has-child-title\" data-type=\"example\">\n<div data-type=\"example\">\n<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul class=\"os-abstract\">\n<li><span class=\"os-abstract-content\">Convert from spherical to rectangular coordinates.<\/span><\/li>\n<li><span class=\"os-abstract-content\">Convert from rectangular to spherical coordinates.<\/span><\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1163723669991\">In the Cartesian coordinate system, the location of a point in space is described using an ordered triple in which each coordinate represents a distance. In the cylindrical coordinate system, location of a point in space is described using two distances ([latex]r[\/latex] and[latex]z[\/latex]) and an angle measure ([latex]\\theta[\/latex]). In the spherical coordinate system, we again use an ordered triple to describe the location of a point in space. In this case, the triple describes one distance and two angles. Spherical coordinates make it simple to describe a sphere, just as cylindrical coordinates make it easy to describe a cylinder. Grid lines for spherical coordinates are based on angle measures, like those for polar coordinates.<\/p>\n<\/div>\n<div data-type=\"example\">\n<div class=\"textbox shaded\">\n<header>\n<h3 class=\"os-title\" style=\"text-align: center;\" data-type=\"title\"><span id=\"2\" class=\"os-title-label\" data-type=\"\">DEFINITION<\/span><\/h3>\n<hr \/>\n<\/header>\n<section>\n<div class=\"os-note-body\">\n<p id=\"fs-id1163723830824\">In the\u00a0<strong><span id=\"293cc618-b0de-4ec9-9029-6efb4eb1b172_term107\" data-type=\"term\">spherical coordinate system<\/span><\/strong>, a point [latex]P[\/latex] in space (Figure 1) is represented by the ordered triple [latex](\\rho,\\theta,\\varphi)[\/latex] where<\/p>\n<ul id=\"fs-id1163723954970\" data-bullet-style=\"bullet\">\n<li>[latex]\\rho[\/latex] (the Greek letter rho) is the distance between [latex]P[\/latex] and the origin [latex](\\rho\\ne0)[\/latex];<\/li>\n<li>[latex]\\theta[\/latex] is the same angle used to describe the location in cylindrical coordinates;<\/li>\n<li>[latex]\\varphi[\/latex] (the Greek letter phi) is the angle formed by the positive [latex]z[\/latex]-axis and line segment [latex]\\overline{OP}[\/latex], where [latex]O[\/latex] is the origin and [latex]0\\leq\\varphi\\leq\\pi[\/latex].<\/li>\n<\/ul>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"attachment_5293\" style=\"width: 447px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5293\" class=\"size-full wp-image-5293\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31230733\/2.97.jpg\" alt=\"This figure is the first quadrant of the 3-dimensional coordinate system. It has a point labeled \u201c(x, y, z) = (rho, theta, phi).\u201d There is a line segment from the origin to the point. It is labeled \u201crho.\u201d The angle between this line segment and the z-axis is phi. There is a line segment in the x y-plane from the origin to the shadow of the point. This segment is labeled \u201cr.\u201d The angle between the x-axis and r is theta.\" width=\"437\" height=\"457\" \/><\/p>\n<p id=\"caption-attachment-5293\" class=\"wp-caption-text\">Figure 1. The relationship among spherical, rectangular, and cylindrical coordinates.<\/p>\n<\/div>\n<p>By convention, the origin is represented as [latex](0, 0, 0)[\/latex] in spherical coordinates.<\/p>\n<div class=\"textbox shaded\">\n<header><\/header>\n<header>\n<h3 class=\"os-title\" style=\"text-align: center;\"><span class=\"os-title-label\">THEOREM: converting among Spherical, cylindrical, and rectangular coordinates<\/span><\/h3>\n<hr \/>\n<\/header>\n<section>\n<div class=\"os-note-body\">\n<p id=\"fs-id1163723772032\">Rectangular coordinates [latex](x, y, z)[\/latex] and spherical coordinates [latex](\\rho,\\theta,\\varphi)[\/latex] of a point are related as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  x&=\\rho\\sin\\varphi\\cos\\theta \\quad \\text{These equations are used to convert from spherical coordinates to rectangular coordinates.} \\\\  y&=\\rho\\sin\\varphi\\sin\\theta \\\\  z&=\\rho\\cos\\varphi \\\\  &\\text{and} \\\\  \\rho^2&=x^2+y^2+z^2 \\quad \\text{These equations are used to convert from rectangular coordinates to spherical coordinates.} \\\\  \\tan\\theta&=\\frac{y}x \\\\  \\varphi&=\\arccos\\left(\\frac{z}{\\sqrt{x^2+y^2+z^2}}\\right)  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1163723968448\">If a point has cylindrical coordinates [latex](r,\\theta,z)[\/latex], then these equations define the relationship between cylindrical and spherical coordinates.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  r&=\\rho\\sin\\varphi \\quad \\text{These equations are used to convert from spherical coordinates to cylindrical coordinates} \\\\  \\theta&=\\theta \\\\  z&=\\rho\\cos\\varphi \\\\  &\\text{and} \\\\  \\rho&=\\sqrt{r^2+z^2} \\quad \\text{These equations are used to convert from cylindrical coordinates to spherical coordinates} \\\\  \\theta&=\\theta \\\\  \\varphi&=\\arccos\\left(\\frac{z}{\\sqrt{r^2+z^2}}\\right)  \\end{aligned}[\/latex]<\/p>\n<div id=\"fs-id1163723563904\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<div class=\"MathJax_Display\"><\/div>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<p id=\"fs-id1163723656797\"><span style=\"font-size: 1rem; text-align: initial;\">The formulas to convert from spherical coordinates to rectangular coordinates may seem complex, but they are straightforward applications of trigonometry. Looking at\u00a0Figure 2, it is easy to see that [latex]r=\\rho\\cos\\varphi[\/latex]. Then, looking at the triangle in the [latex]xy[\/latex]-plane with [latex]r[\/latex] as its hypotenuse, we have [latex]x=r\\cos\\theta=\\rho\\sin\\varphi\\cos\\theta[\/latex]. The derivation of the formula for [latex]y[\/latex] is similar.\u00a0Figure 9 in <a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/cylindrical-coordinates\/\" target=\"_blank\" rel=\"noopener\">Cylindrical Coordinates<\/a>\u00a0also shows that [latex]\\rho^2=r^2+z^2=x^2+y^2+z^2[\/latex] and [latex]z=\\rho\\cos\\varphi[\/latex]. Solving this last equation for [latex]\\varphi[\/latex] and then substituting [latex]\\rho=\\sqrt{r^2+z^2}[\/latex] (from the first equation) yields [latex]\\varphi=\\arccos\\left(\\frac{z}{\\sqrt{r^2+z^2}}\\right)[\/latex]. Also, note that, as before, we must be careful when using the formula [latex]\\tan\\theta=\\frac{y}x[\/latex] to choose the correct value of [latex]\\theta[\/latex].\u00a0<\/span><\/p>\n<div id=\"attachment_5294\" style=\"width: 447px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5294\" class=\"size-full wp-image-5294\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31230837\/2.98.jpg\" alt=\"This figure is the first quadrant of the 3-dimensional coordinate system. It has a point labeled \u201c(x, y, z) = (r, theta, z) = (rho, theta, phi).\u201d There is a line segment from the origin to the point. It is labeled \u201crho.\u201d The angle between this line segment and the z-axis is phi. There is a line segment in the x y-plane from the origin to the shadow of the point. This segment is labeled \u201cr.\u201d The angle between the x-axis and r is theta.The distance from r to the point is labeled \u201cz.\u201d\" width=\"437\" height=\"457\" \/><\/p>\n<p id=\"caption-attachment-5294\" class=\"wp-caption-text\">Figure 2. The equations that convert from one system to another are derived from right-triangle relationships.<\/p>\n<\/div>\n<p>As we did with cylindrical coordinates, let\u2019s consider the surfaces that are generated when each of the coordinates is held constant. Let [latex]c[\/latex] be a constant, and consider surfaces of the form [latex]\\rho=c[\/latex]. Points on these surfaces are at a fixed distance from the origin and form a sphere. The coordinate [latex]\\theta[\/latex] in the spherical coordinate system is the same as in the cylindrical coordinate system, so surfaces of the form [latex]\\theta=c[\/latex] are half-planes, as before. Last, consider surfaces of the form [latex]\\varphi=c[\/latex]. The points on these surfaces are at a fixed angle from the [latex]z[\/latex]-axis and form a half-cone (Figure 3).<\/p>\n<div id=\"attachment_5296\" style=\"width: 633px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5296\" class=\"size-full wp-image-5296\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31230916\/2.99.jpg\" alt=\"This figure has three images. The first image is a sphere centered in the 3-dimensional coordinate system. The second figure is a vertical plane with an edge on the z-axis in the 3-dimensional coordinate system. The third image is an elliptical cone with the center at the origin of the 3-dimensional coordinate system.\" width=\"623\" height=\"262\" \/><\/p>\n<p id=\"caption-attachment-5296\" class=\"wp-caption-text\">Figure 3. In spherical coordinates, surfaces of the form [latex]\\rho=c[\/latex] are spheres of radius [latex]\\rho[\/latex] (a), surfaces of the form [latex]\\theta=c[\/latex] are half-planes at an angle [latex]\\theta[\/latex] from the [latex]x[\/latex]-axis (b), and surfaces of the form [latex]\\varphi=c[\/latex] are half-cones at an angle [latex]\\varphi[\/latex] from the [latex]z[\/latex]-axis (c).<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: converting from spherical coordinates<\/h3>\n<p>Plot the point with spherical coordinates [latex]\\left(8,\\frac{\\pi}3,\\frac{\\pi}6\\right)[\/latex] and express its location in both rectangular and cylindrical coordinates.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q874652471\">Show Solution<\/span><\/p>\n<div id=\"q874652471\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1163723236542\">Use the equations in\u00a0Theorem: Converting among Spherical, Cylindrical, and Rectangular Coordinates\u00a0to translate between spherical and cylindrical coordinates (Figure 4):<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  x&=\\rho\\sin\\varphi\\cos\\theta=8\\sin\\left(\\frac{\\pi}6\\right)\\cos\\left(\\frac{\\pi}3\\right)=8\\left(\\frac12\\right)\\frac12=2 \\\\  y&=\\rho\\sin\\varphi\\sin\\theta=8\\sin\\left(\\frac{\\pi}6\\right)\\sin\\left(\\frac{\\pi}3\\right)=8\\left(\\frac12\\right)\\frac{\\sqrt{3}}2=2\\sqrt3 \\\\  z&=\\rho\\cos\\varphi=8\\cos\\left(\\frac{\\pi}6\\right)=8\\left(\\frac{\\sqrt{3}}2\\right)=4\\sqrt3.  \\end{aligned}[\/latex]<\/p>\n<div id=\"attachment_5304\" style=\"width: 686px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5304\" class=\"size-full wp-image-5304\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31231600\/2.100.jpg\" alt=\"This figure is the first quadrant of the 3-dimensional coordinate system. It has a point labeled \u201c(8, pi\/3, pi\/6).\u201d There is a line segment from the origin to the point. It is labeled \u201crho = 8.\u201d The angle between this line segment and the z-axis is labeled \u201cphi = pi\/6.\u201d There is a line segment in the x y-plane from the origin to the shadow of the point. The angle between the x-axis and r is labeled \u201ctheta = pi\/3.\u201d\" width=\"676\" height=\"538\" \/><\/p>\n<p id=\"caption-attachment-5304\" class=\"wp-caption-text\">Figure 4. The projection of the point in the [latex]xy[\/latex]-plane is [latex]4[\/latex] units from the origin. The line from the origin to the point\u2019s projection forms an angle of [latex]\\frac{\\pi}{3}[\/latex] with the positive [latex]x[\/latex]-axis. The point lies [latex]4\\sqrt3[\/latex] units above the [latex]xy[\/latex]-plane.<\/p>\n<\/div>\n<p id=\"fs-id1163724035807\">The point with spherical coordinates [latex]\\left(8,\\frac{\\pi}3,\\frac{\\pi}6\\right)[\/latex] has rectangular coordinates [latex](2,2\\sqrt3,4\\sqrt3)[\/latex].<\/p>\n<p id=\"fs-id1163723242019\">Finding the values in cylindrical coordinates is equally straightforward:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  r&=\\rho\\sin\\varphi=8\\sin\\frac{\\pi}6=4 \\\\  \\theta&=\\theta \\\\  z&=\\rho\\cos\\varphi=8\\cos\\frac{\\pi}6=4\\sqrt3  \\end{aligned}[\/latex].<\/p>\n<p id=\"fs-id1163723994378\">Thus, cylindrical coordinates for the point are [latex]\\left(4,\\frac{\\pi}3,4\\sqrt3\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Plot the point with spherical coordinates [latex]\\left(2,-\\frac{5\\pi}6,\\frac{\\pi}6\\right)[\/latex] and describe its location in both rectangular and cylindrical coordinates.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q387456286\">Show Solution<\/span><\/p>\n<div id=\"q387456286\" class=\"hidden-answer\" style=\"display: none\">\n<div id=\"attachment_5307\" style=\"width: 667px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5307\" class=\"size-full wp-image-5307\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31231825\/2.58.jpg\" alt=\"This figure is of the 3-dimensional coordinate system. It has a point. There is a line segment from the origin to the point. The angle between this line segment and the z-axis is phi. There is a line segment in the x y-plane from the origin to the shadow of the point.The angle between the x-axis and rho is theta.\" width=\"657\" height=\"392\" \/><\/p>\n<p id=\"caption-attachment-5307\" class=\"wp-caption-text\">Figure 5. The point with spherical coordinates [latex]\\left(2,-\\frac{5\\pi}6,\\frac{\\pi}6\\right)[\/latex]<\/p>\n<\/div>\n<p>Cartesian: [latex]\\left(-\\frac{\\sqrt{3}}2,-\\frac12,\\sqrt3\\right)[\/latex], cylindrical: [latex]\\left(1,-\\frac{5\\pi}6,\\sqrt3\\right)[\/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=7809469&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=yrKGmi_JFbM&amp;video_target=tpm-plugin-f16mh34p-yrKGmi_JFbM\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the transcript for <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP2.58_transcript.html\">\u201cCP 2.58\u201d here (opens in new window).<\/a><\/div>\n<div id=\"fs-id1163723335773\" class=\"ui-has-child-title\" data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: converting from rectangular coordinates<\/h3>\n<p>Convert the rectangular coordinates [latex](-1,1,\\sqrt6)[\/latex] to both spherical and cylindrical coordinates.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q989282734\">Show Solution<\/span><\/p>\n<div id=\"q989282734\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1163723719347\">Start by converting from rectangular to spherical coordinates:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\rho^2=x^2+y^2+z^2=(-1)^2+1^2+\\left(\\sqrt6\\right)^2=8 \\quad \\tan\\theta&=\\frac{1}{-1} \\\\  \\rho&=2\\sqrt2 \\quad \\theta&=\\arctan(-1)=\\frac{3\\pi}4  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1163724120398\">Because [latex](x,y)=(-1,1)[\/latex], then the correct choice for [latex]\\theta[\/latex] is\u00a0[latex]\\frac{3\\pi}4[\/latex].<\/p>\n<p id=\"fs-id1163723727982\">There are actually two ways to identify [latex]\\varphi[\/latex]. We can use the equation [latex]\\varphi=\\arccos\\left(\\frac{z}{\\sqrt{x^2+y^2+z^2}}\\right)[\/latex]. A more simple approach, however, is to use equation [latex]z=\\rho\\cos\\varphi[\/latex]. We know that [latex]z=\\sqrt6[\/latex] and [latex]\\rho=2\\sqrt2[\/latex], so<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt6=2\\sqrt2\\cos\\varphi\\text{, so}\\cos\\varphi=\\frac{\\sqrt6}{2\\sqrt2}=\\frac{\\sqrt3}2[\/latex]<\/p>\n<p id=\"fs-id1163724023218\">and therefore [latex]\\varphi=\\frac{\\pi}6[\/latex]. The spherical coordinates of the point are [latex](2\\sqrt2,\\frac{3\\pi}4,\\frac{\\pi}6)[\/latex].<\/p>\n<p id=\"fs-id1163724081108\">To find the cylindrical coordinates for the point, we need only find\u00a0[latex]r[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]r=\\rho\\sin\\varphi=2\\sqrt2\\sin\\left(\\frac{\\pi}6\\right)=\\sqrt2[\/latex].<\/p>\n<p id=\"fs-id1163723095252\">The cylindrical coordinates for the point are [latex](\\sqrt2,\\frac{3\\pi}4,\\sqrt6)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: identifying surfaces in the spherical coordinate system<\/h3>\n<p id=\"fs-id1163724128608\">Describe the surfaces with the given spherical equations.<\/p>\n<ol id=\"fs-id1163724128611\" type=\"a\">\n<li>[latex]\\theta=\\frac{\\pi}3[\/latex]<\/li>\n<li>[latex]\\varphi=\\frac{5\\pi}6[\/latex]<\/li>\n<li>[latex]\\rho=6[\/latex]<\/li>\n<li>[latex]\\rho=\\sin\\theta\\sin\\varphi[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q746526133\">Show Solution<\/span><\/p>\n<div id=\"q746526133\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1163724069760\" type=\"a\">\n<li>The variable [latex]\\theta[\/latex] represents the measure of the same angle in both the cylindrical and spherical coordinate systems. Points with coordinates [latex](\\rho,\\frac{\\pi}3,\\varphi)[\/latex] lie on the plane that forms angle [latex]\\theta=\\frac{\\pi}3[\/latex] with the positive [latex]x[\/latex]-axis. Because [latex]\\rho>0[\/latex], the surface described by equation [latex]\\theta=\\frac{\\pi}3[\/latex] is the half-plane shown in\u00a0Figure 6.<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"CNX_Calc_Figure_12_07_016\" class=\"os-figure\">\n<div class=\"os-caption-container\">\n<div id=\"attachment_5311\" style=\"width: 289px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5311\" class=\"size-full wp-image-5311\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31232103\/2.101.jpg\" alt=\"This figure is the first quadrant of the 3-dimensional coordinate system. There is a plane attached to the z-axis, dividing the x y-plane with a diagonal line. The angle between the x-axis and this plane is theta = pi\/3.\" width=\"279\" height=\"431\" \/><\/p>\n<p id=\"caption-attachment-5311\" class=\"wp-caption-text\">Figure 6. The surface described by equation [latex]\\theta=\\frac{\\pi}3[\/latex] is a half-plane.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/li>\n<li>Equation [latex]\\varphi=\\frac{5\\pi}6[\/latex] describes all points in the spherical coordinate system that lie on a line from the origin forming an angle measuring [latex]\\frac{5\\pi}6[\/latex] rad with the positive [latex]z[\/latex]-axis. These points form a half-cone (Figure 7). Because there is only one value for [latex]\\varphi[\/latex] that is measured from the positive [latex]z[\/latex]-axis, we do not get the full cone (with two pieces).<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"CNX_Calc_Figure_12_07_017\" class=\"os-figure\">\n<div class=\"os-caption-container\">\n<div id=\"attachment_5312\" style=\"width: 455px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5312\" class=\"size-full wp-image-5312\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31232128\/2.102.jpg\" alt=\"This figure is the upper part of an elliptical cone. The bottom point of the cone is at the origin of the 3-dimensional coordinate system.\" width=\"445\" height=\"360\" \/><\/p>\n<p id=\"caption-attachment-5312\" class=\"wp-caption-text\">Figure 7. The equation [latex]\\varphi=\\frac{5\\pi}6[\/latex] describes a cone.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span>To find the equation in rectangular coordinates, use equation [latex]\\varphi=\\arccos\\left(\\frac{z}{\\sqrt{x^2+y^2+z^2}}\\right)[\/latex].\u00a0<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\frac{5\\pi}6&=\\arccos\\left(\\frac{z}{\\sqrt{x^2+y^2+z^2}}\\right) \\\\  \\cos\\frac{5\\pi}6&=\\frac{z}{\\sqrt{x^2+y^2+z^2}} \\\\  -\\frac{\\sqrt3}2&=\\frac{z}{\\sqrt{x^2+y^2+z^2}} \\\\  \\frac34&=\\frac{z}{\\sqrt{x^2+y^2+z^2}} \\\\  \\frac{3x^2}4+\\frac{3y^2}4+\\frac{3z^2}4&=z^2 \\\\  \\frac{3x^2}4+\\frac{3y^2}4-\\frac{z^2}4&=0.  \\end{aligned}[\/latex]<\/p>\n<p>This is the equation of a cone centered on the [latex]z[\/latex]-axis.<\/li>\n<li>Equation [latex]\\rho=6[\/latex] describes the set of all points [latex]6[\/latex] units away from the origin\u2014a sphere with radius [latex]6[\/latex] (Figure 8).\n<div id=\"attachment_5313\" style=\"width: 471px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5313\" class=\"size-full wp-image-5313\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31232157\/2.103.jpg\" alt=\"This figure is a sphere. The z-axis is vertically through the center and intersects the sphere at (0, 0, 6). The y-axis is horizontally through the center and intersects the sphere at (0, 6, 0).\" width=\"461\" height=\"437\" \/><\/p>\n<p id=\"caption-attachment-5313\" class=\"wp-caption-text\">Figure 8. Equation [latex]\\rho=6[\/latex] describes a sphere with radius [latex]6[\/latex].<\/p>\n<\/div>\n<\/li>\n<li>To identify this surface, convert the equation from spherical to rectangular coordinates, using equations [latex]y=\\rho\\sin\\varphi\\sin\\theta[\/latex] and\u00a0[latex]\\rho^2=x^2+y^2+z^2[\/latex]:\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\rho&=\\sin\\theta\\sin\\varphi \\\\  \\rho^2&=\\rho\\sin\\theta\\sin\\varphi&\\text{Multiply both sides of the equation by }\\varphi \\\\  x^2+y^2+z^2&=y&\\text{Substitute rectangular variables using the equations above.} \\\\  x^2+y^2-y+z^2&=0&\\text{Subtract }y\\text{ from both sides of the equation.} \\\\  x^2+y^2-y+\\frac14+z^2&=\\frac14&\\text{Complete the square.} \\\\  x^2+\\left(y-\\frac12\\right)^2+z^2&=\\frac14&\\text{Rewrite the middle terms as a perfect square.}  \\end{aligned}[\/latex]<\/p>\n<p>The equation describes a sphere centered at point [latex](0,\\frac{1}{2},0)[\/latex] with radius [latex]\\frac{1}{2}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p id=\"fs-id1163723523037\">Describe the surfaces defined by the following equations.<\/p>\n<ol id=\"fs-id1163723523040\" type=\"a\">\n<li>[latex]\\rho=13[\/latex]<\/li>\n<li>[latex]\\theta=\\frac{2\\pi}3[\/latex]<\/li>\n<li>[latex]\\varphi=\\frac{\\pi}4[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q874652711\">Show Solution<\/span><\/p>\n<div id=\"q874652711\" class=\"hidden-answer\" style=\"display: none\">\n<p>a. This is the set of all points [latex]13[\/latex] units from the origin. This set forms a sphere with radius [latex]13[\/latex].<\/p>\n<p>b. This set of points forms a half plane. The angle between the half plane and the positive [latex]x[\/latex]-axis is [latex]\\theta=\\frac{2\\pi}3[\/latex].<\/p>\n<p>c. Let [latex]P[\/latex] be a point on this surface. The position vector of this point forms an angle of [latex]\\varphi=\\frac{\\pi}4[\/latex] with the positive [latex]z[\/latex]-axis, which means that points closer to the origin are closer to the axis. These points form a half-cone.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1163724023236\">Spherical coordinates are useful in analyzing systems that have some degree of symmetry about a point, such as the volume of the space inside a domed stadium or wind speeds in a planet\u2019s atmosphere. A sphere that has Cartesian equation [latex]x^{2}+y^{2}+z^{2}=c^{2}[\/latex] has the simple equation [latex]\\rho=c[\/latex] in spherical coordinates.<\/p>\n<p id=\"fs-id1163724023526\">In geography, latitude and longitude are used to describe locations on Earth\u2019s surface, as shown in\u00a0Figure 9. Although the shape of Earth is not a perfect sphere, we use spherical coordinates to communicate the locations of points on Earth. Let\u2019s assume Earth has the shape of a sphere with radius 4000 mi. We express angle measures in degrees rather than radians because latitude and longitude are measured in degrees.<\/p>\n<div id=\"attachment_5317\" style=\"width: 493px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5317\" class=\"size-full wp-image-5317\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31232343\/2.104.jpg\" alt=\"This figure is an image of the Earth. It has the prime meridian labeled, which is a circle on the surface circumnavigating the Earth vertically through the poles. The equator is also labeled which is a horizontal circle circumnavigating the Earth. Three vectors extend out from the center of Earth. Two of them extend to the equator and indicate a measurement of longitude. Two of them extend to a vertical polar circle and indicate a measurement of latitude.\" width=\"483\" height=\"436\" \/><\/p>\n<p id=\"caption-attachment-5317\" class=\"wp-caption-text\">Figure 9. In the latitude\u2013longitude system, angles describe the location of a point on Earth relative to the equator and the prime meridian.<\/p>\n<\/div>\n<p>Let the center of Earth be the center of the sphere, with the ray from the center through the North Pole representing the positive [latex]z[\/latex]-axis. The prime meridian represents the trace of the surface as it intersects the [latex]xz[\/latex]-plane. The equator is the trace of the sphere intersecting the [latex]xy[\/latex]-plane.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: converting latitude and longitude to spherical coordinates<\/h3>\n<p>The latitude of Columbus, Ohio, is [latex]40^{\\small\\circ}[\/latex] N and the longitude is [latex]83^{\\small\\circ}[\/latex] W, which means that Columbus is [latex]40^{\\small\\circ}[\/latex] north of the equator. Imagine a ray from the center of Earth through Columbus and a ray from the center of Earth through the equator directly south of Columbus. The measure of the angle formed by the rays is [latex]40^{\\small\\circ}[\/latex]. In the same way, measuring from the prime meridian, Columbus lies [latex]83^{\\small\\circ}[\/latex] to the west. Express the location of Columbus in spherical coordinates.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q233355186\">Show Solution<\/span><\/p>\n<div id=\"q233355186\" class=\"hidden-answer\" style=\"display: none\">\n<p>The radius of Earth is [latex]4,000[\/latex] mi, so [latex]\\rho=4,000[\/latex]. The intersection of the prime meridian and the equator lies on the positive [latex]x[\/latex]-axis. Movement to the west is then described with negative angle measures, which shows that [latex]\\theta=-83^{\\small\\circ}[\/latex], Because Columbus lies [latex]40^{\\small\\circ}[\/latex] north of the equator, it lies [latex]50^{\\small\\circ}[\/latex] south of the North Pole, so [latex]\\varphi-50^{\\small\\circ}[\/latex]. In spherical coordinates, Columbus lies at point [latex](4,000,-83^{\\small\\circ},50^{\\small\\circ})[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Sydney, Australia is at [latex]34^{\\small\\circ}[\/latex]S and [latex]151^{\\small\\circ}[\/latex]E. Express Sydney\u2019s location in spherical coordinates.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q467253716\">Show Solution<\/span><\/p>\n<div id=\"q467253716\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex](4,000,151^{\\small\\circ},124^{\\small\\circ})[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Cylindrical and spherical coordinates give us the flexibility to select a coordinate system appropriate to the problem at hand. A thoughtful choice of coordinate system can make a problem much easier to solve, whereas a poor choice can lead to unnecessarily complex calculations. In the following example, we examine several different problems and discuss how to select the best coordinate system for each one.<\/p>\n<\/div>\n<div data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: choosing the best coordinate system<\/h3>\n<p id=\"fs-id1163723217907\">In each of the following situations, we determine which coordinate system is most appropriate and describe how we would orient the coordinate axes. There could be more than one right answer for how the axes should be oriented, but we select an orientation that makes sense in the context of the problem.\u00a0<em data-effect=\"italics\">Note<\/em>: There is not enough information to set up or solve these problems; we simply select the coordinate system (Figure 10).<\/p>\n<ol id=\"fs-id1163723217922\" type=\"a\">\n<li>Find the center of gravity of a bowling ball.<\/li>\n<li>Determine the velocity of a submarine subjected to an ocean current.<\/li>\n<li>Calculate the pressure in a conical water tank.<\/li>\n<li>Find the volume of oil flowing through a pipeline.<\/li>\n<li>Determine the amount of leather required to make a football.<\/li>\n<\/ol>\n<div id=\"attachment_5319\" style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5319\" class=\"size-full wp-image-5319\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31232437\/2.105.jpg\" alt=\"This figure has 5 images. The first image shows bowling balls. The second image is a submarine traveling on an ocean surface. The third image is a traffic cone. The fourth image is a pipeline across some barren land. The fifth image is a football.\" width=\"731\" height=\"492\" \/><\/p>\n<p id=\"caption-attachment-5319\" class=\"wp-caption-text\">Figure 10. (credit: (a) modification of work by scl hua, Wikimedia, (b) modification of work by DVIDSHUB, Flickr, (c) modification of work by Michael Malak, Wikimedia, (d) modification of work by Sean Mack, Wikimedia, (e) modification of work by Elvert Barnes, Flickr)<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q757372734\">Show Solution<\/span><\/p>\n<div id=\"q757372734\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1163723567687\" type=\"a\">\n<li>Clearly, a bowling ball is a sphere, so spherical coordinates would probably work best here. The origin should be located at the physical center of the ball. There is no obvious choice for how the [latex]x[\/latex]-, [latex]y[\/latex]&#8211; and [latex]z[\/latex]-axes should be oriented. Bowling balls normally have a weight block in the center. One possible choice is to align the [latex]z[\/latex]-axis with the axis of symmetry of the weight block.<\/li>\n<li>A submarine generally moves in a straight line. There is no rotational or spherical symmetry that applies in this situation, so rectangular coordinates are a good choice. The [latex]z[\/latex]-axis should probably point upward. The [latex]x[\/latex]&#8211; and [latex]y[\/latex]-axes could be aligned to point east and north, respectively. The origin should be some convenient physical location, such as the starting position of the submarine or the location of a particular port.<\/li>\n<li>A cone has several kinds of symmetry. In cylindrical coordinates, a cone can be represented by equation [latex]z=kr[\/latex], where [latex]k[\/latex] is a constant. In spherical coordinates, we have seen that surfaces of the form [latex]\\varphi=c[\/latex] are half-cones. Last, in rectangular coordinates, elliptic cones are quadric surfaces and can be represented by equations of the form [latex]z^2=\\frac{x^2}{a^2}+\\frac{y^2}{b^2}[\/latex]. In this case, we could choose any of the three. However, the equation for the surface is more complicated in rectangular coordinates than in the other two systems, so we might want to avoid that choice. In addition, we are talking about a water tank, and the depth of the water might come into play at some point in our calculations, so it might be nice to have a component that represents height and depth directly. Based on this reasoning, cylindrical coordinates might be the best choice. Choose the [latex]z[\/latex]-axis to align with the axis of the cone. The orientation of the other two axes is arbitrary. The origin should be the bottom point of the cone.<\/li>\n<li>A pipeline is a cylinder, so cylindrical coordinates would be best the best choice. In this case, however, we would likely choose to orient our [latex]z[\/latex]-axis with the center axis of the pipeline. The [latex]x[\/latex]-axis could be chosen to point straight downward or to some other logical direction. The origin should be chosen based on the problem statement. Note that this puts the [latex]z[\/latex]-axis in a horizontal orientation, which is a little different from what we usually do. It may make sense to choose an unusual orientation for the axes if it makes sense for the problem.<\/li>\n<li>A football has rotational symmetry about a central axis, so cylindrical coordinates would work best. The [latex]z[\/latex]-axis should align with the axis of the ball. The origin could be the center of the ball or perhaps one of the ends. The position of the [latex]x[\/latex]-axis is arbitrary.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Choosing the Best Coordinate System.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7809470&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=1WaXMBJU_j4&amp;video_target=tpm-plugin-9k5unai6-1WaXMBJU_j4\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the transcript for <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/Ex2.67_transcript.html\">\u201cEx 2.67\u201d here (opens in new window).<\/a><\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Which coordinate system is most appropriate for creating a star map, as viewed from Earth (see the following figure)?<\/p>\n<div id=\"attachment_5322\" style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5322\" class=\"size-full wp-image-5322\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/31232545\/2.61.jpg\" alt=\"This figure is a circle with a star chart in the middle.\" width=\"731\" height=\"731\" \/><\/p>\n<p id=\"caption-attachment-5322\" class=\"wp-caption-text\">Figure 11. Star map as viewed from Earth.<\/p>\n<\/div>\n<p>How should we orient the coordinate axes?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q883476235\">Show Solution<\/span><\/p>\n<div id=\"q883476235\" class=\"hidden-answer\" style=\"display: none\">\n<p>Spherical coordinates with the origin located at the center of the earth, the [latex]z[\/latex]-axis aligned with the North Pole, and the [latex]x[\/latex]-axis aligned with the prime meridian.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div data-type=\"example\"><\/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-817\">\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 2.58. <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><li>CP 2.67. <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":349141,"menu_order":30,"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 2.58\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"CP 2.67\",\"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-817","chapter","type-chapter","status-publish","hentry"],"part":20,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/817","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\/349141"}],"version-history":[{"count":60,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/817\/revisions"}],"predecessor-version":[{"id":6443,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/817\/revisions\/6443"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/20"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/817\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=817"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=817"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=817"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=817"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}