{"id":3986,"date":"2022-04-12T18:03:09","date_gmt":"2022-04-12T18:03:09","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=3986"},"modified":"2022-11-01T04:27:28","modified_gmt":"2022-11-01T04:27:28","slug":"triple-integrals","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/triple-integrals\/","title":{"raw":"Triple Integrals","rendered":"Triple Integrals"},"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\">Recognize when a function of three variables is integrable over a rectangular box.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Evaluate a triple integral by expressing it as an iterated integral.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Recognize when a function of three variables is integrable over a closed and bounded region.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Simplify a calculation by changing the order of integration of a triple integral.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<section id=\"fs-id1167793878911\" data-depth=\"1\">\r\n<h2 data-type=\"title\">Integrable Functions of Three Variables<\/h2>\r\n<p id=\"fs-id1167794064648\">We can define a rectangular box [latex]B[\/latex] in [latex]{{\\mathbb{R}}^3}[\/latex] as [latex]{B} = {\\left \\{{(x,y,z)}{\\mid}{a} \\ {\\leq} \\ {x} \\ {\\leq} \\ {b,c} \\ {\\leq} \\ {y} \\ {\\leq} \\ {d,e} \\ {\\leq} \\ {z} \\ {\\leq} \\ {f} \\right \\}}[\/latex]. We follow a similar procedure to what we did in\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/double-integrals-over-rectangular-regions\/\" data-page-slug=\"5-1-double-integrals-over-rectangular-regions\" data-page-uuid=\"d9da6d44-3ba3-4a90-adbc-dec26572e480\" data-page-fragment=\"page_d9da6d44-3ba3-4a90-adbc-dec26572e480\">Double Integrals over Rectangular Regions<\/a>. We divide the interval [latex][a,b][\/latex] into [latex]l[\/latex] subintervals [latex][{{x}_{i-1}},{x_i}][\/latex] of equal length [latex]{{\\Delta}{x}} = {\\frac{{x_i}-{x_{i-1}}}{l}}[\/latex], divide the interval [latex][c,d][\/latex] into [latex]m[\/latex] subintervals [latex][{{y}_{i-1}},{y_i}][\/latex] of equal length [latex]{{\\Delta}{y}} = {\\frac{{y_j}-{y_{j-1}}}{m}}[\/latex], and divide the interval [latex][e,f][\/latex] into [latex]n[\/latex] subintervals [latex][{{z}_{i-1}},{z_i}][\/latex] of equal length [latex]{{\\Delta}{z}} = {\\frac{{z_k}-{z_{k-1}}}{n}}[\/latex]. Then the rectangular box [latex]B[\/latex] is subdivided into [latex]lmn[\/latex] subboxes [latex]{{B}_{ijk}} = [{{x}_{i-1}},{x_i}] \\ {\\times} \\ [{{y}_{i-1}},{y_i}] \\ {\\times} \\ [{{z}_{i-1}},{z_i}] [\/latex], as shown in\u00a0Figure 1.<\/p>\r\n\r\n[caption id=\"attachment_1372\" align=\"aligncenter\" width=\"590\"]<img class=\"size-full wp-image-1372\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25144228\/5-4-1.jpeg\" alt=\"In x y z space, there is a box B with a subbox Bijk with sides of length Delta x, Delta y, and Delta z.\" width=\"590\" height=\"724\" \/> Figure 1.\u00a0A rectangular box in\u00a0[latex]\\mathbb{R}^{3}[\/latex] divided into subboxes by planes parallel to the coordinate planes.[\/caption]\r\n<p id=\"fs-id1167794051785\">For each [latex]i[\/latex], [latex]j[\/latex], and [latex]k[\/latex], consider a sample point [latex]{({{x}^{*}_{ijk}},{{y}^{*}_{ijk}},{{z}^{*}_{ijk}})}[\/latex] in each sub-box [latex]{{B}_{ijk}}[\/latex]. We see that its volume is [latex]{{\\Delta}{V}} = {{\\Delta}{x}}{{\\Delta}{y}}{{\\Delta}{z}}[\/latex]. Form the triple Riemann sum<\/p>\r\n<p style=\"text-align: center;\">[latex]\\Large{\\displaystyle\\sum_{i=1}^{l}}{\\displaystyle\\sum_{j=1}^{m}}{\\displaystyle\\sum_{k=1}^{n}}{f}{({{x}^{*}_{ijk}},{{y}^{*}_{ijk}},{{z}^{*}_{ijk}})}{{\\Delta}{x}}{{\\Delta}{y}}{{\\Delta}{z}}[\/latex].<\/p>\r\nWe define the triple integral in terms of the limit of a triple Riemann sum, as we did for the double integral in terms of a double Riemann sum.\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-id1167793943883\">The\u00a0<strong><span id=\"d140d1cd-af53-466a-8e69-402de1b91259_term219\" data-type=\"term\">triple integral<\/span><\/strong>\u00a0of a function [latex]f(x, y, z)[\/latex] over a rectangular box [latex]B[\/latex] is defined as<\/p>\r\n<p style=\"text-align: center;\">[latex]\\Large{\\displaystyle\\lim_{{l,m,n}{\\rightarrow}{\\infty}}} \\ {\\displaystyle\\sum_{i=1}^{l}}{\\displaystyle\\sum_{j=1}^{m}}{\\displaystyle\\sum_{k=1}^{n}} \\ {f}{({{x}^{*}_{ijk}},{{y}^{*}_{ijk}},{{z}^{*}_{ijk}})}{{\\Delta}{x}}{{\\Delta}{y}}{{\\Delta}{z}} = \\underset{B}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV}[\/latex]<\/p>\r\n<span style=\"font-size: 1rem; text-align: initial;\">if this limit exists.<\/span>\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793449708\">When the triple integral exists on [latex]B[\/latex], the function [latex]f(x, y, z)[\/latex] is said to be integrable on [latex]B[\/latex]. Also, the triple integral exists if [latex]f(x, y, z)[\/latex] is continuous on [latex]B[\/latex]. Therefore, we will use continuous functions for our examples. However, continuity is sufficient but not necessary; in other words, [latex]f[\/latex] is bounded on [latex]B[\/latex] and continuous except possibly on the boundary of [latex]B[\/latex] The sample point [latex]{({{x}^{*}_{ijk}},{{y}^{*}_{ijk}},{{z}^{*}_{ijk}})}[\/latex] can be any point in the rectangular sub-box [latex]{B}_{ijk}[\/latex] and all the properties of a double integral apply to a triple integral. Just as the double integral has many practical applications, the triple integral also has many applications, which we discuss in later sections.<\/p>\r\n<p id=\"fs-id1167794100272\">Now that we have developed the concept of the triple integral, we need to know how to compute it. Just as in the case of the double integral, we can have an iterated triple integral, and consequently, a version of\u00a0<span id=\"d140d1cd-af53-466a-8e69-402de1b91259_term220\" class=\"no-emphasis\" data-type=\"term\">Fubini\u2019s thereom<\/span>\u00a0for triple integrals exists.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">theorem: fubini's theorem for triple integrals<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167794145580\">If [latex]f(x, y, z)[\/latex] is continuous on a rectangular box [latex]{B} = {[a,b]} \\ {\\times} \\ {[c,d]} \\ {\\times} \\ {[e,f]}[\/latex], then<\/p>\r\n<p style=\"text-align: center;\">[latex]\\Large{\\underset{B}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV} = {\\displaystyle\\int_{e}^{f}}{\\displaystyle\\int_{c}^{d}}{\\displaystyle\\int_{a}^{b}}{f}{(x,y,z)}{dx} \\ {dy} \\ {dz}}[\/latex].<\/p>\r\nThis integral is also equal to any of the other five possible orderings for the iterated triple integral.\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793263456\">For [latex]a[\/latex], [latex]b[\/latex], [latex]c[\/latex], [latex]d[\/latex], [latex]e[\/latex], and [latex]f[\/latex] real numbers, the iterated triple integral can be expressed in six different orderings:<\/p>\r\n[latex]\\large{\\begin{align}\r\n\r\n{\\displaystyle\\int_{e}^{f}}{\\displaystyle\\int_{c}^{d}}{\\displaystyle\\int_{a}^{b}}{f}{(x,y,z)}{dx} \\ {dy} \\ {dz}&amp;={\\displaystyle\\int_{e}^{f}}\\left({\\displaystyle\\int_{c}^{d}}\\left({\\displaystyle\\int_{a}^{b}}{f}{(x,y,z)}{dx}\\right){dy}\\right){dz}={\\displaystyle\\int_{c}^{d}}\\left({\\displaystyle\\int_{e}^{f}}\\left({\\displaystyle\\int_{a}^{b}}{f}{(x,y,z)}{dx}\\right){dz}\\right){dy} \\\\\r\n\r\n&amp;={\\displaystyle\\int_{a}^{b}}\\left({\\displaystyle\\int_{e}^{f}}\\left({\\displaystyle\\int_{c}^{d}}{f}{(x,y,z)}{dy}\\right){dz}\\right){dx}={\\displaystyle\\int_{e}^{f}}\\left({\\displaystyle\\int_{a}^{b}}\\left({\\displaystyle\\int_{c}^{d}}{f}{(x,y,z)}{dy}\\right){dx}\\right){dz} \\\\\r\n\r\n&amp;={\\displaystyle\\int_{c}^{e}}\\left({\\displaystyle\\int_{a}^{b}}\\left({\\displaystyle\\int_{e}^{f}}{f}{(x,y,z)}{dz}\\right){dx}\\right){dy}={\\displaystyle\\int_{a}^{b}}\\left({\\displaystyle\\int_{c}^{e}}\\left({\\displaystyle\\int_{e}^{f}}{f}{(x,y,z)}{dz}\\right){dy}\\right){dx}.\r\n\r\n\\end{align}}[\/latex]\r\nFor a rectangular box, the order of integration does not make any significant difference in the level of difficulty in computation. We compute triple integrals using Fubini\u2019s Theorem rather than using the Riemann sum definition. We follow the order of integration in the same way as we did for double integrals (that is, from inside to outside).\r\n<div class=\"textbox exercises\">\r\n<h3>Example: evaluating a triple integral 1<\/h3>\r\nEvaluate the triple integral [latex]{\\displaystyle\\int^{z=1}_{z=0}}{\\displaystyle\\int^{y=4}_{y=2}}{\\displaystyle\\int^{x=5}_{x=-1}}{(x+yz^2)}{dx} \\ {dy} \\ {dz}[\/latex].\r\n\r\n[reveal-answer q=\"390606\"]Show Solution[\/reveal-answer]\r\n\r\n[hidden-answer a=\"390606\"]\r\n\r\nThe order of integration is specified in the problem, so integrate with respect to x first, then y, and then z.\r\n\r\n[latex]\\hspace{3cm}\\begin{align}\r\n\r\n&amp;\\hspace{0.75cm}\\displaystyle\\int^{z=1}_{z=0}\\displaystyle\\int^{y=4}_{y=2}\\displaystyle\\int^{x=5}_{x=-1}(x+yz^2)dx \\ dy \\ dz \\\\\r\n\r\n&amp;=\\displaystyle\\int^{z=1}_{z=0}\\displaystyle\\int^{y=4}_{y=2}\\left[\\frac{x^2}2+xyz^2\\bigg|^{x=5}_{x=-1}\\right]dy \\ dz &amp;\\quad &amp;\\text{Integrate with respect to }x. \\\\\r\n\r\n&amp;=\\displaystyle\\int^{z=1}_{z=0}\\displaystyle\\int^{y=4}_{y=2}[12+6yz^2] \\ dy \\ dz &amp;\\quad &amp;\\text{Evaluate.} \\\\\r\n\r\n&amp;=\\displaystyle\\int^{z=1}_{z=0}\\left[12y+6\\frac{y^2}2x^2\\bigg|^{y=4}_{y=2}\\right]{dz} &amp;\\quad &amp;\\text{Integrate with respect to }y. \\\\\r\n\r\n&amp;={\\displaystyle\\int^{z=1}_{z=0}}[24+36z^2]{dz} &amp;\\quad &amp;\\text{Evaluate}. \\\\\r\n\r\n&amp;=\\left[24z+36\\frac{z^3}3\\right]_{z=0}^{z=1}=36.\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: evaluating a triple integral 2<\/h3>\r\nEvaluate the triple integral [latex]\\underset{B}{\\displaystyle\\iiint}{x^2}{yz}{dV}[\/latex] where\r\n\r\n[latex]{B} = {\\left \\{{(x,y,z)}{\\mid}{-2} \\ {\\leq} \\ {x} \\ {\\leq} \\ {1,0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {3,1} \\ {\\leq} \\ {z} \\ {\\leq} \\ {5} \\right \\}}[\/latex] as shown in the following figure.\r\n\r\n[caption id=\"attachment_1374\" align=\"aligncenter\" width=\"375\"]<img class=\"size-full wp-image-1374\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25144322\/5-4-2.jpeg\" alt=\"In x y z space, there is a box given with corners (1, 0, 5), (1, 0, 1), (1, 3, 1), (1, 3, 5), (negative 2, 0, 5), (negative 2, 0, 1), (negative 2, 3, 1), and (negative 2, 3, 5).\" width=\"375\" height=\"328\" \/> Figure 2.\u00a0Evaluating a triple integral over a given rectangular box.[\/caption]\r\n\r\n[reveal-answer q=\"245801336\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"245801336\"]\r\n<p id=\"fs-id1167793937669\">The order is not specified, but we can use the iterated integral in any order without changing the level of difficulty. Choose, say, to integrate [latex]y[\/latex] first, then\u00a0[latex]x[\/latex], and then\u00a0[latex]z[\/latex].<\/p>\r\n[latex]\\hspace{2cm}\\begin{align}\r\n\r\n\\underset{B}{\\displaystyle\\iiint}{x^2}{yz}{dV}&amp;=\\displaystyle\\int_1^5\\displaystyle\\int_{-2}^1\\displaystyle\\int_0^3[x^2yz]dy \\ dx \\ dz =\\displaystyle\\int_1^5\\displaystyle\\int_{-2}^1\\left[x^2\\frac{y^2}2z\\bigg|_0^3\\right]dx \\ dz \\\\\r\n\r\n&amp;=\\displaystyle\\int_1^5\\displaystyle\\int_{-2}^1\\frac92x^2z \\ dx \\ dz = \\displaystyle\\int_1^5\\left[\\frac92\\frac{x^3}3z\\bigg|_{-2}^1\\right]dz=\\displaystyle\\int_1^5\\frac{27}2\\frac{z^2}2\\bigg|_1^5=162.\r\n\r\n\\end{align}[\/latex]\r\n\r\nNow try to integrate in a different order just to see that we get the same answer. Choose to integrate with respect to [latex]x[\/latex] first, then [latex]z[\/latex], and then [latex]y[\/latex].\r\n\r\n[latex]\\hspace{2cm}\\begin{align}\r\n\r\n\\underset{B}{\\displaystyle\\iiint}{x^2}{yz}{dV}&amp;=\\displaystyle\\int_0^3\\displaystyle\\int_1^5\\displaystyle\\int_{-2}^1[x^2yz]dx \\ dz \\ dy =\\displaystyle\\int_0^3\\displaystyle\\int_1^5\\left[\\frac{x^3}3yz\\bigg|_{-2}^1\\right]dz \\ dy \\\\\r\n\r\n&amp;=\\displaystyle\\int_0^3\\displaystyle\\int_1^53yz \\ dz \\ dy =\\displaystyle\\int_0^3\\left[3y\\frac{z^2}2\\bigg|_1^5\\right]dy=\\displaystyle\\int_0^336y \\ dy = 36\\frac{y^2}2\\bigg|_0^3=18(9-0)=162.\r\n\r\n\\end{align}[\/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\nEvaluate the triple integral [latex]\\underset{B}{\\displaystyle\\iiint}{z} \\ {\\sin \\ x} \\ {\\cos \\ y} \\ {dV}[\/latex] where\r\n\r\n[latex]{B} = {\\left \\{{(x,y,z)}{\\mid}{0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {{\\pi},{\\frac{3{\\pi}}{2}}} \\ {\\leq} \\ {y} \\ {\\leq} \\ {2{\\pi},1} \\ {\\leq} \\ {z} \\ {\\leq} \\ {3} \\right \\}}[\/latex].\r\n\r\n[reveal-answer q=\"159035786\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"159035786\"]\r\n\r\n[latex]\\underset{B}{\\displaystyle\\iiint}z\\sin{x}\\cos{y} \\ dV=8[\/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=8197105&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=csY8RONL3Qs&amp;video_target=tpm-plugin-46801eo2-csY8RONL3Qs\" 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\/CP5.23_transcript.html\">transcript for \u201cCP 5.23\u201d here (opens in new window).<\/a><\/center>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]184576[\/ohm_question]\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">Triple Integrals over a General Bounded Region<\/h2>\r\n<p id=\"fs-id1167794214296\">We now expand the definition of the triple integral to compute a triple integral over a more\u00a0<span id=\"d140d1cd-af53-466a-8e69-402de1b91259_term221\" class=\"no-emphasis\" data-type=\"term\">general bounded region [latex]E[\/latex] in\u00a0[latex]{\\mathbb{R}^3}[\/latex].\u00a0<\/span>The general bounded regions we will consider are of three types. First, let [latex]D[\/latex] be the bounded region that is a projection of [latex]E[\/latex] onto the [latex]xy[\/latex]-plane. Suppose the region [latex]E[\/latex] in [latex]{\\mathbb{R}^3}[\/latex] has the form<\/p>\r\n<p style=\"text-align: center;\">[latex]\\Large{E} = \\left \\{{(x,y,z)}{\\mid}{(x,y)} \\ {\\in} \\ {{D},{u_1}{(x,y)}} \\ {\\leq} \\ {z} \\ {\\leq} \\ {u_2}{(x,y)} \\right \\}[\/latex].<\/p>\r\n<span style=\"font-size: 1rem; text-align: initial;\">For two functions [latex]z=u_1(x, y)[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">and [latex]z=u_2(x,y)[\/latex],\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">such that [latex]{u_1}{(x,y)} \\ {\\leq} \\ {u_2}{(x,y)}[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">for all [latex](x, y)[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">in [latex]D[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">as shown in the following figure.<\/span>\r\n\r\n[caption id=\"attachment_1375\" align=\"aligncenter\" width=\"429\"]<img class=\"size-full wp-image-1375\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25144346\/5-4-3.jpeg\" alt=\"In x y z space, there is a shape E with top surface z = u2(x, y) and bottom surface z = u1(x, y). The bottom projects onto the x y plane as region D.\" width=\"429\" height=\"348\" \/> Figure 3. We can describe region [latex]E[\/latex] as the space between\u00a0[latex]u_{1}(x,y)[\/latex] and\u00a0[latex]u_{2}(x,y)[\/latex] above the projection\u00a0[latex]D[\/latex] of\u00a0[latex]E[\/latex] onto the\u00a0[latex]xy[\/latex]-plane.[\/caption]<\/section>\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">theorem: triple integral over a general region<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793632073\">The triple integral of a continuous function [latex]f(x, y, z)[\/latex] over a general three-dimensional region<\/p>\r\n&nbsp;\r\n<p style=\"text-align: center;\">[latex]\\large{E} = \\left \\{{(x,y,z)}{\\mid}{(x,y)} \\ {\\in} \\ {{D},{u_1}{(x,y)}} \\ {\\leq} \\ {z} \\ {\\leq} \\ {u_2}{(x,y)} \\right \\}[\/latex]<\/p>\r\n&nbsp;\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">in [latex]{{\\mathbb{R}}^3}[\/latex],\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">where [latex]D[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">is the projection of [latex]E[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">onto the [latex]xy[\/latex]-<\/span><span style=\"font-size: 1rem; text-align: initial;\">plane, is<\/span>\r\n<p style=\"text-align: center;\">[latex]\\large{\\underset{E}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV} = \\underset{D}{\\displaystyle\\iint}{\\left [{\\displaystyle\\int^{{u_2}{(x,y)}}_{{u_1}{(x,y)}}}{f}{(x,y,z)}{dz} \\right ]}{dA}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793895376\">Similarly, we can consider a general bounded region [latex]D[\/latex] in the [latex]xy[\/latex]-plane and two functions [latex]{y} = {u_1}{(x,z)}[\/latex] and [latex]{y} = {u_2}{(x,z)}[\/latex] such that [latex]{u_1}{(x,z)} \\ {\\leq} \\ {u_2}{(x,z)}[\/latex] for all [latex](x, z)[\/latex] in [latex]D[\/latex]. Then we can describe the solid region [latex]E[\/latex] in [latex]{{\\mathbb{R}}^3}[\/latex] as<\/p>\r\n<p style=\"text-align: center;\">[latex]{E} = \\left \\{{(x,y,z)}{\\mid}{(x,z)} \\ {\\in} \\ {{D},{u_1}{(x,z)}} \\ {\\leq} \\ {z} \\ {\\leq} \\ {u_2}{(x,z)} \\right \\}[\/latex]<\/p>\r\nwhere [latex]D[\/latex] is the projection of [latex]E[\/latex] onto the [latex]xy[\/latex]-plane and the triple integral is\r\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV} = \\underset{D}{\\displaystyle\\iint}{\\left [{\\displaystyle\\int^{{u_2}{(x,z)}}_{{u_1}{(x,z)}}}{f}{(x,y,z)}{dy} \\right ]}{dA}[\/latex].<\/p>\r\n<span style=\"font-size: 1rem; text-align: initial;\">Finally, if [latex]D[\/latex] is a general bounded region in the [latex]yz[\/latex]-plane and we have two functions [latex]{x} = {u_1}{(y,z)}[\/latex] and [latex]{x} = {u_2}{(y,z)}[\/latex] such that [latex]{u_1}{(y,z)} \\ {\\leq} \\ {u_2}{(y,z)}[\/latex] for all [latex](y, z)[\/latex] in [latex]D[\/latex], then the solid region [latex]E[\/latex] in [latex]{{\\mathbb{R}}^3}[\/latex] can be described as<\/span>\r\n<p style=\"text-align: center;\">[latex]{E} = \\left \\{{(x,y,z)}{\\mid}{(y,z)} \\ {\\in} \\ {{D},{u_1}{(y,z)}} \\ {\\leq} \\ {z} \\ {\\leq} \\ {u_2}{(y,z)} \\right \\}[\/latex]<\/p>\r\nwhere [latex]D[\/latex] is the projection of [latex]E[\/latex] onto the [latex]yz[\/latex]-plane and the triple integral is\r\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV} = \\underset{D}{\\displaystyle\\iint}{\\left [{\\displaystyle\\int^{{u_2}{(y,z)}}_{{u_1}{(y,z)}}}{f}{(x,y,z)}{dx} \\right ]}{dA}[\/latex].<\/p>\r\n<span style=\"font-size: 1rem; text-align: initial;\">Note that the region [latex]D[\/latex] in any of the planes may be of Type I or Type II as described in\u00a0<\/span><a style=\"font-size: 1rem; text-align: initial;\" href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/introduction-to-double-integrals-over-general-regions\/\" data-page-slug=\"5-2-double-integrals-over-general-regions\" data-page-uuid=\"f783ca08-ca2f-4ebb-a559-d1b30c2f9b8c\" data-page-fragment=\"page_f783ca08-ca2f-4ebb-a559-d1b30c2f9b8c\">Double Integrals over General Regions<\/a><span style=\"font-size: 1rem; text-align: initial;\">. If [latex]D[\/latex] in the [latex]xy[\/latex]-plane is of Type I (<\/span>Figure 4<span style=\"font-size: 1rem; text-align: initial;\">), then<\/span>\r\n<p style=\"text-align: center;\">[latex]{E} = \\left \\{{(x,y,z)}{\\mid}{a} \\ {\\leq} \\ {x} \\ {\\leq} \\ {b,{g_1}{(x)}} \\ {\\leq} \\ {y} \\ {\\leq} \\ {g_2{(x)}},{{u_1}{(x,y)}} \\ {\\leq} \\ {z} \\ {\\leq} \\ {u_2}{(x,y)} \\right \\}[\/latex].<\/p>\r\n\r\n<section data-depth=\"1\"><\/section><section data-depth=\"1\">[caption id=\"attachment_1376\" align=\"aligncenter\" width=\"571\"]<img class=\"wp-image-1376 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25144428\/5-4-4.jpeg\" alt=\"In x y z space, there is a complex shape E with top surface z = u2(x, y) and bottom surface z = u1(x, y). The bottom projects onto the xy plane as region D with boundaries x = a, x = b, y = g1(x), and y = g2(x).\" width=\"571\" height=\"443\" \/> Figure 4. A box\u00a0[latex]E[\/latex] where the projection\u00a0[latex]D[\/latex] in the\u00a0[latex]xy[\/latex]-plane is of Type I.[\/caption]Then the triple integral becomes\r\n<p style=\"text-align: center;\"><span style=\"font-size: 1em;\">[latex]\\underset{E}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV} = {\\displaystyle\\int_{a}^{b}} \\ {\\displaystyle\\int_{{g_1}{(x)}}^{{g_2}{(x)}}} \\ {\\displaystyle\\int_{{u_1}{(x,y)}}^{{u_2}{(x,y)}}}{f}{(x,y,z)}{dz}{dy}{dx}[\/latex].<\/span><\/p>\r\n<p style=\"text-align: left;\"><span style=\"font-size: 1em;\">If\u00a0[latex]D[\/latex] in the\u00a0[latex]xy[\/latex]-plane is of Type II (Figure 5), then<\/span><\/p>\r\n<p style=\"text-align: center;\"><span style=\"font-size: 1rem; text-align: initial;\">[latex]{E} = \\left \\{{(x,y,z)}{\\mid}{c} \\ {\\leq} \\ {x} \\ {\\leq} \\ {d,{h_1}{(x)}} \\ {\\leq} \\ {y} \\ {\\leq} \\ {h_2{(x)}},{{u_1}{(x,y)}} \\ {\\leq} \\ {z} \\ {\\leq} \\ {u_2}{(x,y)} \\right \\}[\/latex].<\/span><\/p>\r\n\r\n<\/section>[caption id=\"attachment_1377\" align=\"aligncenter\" width=\"444\"]<img class=\"size-full wp-image-1377\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25144556\/5-4-5.jpeg\" alt=\"In x y z space, there is a complex shape E with top surface z = u2(x, y) and bottom surface z = u1(x, y). The bottom projects onto the xy plane as region D with boundaries y = c, y = d, x = h1(y), and x = h2(y).\" width=\"444\" height=\"397\" \/> Figure 5.\u00a0A box\u00a0[latex]E[\/latex] where the projection\u00a0[latex]D[\/latex] in the\u00a0[latex]xy[\/latex]-plane is of Type II.[\/caption]Then the triple integral becomes\r\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV} = {\\displaystyle\\int^{y=d}_{y=c}}{\\displaystyle\\int^{x={{h_2}{(y)}}}_{x={h_1}{(y)}}}{\\displaystyle\\int^{z={{u_2}{(x,y)}}}_{z={u_1}{(x,y)}}}{f}{(x,y,z)}{dz}{dx}{dy}[\/latex].<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: evaluating a triple integral over a general bounded region<\/h3>\r\nEvaluate the triple integral of the function [latex]f(x, y, z)=5x-3y[\/latex] over the solid tetrahedron bounded by the planes\u00a0[latex]x=0, \\ y=0, \\ z=0,[\/latex] and\u00a0[latex]x+y+z=1[\/latex].\r\n\r\n[reveal-answer q=\"331457700\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"331457700\"]\r\n\r\nFigure 6\u00a0shows the solid tetrahedron [latex]E[\/latex] and its projection [latex]D[\/latex] on the [latex]xy[\/latex]-plane.\r\n\r\n[caption id=\"attachment_1378\" align=\"aligncenter\" width=\"685\"]<img class=\"size-full wp-image-1378\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25144633\/5-4-6.jpeg\" alt=\"In x y z space, there is a solid E with boundaries being the x y, z y, and x z planes and z = 1 minus x minus y. The points are the origin, (1, 0, 0), (0, 0, 1), and (0, 1, 0). Its surface on the x y plane is shown as being a rectangle marked D with line y = 1 minus x. Additionally, there is a vertical line shown on D.\" width=\"685\" height=\"357\" \/> Figure 6.\u00a0The solid [latex]E[\/latex] has a projection\u00a0[latex]D[\/latex] on the\u00a0[latex]xy[\/latex]-plane of Type I.[\/caption]\r\n<p id=\"fs-id1167794051980\">We can describe the solid region tetrahedron as<\/p>\r\n<p style=\"text-align: center;\">[latex]{E} = \\left \\{{(x,y,z)}{\\mid}{0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {1,0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {1-x},{0} \\ {\\leq} \\ {z} \\ {\\leq} \\ {1-x-y} \\right \\}[\/latex].<\/p>\r\n<p id=\"fs-id1167793887543\">Hence, the triple integral is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV} = {\\displaystyle\\int^{x=1}_{x=0}}{\\displaystyle\\int^{y={1-x}}_{y=0}}{\\displaystyle\\int^{z={1-x-y}}_{z=0}}{(5x-3y)}{dz}{dy}{dx}[\/latex].<\/p>\r\n<p id=\"fs-id1167793423320\">To simplify the calculation, first evaluate the integral [latex]{\\displaystyle\\int^{z=1-x-y}_{z=0}}{(5x-3y)}{dz}[\/latex]. We have<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{z=1-x-y}_{z=0}}{(5x-3y)}{dz} = {(5x-3y)}{(1-x-y)}[\/latex].<\/p>\r\n<p id=\"fs-id1167793421147\">Now evaluate the integral [latex]{\\displaystyle\\int^{y=1-x}_{y=0}}{(5x-3y)}{(1-x-y)}{dy}[\/latex], obtaining<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{y=1-x}_{y=0}}{(5x-3y)}{(1-x-y)}{dy} = {\\frac{1}{2}}{(x-1)^2}{(6x-1)}[\/latex].<\/p>\r\n<p id=\"fs-id1167794045783\">Finally, evaluate<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{x=1}_{x=0}}{\\frac{1}{2}}{(x-1)^2}{(6x-1)}{dx} = {\\frac{1}{12}}[\/latex].<\/p>\r\n<span style=\"font-size: 1rem; text-align: initial;\">Putting it all together, we have<\/span>\r\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV} = {\\displaystyle\\int^{x=1}_{x=0}}{\\displaystyle\\int^{y={1-x}}_{y=0}}{\\displaystyle\\int^{z={1-x-y}}_{z=0}}{(5x-3y)}{dz}{dy}{dx} = {\\frac{1}{12}}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nJust as we used the double integral [latex]\\underset{D}{\\displaystyle\\iint}{1}{dA}[\/latex] to find the area of a general bounded region [latex]D[\/latex], we can use [latex]\\underset{E}{\\displaystyle\\iiint}{1}{dV}[\/latex] to find the volume of a general solid bounded region [latex]E[\/latex]. The next example illustrates the method.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding a volume by evaluating a triple integral<\/h3>\r\n<p id=\"fs-id1167793504416\">Find the volume of a right pyramid that has the square base in the [latex]xy[\/latex]-plane [latex][-1,1] \\ {\\times} \\ [-1,1][\/latex] and vertex at the point [latex](0, 0, 1)[\/latex] as shown in the following figure.<\/p>\r\n\r\n\r\n[caption id=\"attachment_1379\" align=\"aligncenter\" width=\"543\"]<img class=\"size-full wp-image-1379\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25144721\/5-4-7.jpeg\" alt=\"In x y z space, there is a pyramid with a square base centered at the origin. The apex of the pyramid is (0, 0, 1).\" width=\"543\" height=\"527\" \/> Figure 7.\u00a0Finding the volume of a pyramid with a square base.[\/caption]\r\n\r\n[reveal-answer q=\"425689014\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"425689014\"]\r\n<p id=\"fs-id1167794005598\">In this pyramid the value of [latex]z[\/latex] changes from 0 to 1 and at each height [latex]z[\/latex], the cross section of the pyramid for any value of [latex]z[\/latex] is the square [latex][-1+z,1-z] \\ {\\times} \\ [-1+z,1-z][\/latex]. Hence, the volume of the pyramid is [latex]\\underset{E}{\\displaystyle\\iiint}{1}{dV}[\/latex] where<\/p>\r\n<p style=\"text-align: center;\">[latex]{E} = \\left \\{{(x,y,z)}{\\mid}{0} \\ {\\leq} \\ {z} \\ {\\leq} \\ {1,-1+z} \\ {\\leq} \\ {y} \\ {\\leq} \\ {1-z,-1+z} \\ {\\leq} \\ {x} \\ {\\leq} \\ {1-z} \\right \\}[\/latex].<\/p>\r\n<p id=\"fs-id1167793442473\">Thus, we have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{1}{dV} = {\\displaystyle\\int^{z=1}_{z=0}}{\\displaystyle\\int^{y={1-z}}_{y=-1+z}}{\\displaystyle\\int^{x={1-z}}_{x=1+z}}{1}{dx}{dy}{dz} = {\\displaystyle\\int^{z=1}_{z=0}}{\\displaystyle\\int^{y={1-z}}_{y=-1+z}}{(2-2z)}{dy}{dz} = {\\displaystyle\\int^{z=1}_{z=0}}{(2-2z)^2}{dz} = {\\frac{4}{3}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793637795\">Hence, the volume of the pyramid is\u00a0[latex]{\\frac{4}{3}}[\/latex] cubic units.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nConsider the solid sphere [latex]{E} = {(x,y,z)}{\\mid}\\left \\{{{x^2}+{y^2}+{z^2}} = {9} \\right \\}[\/latex]. Write the triple integral [latex]\\underset{E}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV}[\/latex] for an arbitrary function [latex]f[\/latex] as an iterated integral. Then evaluate this triple integral with [latex]f(x, y, z)=1[\/latex]. Notice that this gives the volume of a sphere using a triple integral.\r\n\r\n[reveal-answer q=\"178836902\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"178836902\"]\r\n\r\n[latex]\\underset{E}{\\displaystyle\\iiint}1 \\ dV=8\\displaystyle\\int_{x=-3}^{x=3}\\displaystyle\\int_{y=-\\sqrt{9-x^2}}^{y=\\sqrt{9-x^2}}\\displaystyle\\int_{z=-\\sqrt{9-x^2-y^2}}^{z=\\sqrt{9-x^2-y^2}}1 \\ dz \\ dy \\ dx = 36\\pi.[\/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=8197106&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=yINls9G1Yek&amp;video_target=tpm-plugin-q6r3g93d-yINls9G1Yek\" 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\/CP5.24_transcript.html\">transcript for \u201cCP 5.24\u201d here (opens in new window).<\/a><\/center>\r\n<h2 data-type=\"title\">Changing the Order of Integration<\/h2>\r\n<p id=\"fs-id1167793522391\">As we have already seen in double integrals over general bounded regions, changing the order of the integration is done quite often to simplify the computation. With a triple integral over a rectangular box, the order of integration does not change the level of difficulty of the calculation. However, with a triple integral over a general bounded region, choosing an appropriate order of integration can simplify the computation quite a bit. Sometimes making the change to polar coordinates can also be very helpful. We demonstrate two examples here.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: changing the order of integration<\/h3>\r\n<p id=\"fs-id1167794296751\">Consider the iterated integral<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{x=1}_{x=0}} \\ {\\displaystyle\\int^{y=x^2}_{y=0}} \\ {\\displaystyle\\int^{z=y^2}_{z=0}}{f}{(x,y,z)}{dz}{dy}{dx}[\/latex].<\/p>\r\n<p id=\"fs-id1167793544547\">The order of integration here is first with respect to [latex]z[\/latex], then [latex]y[\/latex], and then [latex]x[\/latex]. Express this integral by changing the order of integration to be first with respect to [latex]x[\/latex], then [latex]z[\/latex], and then [latex]y[\/latex]. Verify that the value of the integral is the same if we let\u00a0[latex]f(x, y, z)=xyz[\/latex].<\/p>\r\n[reveal-answer q=\"567723190\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"567723190\"]\r\n<p id=\"fs-id1167793612495\">The best way to do this is to sketch the region [latex]E[\/latex] and its projections onto each of the three coordinate planes. Thus, let<\/p>\r\n<p style=\"text-align: center;\">[latex]{E} = \\left \\{{(x,y,z)}{\\mid}{0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {1,0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {x^2,0} \\ {\\leq} \\ {z} \\ {\\leq} \\ {y^2} \\right \\}[\/latex].<\/p>\r\n<p id=\"fs-id1167793432448\">and<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{x=1}_{x=0}} \\ {\\displaystyle\\int^{y=x^2}_{y=0}} \\ {\\displaystyle\\int^{z=y^2}_{z=0}}{f}{(x,y,z)}{dz}{dy}{dx} = \\underset{E}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV}[\/latex].<\/p>\r\n<p id=\"fs-id1167793424343\">We need to express this triple integral as<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{y=d}_{y=c}} \\ {\\displaystyle\\int^{z={{v_2}{(y)}}}_{z={{v_1}{(y)}}}} \\ {\\displaystyle\\int^{x={{u_2}{(y,z)}}}_{x={{u_1}{(y,z)}}}}{f}{(x,y,z)}{dx}{dz}{dy}[\/latex].<\/p>\r\n<p id=\"fs-id1167793384512\">Knowing the region [latex]E[\/latex] we can draw the following projections (Figure 8):<\/p>\r\n<p id=\"fs-id1167793384524\">on the [latex]xy[\/latex]-plane is [latex]{D_1} = {\\left \\{{(x,y)}{\\mid}{0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {1,0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {x^2} \\right \\}} = {\\left \\{{(x,y)}{\\mid}{0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {1,{\\sqrt{y}}} \\ {\\leq} \\ {x} \\ {\\leq} \\ {1} \\right \\}}[\/latex],<\/p>\r\non the [latex]yz[\/latex]-plane is [latex]{D_2} = {\\left \\{{(y,z)}{\\mid}{0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {1,0} \\ {\\leq} \\ {z} \\ {\\leq} \\ {y^2} \\right \\}}[\/latex], and\r\n\r\non the [latex]xz[\/latex]-plane is [latex]{D_3} = {\\left \\{{(x,z)}{\\mid}{0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {1,0} \\ {\\leq} \\ {z} \\ {\\leq} \\ {x^2} \\right \\}}[\/latex].\r\n\r\n[caption id=\"attachment_1380\" align=\"aligncenter\" width=\"899\"]<img class=\"size-full wp-image-1380\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25144759\/5-4-8.jpeg\" alt=\"Three similar versions of the following graph are shown: In the x y plane, a region D1 is bounded by the x axis, the line x = 1, and the curve y = x squared. In the second version, region D2 on the z y plane is shown with equation z = y squared. And in the third version, region D3 on the x z plane is shown with equation z = x cubed.\" width=\"899\" height=\"273\" \/> Figure 8.\u00a0The three cross sections of\u00a0[latex]E[\/latex] on the three coordinate planes.[\/caption]Now we can describe the same region [latex]E[\/latex] as [latex]{\\left \\{{(x,y,z)}{\\mid}{0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {1,0} \\ {\\leq} \\ {z} \\ {\\leq} \\ {{y^2},{\\sqrt{y}}} \\ {\\leq} \\ {x} \\ {\\leq} \\ {1} \\right \\}}[\/latex], and consequently, the triple integral becomes\r\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{y=d}_{y=c}} \\ {\\displaystyle\\int^{z={{v_2}{(y)}}}_{z={v_1}{(y)}}} \\ {\\displaystyle\\int^{x={{u_2}{(y,z)}}}_{x={u_1}{(y,z)}}}{f}{(x,y,z)}{dx}{dz}{dy}= {\\displaystyle\\int^{y=1}_{y=0}} \\ {\\displaystyle\\int^{z={x^2}}_{z=0}} \\ {\\displaystyle\\int^{x=1}_{x={\\sqrt{y}}}}{f}{(x,y,z)}{dx}{dz}{dy}[\/latex].<\/p>\r\nNow assume that [latex]f(x, y, z)=xyz[\/latex] in each of the integrals. Then we have\r\n\r\n<center>[latex]\\begin{align}<\/center>\\displaystyle\\int^{x=1}_{x=0}\\displaystyle\\int^{y=x^2}_{y=0}\\displaystyle\\int^{z=y^2}_{z=0}xyz{dz}{dy}{dx}&amp;=\\displaystyle\\int^{x=1}_{x=0}\\displaystyle\\int^{y=x^2}_{y=0}\\left[xy\\frac{z^2}2\\bigg|_{z=0}^{z=y^2}\\right]{dy}{dx} \\\\\r\n\r\n&amp;=\\displaystyle\\int^{x=1}_{x=0}\\displaystyle\\int^{y=x^2}_{y=0}\\left(x\\frac{y^5}2\\right)dy \\ dx \\\\\r\n\r\n&amp;=\\displaystyle\\int^{x=1}_{x=0}\\left[x\\frac{y^6}{12}\\bigg|^{y=x^2}_{y=0}\\right]dx \\\\\r\n&amp; =\\displaystyle\\int^{x=1}_{x=0}\\frac{x^{13}}{12}dx =\\frac1{168}, \\\\\r\n\r\n\\displaystyle\\int^{y=1}_{y=0}\\displaystyle\\int^{z=y^2}_{z=0}\\displaystyle\\int^{x=1}_{x=\\sqrt{y}}xyz{dx}{dz}{dy} &amp;=\\displaystyle\\int^{y=1}_{y=0}\\displaystyle\\int^{z=y^2}_{z=0}\\left[yz\\frac{x^2}2\\bigg|^{1}_{\\sqrt{y}}\\right]dz \\ dy\\\\\r\n\r\n&amp;=\\displaystyle\\int^{y=1}_{y=0}\\displaystyle\\int^{z=y^2}_{z=0}\\left(\\frac{yz}2-\\frac{y^2z}2\\right)dz \\ dy \\\\\r\n\r\n&amp;=\\displaystyle\\int^{y=1}_{y=0}\\left[\\frac{yz^2}4-\\frac{y^2z^2}4\\bigg|^{z=y^2}_{z=0}\\right]dy \\\\\r\n\r\n&amp;=\\displaystyle\\int^{y=1}_{y=0}\\left(\\frac{y^5}4-\\frac{y^6}4\\right)dy=\\frac1{168}.\r\n\r\n\\end{align}[\/latex]\r\n<p id=\"fs-id1167793617989\">The answers match.<\/p>\r\n[\/hidden-answer]\r\n\r\n<section data-depth=\"1\"><\/section><\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nWrite five different iterated integrals equal to the given integral\u00a0[latex]{\\displaystyle\\int^{z=4}_{z=0}} \\ {\\displaystyle\\int^{y={4-z}}_{y=0}} \\ {\\displaystyle\\int^{x={\\sqrt{y}}}_{x=0}}{f}{(x,y,z)}{dx}{dy}{dz}[\/latex].\r\n\r\n[reveal-answer q=\"297541139\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"297541139\"]\r\n\r\n(i)\u00a0[latex]\\displaystyle\\int_{z=0}^{z=4}\\displaystyle\\int_{x=0}^{x=\\sqrt{x-4}}\\displaystyle\\int_{y=x^2}^{y=4-z}f(x,y,z)dy \\ dx \\ dz[\/latex],\r\n\r\n(ii)[latex] \\displaystyle\\int_{y=0}^{y=4}\\displaystyle\\int_{z=0}^{z=4-y}\\displaystyle\\int_{x=0}^{x=\\sqrt{y}}f(x,y,z)dx \\ dz \\ dy[\/latex],\r\n\r\n(iii)[latex] \\displaystyle\\int_{y=0}^{y=4}\\displaystyle\\int_{x=0}^{x=\\sqrt{y}}\\displaystyle\\int_{z=0}^{z=4-y}f(x,y,z)dz \\ dx \\ dy[\/latex]\u00a0,\r\n\r\n(iv)\u00a0[latex]\\displaystyle\\int_{x=0}^{x=2}\\displaystyle\\int_{y=x^2}^{y=4}\\displaystyle\\int_{z=0}^{z=4-y}f(x,y,z)dz \\ dy \\ dx[\/latex],\r\n\r\n(v)\u00a0[latex]\\displaystyle\\int_{x=0}^{x=2}\\displaystyle\\int_{z=0}^{z=4-x^2}\\displaystyle\\int_{y=x^2}^{y=4-y}f(x,y,z)dy \\ dz \\ dx[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: changing integration order and coordinate systems<\/h3>\r\nEvaluate the triple integral [latex]\\underset{E}{\\displaystyle\\iiint}{\\sqrt{{x^2}+{z^2}}}{dV}[\/latex], where [latex]E[\/latex] is the region bounded by the paraboloid [latex]y=x^{2}+z^{2}[\/latex] (Figure 9) and the plane [latex]y=4[\/latex].\r\n\r\n[caption id=\"attachment_1382\" align=\"aligncenter\" width=\"577\"]<img class=\"size-full wp-image-1382\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25144936\/5-4-9.jpeg\" alt=\"The paraboloid y = x squared + z squared is shown opening up along the y axis to y = 4.\" width=\"577\" height=\"327\" \/> Figure 9.\u00a0Integrating a triple integral over a paraboloid.[\/caption]\r\n\r\n[reveal-answer q=\"843152286\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"843152286\"]\r\n\r\nThe projection of the solid region [latex]E[\/latex] onto the [latex]xy[\/latex]-plane is the region bounded above by [latex]y=4[\/latex] and below by the parabola [latex]y=x^{2}[\/latex] as shown.\r\n\r\n[caption id=\"attachment_1383\" align=\"aligncenter\" width=\"417\"]<img class=\"size-full wp-image-1383\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25145013\/5-4-10.jpeg\" alt=\"In the x y plane, the graph of y = x squared is shown with the line y = 4 intersecting the graph at (negative 2, 4) and (2, 4).\" width=\"417\" height=\"272\" \/> Figure 10.\u00a0Cross section in the [latex]xy[\/latex]-plane of the paraboloid in Figure 9.[\/caption]\r\n<p id=\"fs-id1167793285294\">Thus, we have<\/p>\r\n<p style=\"text-align: center;\">[latex]{E} = {\\left \\{ {(x,y,z)}{\\mid}{-2} \\ {\\leq} \\ {x} \\ {\\leq} \\ {2,{x^2}} \\ {\\leq} \\ {y} \\ {\\leq} \\ {{4},-{\\sqrt{y-x^2}}} \\ {\\leq} \\ {z} \\ {\\leq} \\ {\\sqrt{y-x^2}} \\right \\}}[\/latex].<\/p>\r\n<p id=\"fs-id1167794200516\">The triple integral becomes<\/p>\r\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{\\sqrt{x^2+z^2}dV} = {\\displaystyle\\int^{x=2}_{x=-2}} \\ {\\displaystyle\\int^{y=4}_{y=x^2}} \\ {\\displaystyle\\int^{z={\\sqrt{y-x^2}}}_{z=-{\\sqrt{y-x^2}}}}{\\sqrt{{x^2}+{z^2}}}{dz}{dy}{dx}[\/latex].<\/p>\r\n<p id=\"fs-id1167793607349\">This expression is difficult to compute, so consider the projection of [latex]E[\/latex] onto the [latex]xz[\/latex]-plane. This is a circular disc [latex]{x^2} + {z^2} \\ {\\leq} \\ 4[\/latex]. So we obtain<\/p>\r\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{\\sqrt{x^2+z^2}dV} = {\\displaystyle\\int^{x=2}_{x=-2}} \\ {\\displaystyle\\int^{y=4}_{y=x^2}} \\ {\\displaystyle\\int^{z={\\sqrt{y-x^2}}}_{z=-{\\sqrt{y-x^2}}}}{\\sqrt{{x^2}+{z^2}}}{dz}{dy}{dx} = {\\displaystyle\\int^{x=2}_{x=-2}} \\ {\\displaystyle\\int^{z={\\sqrt{4-x^2}}}_{z=-{\\sqrt{4-x^2}}}} \\ {\\displaystyle\\int^{y=4}_{y=x^2+z^2}}{\\sqrt{{x^2}+{z^2}}}{dy}{dz}{dx}[\/latex].<\/p>\r\n<p id=\"fs-id1167794246456\">Here the order of integration changes from being first with respect to [latex]z[\/latex], then [latex]y[\/latex] and then [latex]x[\/latex] to being first with respect to [latex]y[\/latex], then to [latex]z[\/latex], and then to [latex]x[\/latex]. It will soon be clear how this change can be beneficial for computation. We have<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{x=2}_{x=-2}} \\ {\\displaystyle\\int^{z={\\sqrt{4-x^2}}}_{z=-{\\sqrt{4-x^2}}}} \\ {\\displaystyle\\int^{y=4}_{y=x^2+z^2}}{\\sqrt{{x^2}+{z^2}}}{dy}{dz}{dx} = {\\displaystyle\\int^{x=2}_{x=-2}} \\ {\\displaystyle\\int^{z={\\sqrt{4-x^2}}}_{z=-{\\sqrt{4-x^2}}}}{(4-x^2-z^2)}{\\sqrt{{x^2}+{z^2}}}{dz}{dx}[\/latex].<\/p>\r\n<p id=\"fs-id1167793485217\">Now use the polar substitution [latex]{{x} = {r}{\\cos}{\\theta}},{{z} = {r}{\\sin}{\\theta}}[\/latex], and [latex]{dz}{dx} = {r}{dr}{{d}{\\theta}}[\/latex] in the [latex]xz[\/latex]-plane. This is essentially the same thing as when we used polar coordinates in the [latex]xy[\/latex]-plane, except we are replacing [latex]y[\/latex] by [latex]z[\/latex]. Consequently the limits of integration change and we have, by using [latex]r^{2}=x^{2}+z^{2}[\/latex],<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{x=2}_{x=-2}} \\ {\\displaystyle\\int^{z={\\sqrt{4-x^2}}}_{z=-{\\sqrt{4-x^2}}}}{(4-x^2-z^2)}{\\sqrt{{x^2}+{z^2}}}{dz}{dx} = {\\displaystyle\\int^{{\\theta} = 2{\\pi}}_{{\\theta} = 0}} \\ {\\displaystyle\\int^{r=2}_{r=0}}{(4-{r^2})}{rr}{dr}{{d}{\\theta}}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]= {\\displaystyle\\int^{2{\\pi}}_{0}} \\ {\\left [ {\\frac{4r^3}{3}}-{\\frac{r^5}{5}}{{\\bigg\\vert}^2_0} \\right ]} \\ {{d}{\\theta}} = {\\displaystyle\\int^{2{\\pi}}_{0}} \\ {\\frac{64}{15}}{{d}{\\theta}} = {\\frac{{128}{\\pi}}{15}}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\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\">Recognize when a function of three variables is integrable over a rectangular box.<\/span><\/li>\n<li><span class=\"os-abstract-content\">Evaluate a triple integral by expressing it as an iterated integral.<\/span><\/li>\n<li><span class=\"os-abstract-content\">Recognize when a function of three variables is integrable over a closed and bounded region.<\/span><\/li>\n<li><span class=\"os-abstract-content\">Simplify a calculation by changing the order of integration of a triple integral.<\/span><\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-id1167793878911\" data-depth=\"1\">\n<h2 data-type=\"title\">Integrable Functions of Three Variables<\/h2>\n<p id=\"fs-id1167794064648\">We can define a rectangular box [latex]B[\/latex] in [latex]{{\\mathbb{R}}^3}[\/latex] as [latex]{B} = {\\left \\{{(x,y,z)}{\\mid}{a} \\ {\\leq} \\ {x} \\ {\\leq} \\ {b,c} \\ {\\leq} \\ {y} \\ {\\leq} \\ {d,e} \\ {\\leq} \\ {z} \\ {\\leq} \\ {f} \\right \\}}[\/latex]. We follow a similar procedure to what we did in\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/double-integrals-over-rectangular-regions\/\" data-page-slug=\"5-1-double-integrals-over-rectangular-regions\" data-page-uuid=\"d9da6d44-3ba3-4a90-adbc-dec26572e480\" data-page-fragment=\"page_d9da6d44-3ba3-4a90-adbc-dec26572e480\">Double Integrals over Rectangular Regions<\/a>. We divide the interval [latex][a,b][\/latex] into [latex]l[\/latex] subintervals [latex][{{x}_{i-1}},{x_i}][\/latex] of equal length [latex]{{\\Delta}{x}} = {\\frac{{x_i}-{x_{i-1}}}{l}}[\/latex], divide the interval [latex][c,d][\/latex] into [latex]m[\/latex] subintervals [latex][{{y}_{i-1}},{y_i}][\/latex] of equal length [latex]{{\\Delta}{y}} = {\\frac{{y_j}-{y_{j-1}}}{m}}[\/latex], and divide the interval [latex][e,f][\/latex] into [latex]n[\/latex] subintervals [latex][{{z}_{i-1}},{z_i}][\/latex] of equal length [latex]{{\\Delta}{z}} = {\\frac{{z_k}-{z_{k-1}}}{n}}[\/latex]. Then the rectangular box [latex]B[\/latex] is subdivided into [latex]lmn[\/latex] subboxes [latex]{{B}_{ijk}} = [{{x}_{i-1}},{x_i}] \\ {\\times} \\ [{{y}_{i-1}},{y_i}] \\ {\\times} \\ [{{z}_{i-1}},{z_i}][\/latex], as shown in\u00a0Figure 1.<\/p>\n<div id=\"attachment_1372\" style=\"width: 600px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1372\" class=\"size-full wp-image-1372\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25144228\/5-4-1.jpeg\" alt=\"In x y z space, there is a box B with a subbox Bijk with sides of length Delta x, Delta y, and Delta z.\" width=\"590\" height=\"724\" \/><\/p>\n<p id=\"caption-attachment-1372\" class=\"wp-caption-text\">Figure 1.\u00a0A rectangular box in\u00a0[latex]\\mathbb{R}^{3}[\/latex] divided into subboxes by planes parallel to the coordinate planes.<\/p>\n<\/div>\n<p id=\"fs-id1167794051785\">For each [latex]i[\/latex], [latex]j[\/latex], and [latex]k[\/latex], consider a sample point [latex]{({{x}^{*}_{ijk}},{{y}^{*}_{ijk}},{{z}^{*}_{ijk}})}[\/latex] in each sub-box [latex]{{B}_{ijk}}[\/latex]. We see that its volume is [latex]{{\\Delta}{V}} = {{\\Delta}{x}}{{\\Delta}{y}}{{\\Delta}{z}}[\/latex]. Form the triple Riemann sum<\/p>\n<p style=\"text-align: center;\">[latex]\\Large{\\displaystyle\\sum_{i=1}^{l}}{\\displaystyle\\sum_{j=1}^{m}}{\\displaystyle\\sum_{k=1}^{n}}{f}{({{x}^{*}_{ijk}},{{y}^{*}_{ijk}},{{z}^{*}_{ijk}})}{{\\Delta}{x}}{{\\Delta}{y}}{{\\Delta}{z}}[\/latex].<\/p>\n<p>We define the triple integral in terms of the limit of a triple Riemann sum, as we did for the double integral in terms of a double Riemann sum.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">definition<\/h3>\n<hr \/>\n<p id=\"fs-id1167793943883\">The\u00a0<strong><span id=\"d140d1cd-af53-466a-8e69-402de1b91259_term219\" data-type=\"term\">triple integral<\/span><\/strong>\u00a0of a function [latex]f(x, y, z)[\/latex] over a rectangular box [latex]B[\/latex] is defined as<\/p>\n<p style=\"text-align: center;\">[latex]\\Large{\\displaystyle\\lim_{{l,m,n}{\\rightarrow}{\\infty}}} \\ {\\displaystyle\\sum_{i=1}^{l}}{\\displaystyle\\sum_{j=1}^{m}}{\\displaystyle\\sum_{k=1}^{n}} \\ {f}{({{x}^{*}_{ijk}},{{y}^{*}_{ijk}},{{z}^{*}_{ijk}})}{{\\Delta}{x}}{{\\Delta}{y}}{{\\Delta}{z}} = \\underset{B}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV}[\/latex]<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">if this limit exists.<\/span><\/p>\n<\/div>\n<p id=\"fs-id1167793449708\">When the triple integral exists on [latex]B[\/latex], the function [latex]f(x, y, z)[\/latex] is said to be integrable on [latex]B[\/latex]. Also, the triple integral exists if [latex]f(x, y, z)[\/latex] is continuous on [latex]B[\/latex]. Therefore, we will use continuous functions for our examples. However, continuity is sufficient but not necessary; in other words, [latex]f[\/latex] is bounded on [latex]B[\/latex] and continuous except possibly on the boundary of [latex]B[\/latex] The sample point [latex]{({{x}^{*}_{ijk}},{{y}^{*}_{ijk}},{{z}^{*}_{ijk}})}[\/latex] can be any point in the rectangular sub-box [latex]{B}_{ijk}[\/latex] and all the properties of a double integral apply to a triple integral. Just as the double integral has many practical applications, the triple integral also has many applications, which we discuss in later sections.<\/p>\n<p id=\"fs-id1167794100272\">Now that we have developed the concept of the triple integral, we need to know how to compute it. Just as in the case of the double integral, we can have an iterated triple integral, and consequently, a version of\u00a0<span id=\"d140d1cd-af53-466a-8e69-402de1b91259_term220\" class=\"no-emphasis\" data-type=\"term\">Fubini\u2019s thereom<\/span>\u00a0for triple integrals exists.<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">theorem: fubini&#8217;s theorem for triple integrals<\/h3>\n<hr \/>\n<p id=\"fs-id1167794145580\">If [latex]f(x, y, z)[\/latex] is continuous on a rectangular box [latex]{B} = {[a,b]} \\ {\\times} \\ {[c,d]} \\ {\\times} \\ {[e,f]}[\/latex], then<\/p>\n<p style=\"text-align: center;\">[latex]\\Large{\\underset{B}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV} = {\\displaystyle\\int_{e}^{f}}{\\displaystyle\\int_{c}^{d}}{\\displaystyle\\int_{a}^{b}}{f}{(x,y,z)}{dx} \\ {dy} \\ {dz}}[\/latex].<\/p>\n<p>This integral is also equal to any of the other five possible orderings for the iterated triple integral.<\/p>\n<\/div>\n<p id=\"fs-id1167793263456\">For [latex]a[\/latex], [latex]b[\/latex], [latex]c[\/latex], [latex]d[\/latex], [latex]e[\/latex], and [latex]f[\/latex] real numbers, the iterated triple integral can be expressed in six different orderings:<\/p>\n<p>[latex]\\large{\\begin{align}    {\\displaystyle\\int_{e}^{f}}{\\displaystyle\\int_{c}^{d}}{\\displaystyle\\int_{a}^{b}}{f}{(x,y,z)}{dx} \\ {dy} \\ {dz}&={\\displaystyle\\int_{e}^{f}}\\left({\\displaystyle\\int_{c}^{d}}\\left({\\displaystyle\\int_{a}^{b}}{f}{(x,y,z)}{dx}\\right){dy}\\right){dz}={\\displaystyle\\int_{c}^{d}}\\left({\\displaystyle\\int_{e}^{f}}\\left({\\displaystyle\\int_{a}^{b}}{f}{(x,y,z)}{dx}\\right){dz}\\right){dy} \\\\    &={\\displaystyle\\int_{a}^{b}}\\left({\\displaystyle\\int_{e}^{f}}\\left({\\displaystyle\\int_{c}^{d}}{f}{(x,y,z)}{dy}\\right){dz}\\right){dx}={\\displaystyle\\int_{e}^{f}}\\left({\\displaystyle\\int_{a}^{b}}\\left({\\displaystyle\\int_{c}^{d}}{f}{(x,y,z)}{dy}\\right){dx}\\right){dz} \\\\    &={\\displaystyle\\int_{c}^{e}}\\left({\\displaystyle\\int_{a}^{b}}\\left({\\displaystyle\\int_{e}^{f}}{f}{(x,y,z)}{dz}\\right){dx}\\right){dy}={\\displaystyle\\int_{a}^{b}}\\left({\\displaystyle\\int_{c}^{e}}\\left({\\displaystyle\\int_{e}^{f}}{f}{(x,y,z)}{dz}\\right){dy}\\right){dx}.    \\end{align}}[\/latex]<br \/>\nFor a rectangular box, the order of integration does not make any significant difference in the level of difficulty in computation. We compute triple integrals using Fubini\u2019s Theorem rather than using the Riemann sum definition. We follow the order of integration in the same way as we did for double integrals (that is, from inside to outside).<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: evaluating a triple integral 1<\/h3>\n<p>Evaluate the triple integral [latex]{\\displaystyle\\int^{z=1}_{z=0}}{\\displaystyle\\int^{y=4}_{y=2}}{\\displaystyle\\int^{x=5}_{x=-1}}{(x+yz^2)}{dx} \\ {dy} \\ {dz}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q390606\">Show Solution<\/span><\/p>\n<div id=\"q390606\" class=\"hidden-answer\" style=\"display: none\">\n<p>The order of integration is specified in the problem, so integrate with respect to x first, then y, and then z.<\/p>\n<p>[latex]\\hspace{3cm}\\begin{align}    &\\hspace{0.75cm}\\displaystyle\\int^{z=1}_{z=0}\\displaystyle\\int^{y=4}_{y=2}\\displaystyle\\int^{x=5}_{x=-1}(x+yz^2)dx \\ dy \\ dz \\\\    &=\\displaystyle\\int^{z=1}_{z=0}\\displaystyle\\int^{y=4}_{y=2}\\left[\\frac{x^2}2+xyz^2\\bigg|^{x=5}_{x=-1}\\right]dy \\ dz &\\quad &\\text{Integrate with respect to }x. \\\\    &=\\displaystyle\\int^{z=1}_{z=0}\\displaystyle\\int^{y=4}_{y=2}[12+6yz^2] \\ dy \\ dz &\\quad &\\text{Evaluate.} \\\\    &=\\displaystyle\\int^{z=1}_{z=0}\\left[12y+6\\frac{y^2}2x^2\\bigg|^{y=4}_{y=2}\\right]{dz} &\\quad &\\text{Integrate with respect to }y. \\\\    &={\\displaystyle\\int^{z=1}_{z=0}}[24+36z^2]{dz} &\\quad &\\text{Evaluate}. \\\\    &=\\left[24z+36\\frac{z^3}3\\right]_{z=0}^{z=1}=36.    \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: evaluating a triple integral 2<\/h3>\n<p>Evaluate the triple integral [latex]\\underset{B}{\\displaystyle\\iiint}{x^2}{yz}{dV}[\/latex] where<\/p>\n<p>[latex]{B} = {\\left \\{{(x,y,z)}{\\mid}{-2} \\ {\\leq} \\ {x} \\ {\\leq} \\ {1,0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {3,1} \\ {\\leq} \\ {z} \\ {\\leq} \\ {5} \\right \\}}[\/latex] as shown in the following figure.<\/p>\n<div id=\"attachment_1374\" style=\"width: 385px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1374\" class=\"size-full wp-image-1374\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25144322\/5-4-2.jpeg\" alt=\"In x y z space, there is a box given with corners (1, 0, 5), (1, 0, 1), (1, 3, 1), (1, 3, 5), (negative 2, 0, 5), (negative 2, 0, 1), (negative 2, 3, 1), and (negative 2, 3, 5).\" width=\"375\" height=\"328\" \/><\/p>\n<p id=\"caption-attachment-1374\" class=\"wp-caption-text\">Figure 2.\u00a0Evaluating a triple integral over a given rectangular box.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q245801336\">Show Solution<\/span><\/p>\n<div id=\"q245801336\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793937669\">The order is not specified, but we can use the iterated integral in any order without changing the level of difficulty. Choose, say, to integrate [latex]y[\/latex] first, then\u00a0[latex]x[\/latex], and then\u00a0[latex]z[\/latex].<\/p>\n<p>[latex]\\hspace{2cm}\\begin{align}    \\underset{B}{\\displaystyle\\iiint}{x^2}{yz}{dV}&=\\displaystyle\\int_1^5\\displaystyle\\int_{-2}^1\\displaystyle\\int_0^3[x^2yz]dy \\ dx \\ dz =\\displaystyle\\int_1^5\\displaystyle\\int_{-2}^1\\left[x^2\\frac{y^2}2z\\bigg|_0^3\\right]dx \\ dz \\\\    &=\\displaystyle\\int_1^5\\displaystyle\\int_{-2}^1\\frac92x^2z \\ dx \\ dz = \\displaystyle\\int_1^5\\left[\\frac92\\frac{x^3}3z\\bigg|_{-2}^1\\right]dz=\\displaystyle\\int_1^5\\frac{27}2\\frac{z^2}2\\bigg|_1^5=162.    \\end{align}[\/latex]<\/p>\n<p>Now try to integrate in a different order just to see that we get the same answer. Choose to integrate with respect to [latex]x[\/latex] first, then [latex]z[\/latex], and then [latex]y[\/latex].<\/p>\n<p>[latex]\\hspace{2cm}\\begin{align}    \\underset{B}{\\displaystyle\\iiint}{x^2}{yz}{dV}&=\\displaystyle\\int_0^3\\displaystyle\\int_1^5\\displaystyle\\int_{-2}^1[x^2yz]dx \\ dz \\ dy =\\displaystyle\\int_0^3\\displaystyle\\int_1^5\\left[\\frac{x^3}3yz\\bigg|_{-2}^1\\right]dz \\ dy \\\\    &=\\displaystyle\\int_0^3\\displaystyle\\int_1^53yz \\ dz \\ dy =\\displaystyle\\int_0^3\\left[3y\\frac{z^2}2\\bigg|_1^5\\right]dy=\\displaystyle\\int_0^336y \\ dy = 36\\frac{y^2}2\\bigg|_0^3=18(9-0)=162.    \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Evaluate the triple integral [latex]\\underset{B}{\\displaystyle\\iiint}{z} \\ {\\sin \\ x} \\ {\\cos \\ y} \\ {dV}[\/latex] where<\/p>\n<p>[latex]{B} = {\\left \\{{(x,y,z)}{\\mid}{0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {{\\pi},{\\frac{3{\\pi}}{2}}} \\ {\\leq} \\ {y} \\ {\\leq} \\ {2{\\pi},1} \\ {\\leq} \\ {z} \\ {\\leq} \\ {3} \\right \\}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q159035786\">Show Solution<\/span><\/p>\n<div id=\"q159035786\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\underset{B}{\\displaystyle\\iiint}z\\sin{x}\\cos{y} \\ dV=8[\/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=8197105&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=csY8RONL3Qs&amp;video_target=tpm-plugin-46801eo2-csY8RONL3Qs\" 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\/CP5.23_transcript.html\">transcript for \u201cCP 5.23\u201d here (opens in new window).<\/a><\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm184576\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=184576&theme=oea&iframe_resize_id=ohm184576&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2 data-type=\"title\">Triple Integrals over a General Bounded Region<\/h2>\n<p id=\"fs-id1167794214296\">We now expand the definition of the triple integral to compute a triple integral over a more\u00a0<span id=\"d140d1cd-af53-466a-8e69-402de1b91259_term221\" class=\"no-emphasis\" data-type=\"term\">general bounded region [latex]E[\/latex] in\u00a0[latex]{\\mathbb{R}^3}[\/latex].\u00a0<\/span>The general bounded regions we will consider are of three types. First, let [latex]D[\/latex] be the bounded region that is a projection of [latex]E[\/latex] onto the [latex]xy[\/latex]-plane. Suppose the region [latex]E[\/latex] in [latex]{\\mathbb{R}^3}[\/latex] has the form<\/p>\n<p style=\"text-align: center;\">[latex]\\Large{E} = \\left \\{{(x,y,z)}{\\mid}{(x,y)} \\ {\\in} \\ {{D},{u_1}{(x,y)}} \\ {\\leq} \\ {z} \\ {\\leq} \\ {u_2}{(x,y)} \\right \\}[\/latex].<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">For two functions [latex]z=u_1(x, y)[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">and [latex]z=u_2(x,y)[\/latex],\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">such that [latex]{u_1}{(x,y)} \\ {\\leq} \\ {u_2}{(x,y)}[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">for all [latex](x, y)[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">in [latex]D[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">as shown in the following figure.<\/span><\/p>\n<div id=\"attachment_1375\" style=\"width: 439px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1375\" class=\"size-full wp-image-1375\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25144346\/5-4-3.jpeg\" alt=\"In x y z space, there is a shape E with top surface z = u2(x, y) and bottom surface z = u1(x, y). The bottom projects onto the x y plane as region D.\" width=\"429\" height=\"348\" \/><\/p>\n<p id=\"caption-attachment-1375\" class=\"wp-caption-text\">Figure 3. We can describe region [latex]E[\/latex] as the space between\u00a0[latex]u_{1}(x,y)[\/latex] and\u00a0[latex]u_{2}(x,y)[\/latex] above the projection\u00a0[latex]D[\/latex] of\u00a0[latex]E[\/latex] onto the\u00a0[latex]xy[\/latex]-plane.<\/p>\n<\/div>\n<\/section>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">theorem: triple integral over a general region<\/h3>\n<hr \/>\n<p id=\"fs-id1167793632073\">The triple integral of a continuous function [latex]f(x, y, z)[\/latex] over a general three-dimensional region<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]\\large{E} = \\left \\{{(x,y,z)}{\\mid}{(x,y)} \\ {\\in} \\ {{D},{u_1}{(x,y)}} \\ {\\leq} \\ {z} \\ {\\leq} \\ {u_2}{(x,y)} \\right \\}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">in [latex]{{\\mathbb{R}}^3}[\/latex],\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">where [latex]D[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">is the projection of [latex]E[\/latex]\u00a0<\/span><span style=\"font-size: 1rem; text-align: initial;\">onto the [latex]xy[\/latex]&#8211;<\/span><span style=\"font-size: 1rem; text-align: initial;\">plane, is<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\underset{E}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV} = \\underset{D}{\\displaystyle\\iint}{\\left [{\\displaystyle\\int^{{u_2}{(x,y)}}_{{u_1}{(x,y)}}}{f}{(x,y,z)}{dz} \\right ]}{dA}}[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1167793895376\">Similarly, we can consider a general bounded region [latex]D[\/latex] in the [latex]xy[\/latex]-plane and two functions [latex]{y} = {u_1}{(x,z)}[\/latex] and [latex]{y} = {u_2}{(x,z)}[\/latex] such that [latex]{u_1}{(x,z)} \\ {\\leq} \\ {u_2}{(x,z)}[\/latex] for all [latex](x, z)[\/latex] in [latex]D[\/latex]. Then we can describe the solid region [latex]E[\/latex] in [latex]{{\\mathbb{R}}^3}[\/latex] as<\/p>\n<p style=\"text-align: center;\">[latex]{E} = \\left \\{{(x,y,z)}{\\mid}{(x,z)} \\ {\\in} \\ {{D},{u_1}{(x,z)}} \\ {\\leq} \\ {z} \\ {\\leq} \\ {u_2}{(x,z)} \\right \\}[\/latex]<\/p>\n<p>where [latex]D[\/latex] is the projection of [latex]E[\/latex] onto the [latex]xy[\/latex]-plane and the triple integral is<\/p>\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV} = \\underset{D}{\\displaystyle\\iint}{\\left [{\\displaystyle\\int^{{u_2}{(x,z)}}_{{u_1}{(x,z)}}}{f}{(x,y,z)}{dy} \\right ]}{dA}[\/latex].<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">Finally, if [latex]D[\/latex] is a general bounded region in the [latex]yz[\/latex]-plane and we have two functions [latex]{x} = {u_1}{(y,z)}[\/latex] and [latex]{x} = {u_2}{(y,z)}[\/latex] such that [latex]{u_1}{(y,z)} \\ {\\leq} \\ {u_2}{(y,z)}[\/latex] for all [latex](y, z)[\/latex] in [latex]D[\/latex], then the solid region [latex]E[\/latex] in [latex]{{\\mathbb{R}}^3}[\/latex] can be described as<\/span><\/p>\n<p style=\"text-align: center;\">[latex]{E} = \\left \\{{(x,y,z)}{\\mid}{(y,z)} \\ {\\in} \\ {{D},{u_1}{(y,z)}} \\ {\\leq} \\ {z} \\ {\\leq} \\ {u_2}{(y,z)} \\right \\}[\/latex]<\/p>\n<p>where [latex]D[\/latex] is the projection of [latex]E[\/latex] onto the [latex]yz[\/latex]-plane and the triple integral is<\/p>\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV} = \\underset{D}{\\displaystyle\\iint}{\\left [{\\displaystyle\\int^{{u_2}{(y,z)}}_{{u_1}{(y,z)}}}{f}{(x,y,z)}{dx} \\right ]}{dA}[\/latex].<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">Note that the region [latex]D[\/latex] in any of the planes may be of Type I or Type II as described in\u00a0<\/span><a style=\"font-size: 1rem; text-align: initial;\" href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/introduction-to-double-integrals-over-general-regions\/\" data-page-slug=\"5-2-double-integrals-over-general-regions\" data-page-uuid=\"f783ca08-ca2f-4ebb-a559-d1b30c2f9b8c\" data-page-fragment=\"page_f783ca08-ca2f-4ebb-a559-d1b30c2f9b8c\">Double Integrals over General Regions<\/a><span style=\"font-size: 1rem; text-align: initial;\">. If [latex]D[\/latex] in the [latex]xy[\/latex]-plane is of Type I (<\/span>Figure 4<span style=\"font-size: 1rem; text-align: initial;\">), then<\/span><\/p>\n<p style=\"text-align: center;\">[latex]{E} = \\left \\{{(x,y,z)}{\\mid}{a} \\ {\\leq} \\ {x} \\ {\\leq} \\ {b,{g_1}{(x)}} \\ {\\leq} \\ {y} \\ {\\leq} \\ {g_2{(x)}},{{u_1}{(x,y)}} \\ {\\leq} \\ {z} \\ {\\leq} \\ {u_2}{(x,y)} \\right \\}[\/latex].<\/p>\n<section data-depth=\"1\"><\/section>\n<section data-depth=\"1\">\n<div id=\"attachment_1376\" style=\"width: 581px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1376\" class=\"wp-image-1376 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25144428\/5-4-4.jpeg\" alt=\"In x y z space, there is a complex shape E with top surface z = u2(x, y) and bottom surface z = u1(x, y). The bottom projects onto the xy plane as region D with boundaries x = a, x = b, y = g1(x), and y = g2(x).\" width=\"571\" height=\"443\" \/><\/p>\n<p id=\"caption-attachment-1376\" class=\"wp-caption-text\">Figure 4. A box\u00a0[latex]E[\/latex] where the projection\u00a0[latex]D[\/latex] in the\u00a0[latex]xy[\/latex]-plane is of Type I.<\/p>\n<\/div>\n<p>Then the triple integral becomes<\/p>\n<p style=\"text-align: center;\"><span style=\"font-size: 1em;\">[latex]\\underset{E}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV} = {\\displaystyle\\int_{a}^{b}} \\ {\\displaystyle\\int_{{g_1}{(x)}}^{{g_2}{(x)}}} \\ {\\displaystyle\\int_{{u_1}{(x,y)}}^{{u_2}{(x,y)}}}{f}{(x,y,z)}{dz}{dy}{dx}[\/latex].<\/span><\/p>\n<p style=\"text-align: left;\"><span style=\"font-size: 1em;\">If\u00a0[latex]D[\/latex] in the\u00a0[latex]xy[\/latex]-plane is of Type II (Figure 5), then<\/span><\/p>\n<p style=\"text-align: center;\"><span style=\"font-size: 1rem; text-align: initial;\">[latex]{E} = \\left \\{{(x,y,z)}{\\mid}{c} \\ {\\leq} \\ {x} \\ {\\leq} \\ {d,{h_1}{(x)}} \\ {\\leq} \\ {y} \\ {\\leq} \\ {h_2{(x)}},{{u_1}{(x,y)}} \\ {\\leq} \\ {z} \\ {\\leq} \\ {u_2}{(x,y)} \\right \\}[\/latex].<\/span><\/p>\n<\/section>\n<div id=\"attachment_1377\" style=\"width: 454px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1377\" class=\"size-full wp-image-1377\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25144556\/5-4-5.jpeg\" alt=\"In x y z space, there is a complex shape E with top surface z = u2(x, y) and bottom surface z = u1(x, y). The bottom projects onto the xy plane as region D with boundaries y = c, y = d, x = h1(y), and x = h2(y).\" width=\"444\" height=\"397\" \/><\/p>\n<p id=\"caption-attachment-1377\" class=\"wp-caption-text\">Figure 5.\u00a0A box\u00a0[latex]E[\/latex] where the projection\u00a0[latex]D[\/latex] in the\u00a0[latex]xy[\/latex]-plane is of Type II.<\/p>\n<\/div>\n<p>Then the triple integral becomes<\/p>\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV} = {\\displaystyle\\int^{y=d}_{y=c}}{\\displaystyle\\int^{x={{h_2}{(y)}}}_{x={h_1}{(y)}}}{\\displaystyle\\int^{z={{u_2}{(x,y)}}}_{z={u_1}{(x,y)}}}{f}{(x,y,z)}{dz}{dx}{dy}[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: evaluating a triple integral over a general bounded region<\/h3>\n<p>Evaluate the triple integral of the function [latex]f(x, y, z)=5x-3y[\/latex] over the solid tetrahedron bounded by the planes\u00a0[latex]x=0, \\ y=0, \\ z=0,[\/latex] and\u00a0[latex]x+y+z=1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q331457700\">Show Solution<\/span><\/p>\n<div id=\"q331457700\" class=\"hidden-answer\" style=\"display: none\">\n<p>Figure 6\u00a0shows the solid tetrahedron [latex]E[\/latex] and its projection [latex]D[\/latex] on the [latex]xy[\/latex]-plane.<\/p>\n<div id=\"attachment_1378\" style=\"width: 695px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1378\" class=\"size-full wp-image-1378\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25144633\/5-4-6.jpeg\" alt=\"In x y z space, there is a solid E with boundaries being the x y, z y, and x z planes and z = 1 minus x minus y. The points are the origin, (1, 0, 0), (0, 0, 1), and (0, 1, 0). Its surface on the x y plane is shown as being a rectangle marked D with line y = 1 minus x. Additionally, there is a vertical line shown on D.\" width=\"685\" height=\"357\" \/><\/p>\n<p id=\"caption-attachment-1378\" class=\"wp-caption-text\">Figure 6.\u00a0The solid [latex]E[\/latex] has a projection\u00a0[latex]D[\/latex] on the\u00a0[latex]xy[\/latex]-plane of Type I.<\/p>\n<\/div>\n<p id=\"fs-id1167794051980\">We can describe the solid region tetrahedron as<\/p>\n<p style=\"text-align: center;\">[latex]{E} = \\left \\{{(x,y,z)}{\\mid}{0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {1,0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {1-x},{0} \\ {\\leq} \\ {z} \\ {\\leq} \\ {1-x-y} \\right \\}[\/latex].<\/p>\n<p id=\"fs-id1167793887543\">Hence, the triple integral is<\/p>\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV} = {\\displaystyle\\int^{x=1}_{x=0}}{\\displaystyle\\int^{y={1-x}}_{y=0}}{\\displaystyle\\int^{z={1-x-y}}_{z=0}}{(5x-3y)}{dz}{dy}{dx}[\/latex].<\/p>\n<p id=\"fs-id1167793423320\">To simplify the calculation, first evaluate the integral [latex]{\\displaystyle\\int^{z=1-x-y}_{z=0}}{(5x-3y)}{dz}[\/latex]. We have<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{z=1-x-y}_{z=0}}{(5x-3y)}{dz} = {(5x-3y)}{(1-x-y)}[\/latex].<\/p>\n<p id=\"fs-id1167793421147\">Now evaluate the integral [latex]{\\displaystyle\\int^{y=1-x}_{y=0}}{(5x-3y)}{(1-x-y)}{dy}[\/latex], obtaining<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{y=1-x}_{y=0}}{(5x-3y)}{(1-x-y)}{dy} = {\\frac{1}{2}}{(x-1)^2}{(6x-1)}[\/latex].<\/p>\n<p id=\"fs-id1167794045783\">Finally, evaluate<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{x=1}_{x=0}}{\\frac{1}{2}}{(x-1)^2}{(6x-1)}{dx} = {\\frac{1}{12}}[\/latex].<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">Putting it all together, we have<\/span><\/p>\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV} = {\\displaystyle\\int^{x=1}_{x=0}}{\\displaystyle\\int^{y={1-x}}_{y=0}}{\\displaystyle\\int^{z={1-x-y}}_{z=0}}{(5x-3y)}{dz}{dy}{dx} = {\\frac{1}{12}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Just as we used the double integral [latex]\\underset{D}{\\displaystyle\\iint}{1}{dA}[\/latex] to find the area of a general bounded region [latex]D[\/latex], we can use [latex]\\underset{E}{\\displaystyle\\iiint}{1}{dV}[\/latex] to find the volume of a general solid bounded region [latex]E[\/latex]. The next example illustrates the method.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: finding a volume by evaluating a triple integral<\/h3>\n<p id=\"fs-id1167793504416\">Find the volume of a right pyramid that has the square base in the [latex]xy[\/latex]-plane [latex][-1,1] \\ {\\times} \\ [-1,1][\/latex] and vertex at the point [latex](0, 0, 1)[\/latex] as shown in the following figure.<\/p>\n<div id=\"attachment_1379\" style=\"width: 553px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1379\" class=\"size-full wp-image-1379\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25144721\/5-4-7.jpeg\" alt=\"In x y z space, there is a pyramid with a square base centered at the origin. The apex of the pyramid is (0, 0, 1).\" width=\"543\" height=\"527\" \/><\/p>\n<p id=\"caption-attachment-1379\" class=\"wp-caption-text\">Figure 7.\u00a0Finding the volume of a pyramid with a square base.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q425689014\">Show Solution<\/span><\/p>\n<div id=\"q425689014\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794005598\">In this pyramid the value of [latex]z[\/latex] changes from 0 to 1 and at each height [latex]z[\/latex], the cross section of the pyramid for any value of [latex]z[\/latex] is the square [latex][-1+z,1-z] \\ {\\times} \\ [-1+z,1-z][\/latex]. Hence, the volume of the pyramid is [latex]\\underset{E}{\\displaystyle\\iiint}{1}{dV}[\/latex] where<\/p>\n<p style=\"text-align: center;\">[latex]{E} = \\left \\{{(x,y,z)}{\\mid}{0} \\ {\\leq} \\ {z} \\ {\\leq} \\ {1,-1+z} \\ {\\leq} \\ {y} \\ {\\leq} \\ {1-z,-1+z} \\ {\\leq} \\ {x} \\ {\\leq} \\ {1-z} \\right \\}[\/latex].<\/p>\n<p id=\"fs-id1167793442473\">Thus, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{1}{dV} = {\\displaystyle\\int^{z=1}_{z=0}}{\\displaystyle\\int^{y={1-z}}_{y=-1+z}}{\\displaystyle\\int^{x={1-z}}_{x=1+z}}{1}{dx}{dy}{dz} = {\\displaystyle\\int^{z=1}_{z=0}}{\\displaystyle\\int^{y={1-z}}_{y=-1+z}}{(2-2z)}{dy}{dz} = {\\displaystyle\\int^{z=1}_{z=0}}{(2-2z)^2}{dz} = {\\frac{4}{3}}[\/latex].<\/p>\n<p id=\"fs-id1167793637795\">Hence, the volume of the pyramid is\u00a0[latex]{\\frac{4}{3}}[\/latex] cubic units.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Consider the solid sphere [latex]{E} = {(x,y,z)}{\\mid}\\left \\{{{x^2}+{y^2}+{z^2}} = {9} \\right \\}[\/latex]. Write the triple integral [latex]\\underset{E}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV}[\/latex] for an arbitrary function [latex]f[\/latex] as an iterated integral. Then evaluate this triple integral with [latex]f(x, y, z)=1[\/latex]. Notice that this gives the volume of a sphere using a triple integral.<\/p>\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>[latex]\\underset{E}{\\displaystyle\\iiint}1 \\ dV=8\\displaystyle\\int_{x=-3}^{x=3}\\displaystyle\\int_{y=-\\sqrt{9-x^2}}^{y=\\sqrt{9-x^2}}\\displaystyle\\int_{z=-\\sqrt{9-x^2-y^2}}^{z=\\sqrt{9-x^2-y^2}}1 \\ dz \\ dy \\ dx = 36\\pi.[\/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=8197106&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=yINls9G1Yek&amp;video_target=tpm-plugin-q6r3g93d-yINls9G1Yek\" 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\/CP5.24_transcript.html\">transcript for \u201cCP 5.24\u201d here (opens in new window).<\/a><\/div>\n<h2 data-type=\"title\">Changing the Order of Integration<\/h2>\n<p id=\"fs-id1167793522391\">As we have already seen in double integrals over general bounded regions, changing the order of the integration is done quite often to simplify the computation. With a triple integral over a rectangular box, the order of integration does not change the level of difficulty of the calculation. However, with a triple integral over a general bounded region, choosing an appropriate order of integration can simplify the computation quite a bit. Sometimes making the change to polar coordinates can also be very helpful. We demonstrate two examples here.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: changing the order of integration<\/h3>\n<p id=\"fs-id1167794296751\">Consider the iterated integral<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{x=1}_{x=0}} \\ {\\displaystyle\\int^{y=x^2}_{y=0}} \\ {\\displaystyle\\int^{z=y^2}_{z=0}}{f}{(x,y,z)}{dz}{dy}{dx}[\/latex].<\/p>\n<p id=\"fs-id1167793544547\">The order of integration here is first with respect to [latex]z[\/latex], then [latex]y[\/latex], and then [latex]x[\/latex]. Express this integral by changing the order of integration to be first with respect to [latex]x[\/latex], then [latex]z[\/latex], and then [latex]y[\/latex]. Verify that the value of the integral is the same if we let\u00a0[latex]f(x, y, z)=xyz[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q567723190\">Show Solution<\/span><\/p>\n<div id=\"q567723190\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793612495\">The best way to do this is to sketch the region [latex]E[\/latex] and its projections onto each of the three coordinate planes. Thus, let<\/p>\n<p style=\"text-align: center;\">[latex]{E} = \\left \\{{(x,y,z)}{\\mid}{0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {1,0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {x^2,0} \\ {\\leq} \\ {z} \\ {\\leq} \\ {y^2} \\right \\}[\/latex].<\/p>\n<p id=\"fs-id1167793432448\">and<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{x=1}_{x=0}} \\ {\\displaystyle\\int^{y=x^2}_{y=0}} \\ {\\displaystyle\\int^{z=y^2}_{z=0}}{f}{(x,y,z)}{dz}{dy}{dx} = \\underset{E}{\\displaystyle\\iiint}{f}{(x,y,z)}{dV}[\/latex].<\/p>\n<p id=\"fs-id1167793424343\">We need to express this triple integral as<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{y=d}_{y=c}} \\ {\\displaystyle\\int^{z={{v_2}{(y)}}}_{z={{v_1}{(y)}}}} \\ {\\displaystyle\\int^{x={{u_2}{(y,z)}}}_{x={{u_1}{(y,z)}}}}{f}{(x,y,z)}{dx}{dz}{dy}[\/latex].<\/p>\n<p id=\"fs-id1167793384512\">Knowing the region [latex]E[\/latex] we can draw the following projections (Figure 8):<\/p>\n<p id=\"fs-id1167793384524\">on the [latex]xy[\/latex]-plane is [latex]{D_1} = {\\left \\{{(x,y)}{\\mid}{0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {1,0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {x^2} \\right \\}} = {\\left \\{{(x,y)}{\\mid}{0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {1,{\\sqrt{y}}} \\ {\\leq} \\ {x} \\ {\\leq} \\ {1} \\right \\}}[\/latex],<\/p>\n<p>on the [latex]yz[\/latex]-plane is [latex]{D_2} = {\\left \\{{(y,z)}{\\mid}{0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {1,0} \\ {\\leq} \\ {z} \\ {\\leq} \\ {y^2} \\right \\}}[\/latex], and<\/p>\n<p>on the [latex]xz[\/latex]-plane is [latex]{D_3} = {\\left \\{{(x,z)}{\\mid}{0} \\ {\\leq} \\ {x} \\ {\\leq} \\ {1,0} \\ {\\leq} \\ {z} \\ {\\leq} \\ {x^2} \\right \\}}[\/latex].<\/p>\n<div id=\"attachment_1380\" style=\"width: 909px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1380\" class=\"size-full wp-image-1380\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25144759\/5-4-8.jpeg\" alt=\"Three similar versions of the following graph are shown: In the x y plane, a region D1 is bounded by the x axis, the line x = 1, and the curve y = x squared. In the second version, region D2 on the z y plane is shown with equation z = y squared. And in the third version, region D3 on the x z plane is shown with equation z = x cubed.\" width=\"899\" height=\"273\" \/><\/p>\n<p id=\"caption-attachment-1380\" class=\"wp-caption-text\">Figure 8.\u00a0The three cross sections of\u00a0[latex]E[\/latex] on the three coordinate planes.<\/p>\n<\/div>\n<p>Now we can describe the same region [latex]E[\/latex] as [latex]{\\left \\{{(x,y,z)}{\\mid}{0} \\ {\\leq} \\ {y} \\ {\\leq} \\ {1,0} \\ {\\leq} \\ {z} \\ {\\leq} \\ {{y^2},{\\sqrt{y}}} \\ {\\leq} \\ {x} \\ {\\leq} \\ {1} \\right \\}}[\/latex], and consequently, the triple integral becomes<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{y=d}_{y=c}} \\ {\\displaystyle\\int^{z={{v_2}{(y)}}}_{z={v_1}{(y)}}} \\ {\\displaystyle\\int^{x={{u_2}{(y,z)}}}_{x={u_1}{(y,z)}}}{f}{(x,y,z)}{dx}{dz}{dy}= {\\displaystyle\\int^{y=1}_{y=0}} \\ {\\displaystyle\\int^{z={x^2}}_{z=0}} \\ {\\displaystyle\\int^{x=1}_{x={\\sqrt{y}}}}{f}{(x,y,z)}{dx}{dz}{dy}[\/latex].<\/p>\n<p>Now assume that [latex]f(x, y, z)=xyz[\/latex] in each of the integrals. Then we have<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{align}<\/div>\n<p>\\displaystyle\\int^{x=1}_{x=0}\\displaystyle\\int^{y=x^2}_{y=0}\\displaystyle\\int^{z=y^2}_{z=0}xyz{dz}{dy}{dx}&=\\displaystyle\\int^{x=1}_{x=0}\\displaystyle\\int^{y=x^2}_{y=0}\\left[xy\\frac{z^2}2\\bigg|_{z=0}^{z=y^2}\\right]{dy}{dx} \\\\    &=\\displaystyle\\int^{x=1}_{x=0}\\displaystyle\\int^{y=x^2}_{y=0}\\left(x\\frac{y^5}2\\right)dy \\ dx \\\\    &=\\displaystyle\\int^{x=1}_{x=0}\\left[x\\frac{y^6}{12}\\bigg|^{y=x^2}_{y=0}\\right]dx \\\\  & =\\displaystyle\\int^{x=1}_{x=0}\\frac{x^{13}}{12}dx =\\frac1{168}, \\\\    \\displaystyle\\int^{y=1}_{y=0}\\displaystyle\\int^{z=y^2}_{z=0}\\displaystyle\\int^{x=1}_{x=\\sqrt{y}}xyz{dx}{dz}{dy} &=\\displaystyle\\int^{y=1}_{y=0}\\displaystyle\\int^{z=y^2}_{z=0}\\left[yz\\frac{x^2}2\\bigg|^{1}_{\\sqrt{y}}\\right]dz \\ dy\\\\    &=\\displaystyle\\int^{y=1}_{y=0}\\displaystyle\\int^{z=y^2}_{z=0}\\left(\\frac{yz}2-\\frac{y^2z}2\\right)dz \\ dy \\\\    &=\\displaystyle\\int^{y=1}_{y=0}\\left[\\frac{yz^2}4-\\frac{y^2z^2}4\\bigg|^{z=y^2}_{z=0}\\right]dy \\\\    &=\\displaystyle\\int^{y=1}_{y=0}\\left(\\frac{y^5}4-\\frac{y^6}4\\right)dy=\\frac1{168}.    \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1167793617989\">The answers match.<\/p>\n<\/div>\n<\/div>\n<section data-depth=\"1\"><\/section>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Write five different iterated integrals equal to the given integral\u00a0[latex]{\\displaystyle\\int^{z=4}_{z=0}} \\ {\\displaystyle\\int^{y={4-z}}_{y=0}} \\ {\\displaystyle\\int^{x={\\sqrt{y}}}_{x=0}}{f}{(x,y,z)}{dx}{dy}{dz}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q297541139\">Show Solution<\/span><\/p>\n<div id=\"q297541139\" class=\"hidden-answer\" style=\"display: none\">\n<p>(i)\u00a0[latex]\\displaystyle\\int_{z=0}^{z=4}\\displaystyle\\int_{x=0}^{x=\\sqrt{x-4}}\\displaystyle\\int_{y=x^2}^{y=4-z}f(x,y,z)dy \\ dx \\ dz[\/latex],<\/p>\n<p>(ii)[latex]\\displaystyle\\int_{y=0}^{y=4}\\displaystyle\\int_{z=0}^{z=4-y}\\displaystyle\\int_{x=0}^{x=\\sqrt{y}}f(x,y,z)dx \\ dz \\ dy[\/latex],<\/p>\n<p>(iii)[latex]\\displaystyle\\int_{y=0}^{y=4}\\displaystyle\\int_{x=0}^{x=\\sqrt{y}}\\displaystyle\\int_{z=0}^{z=4-y}f(x,y,z)dz \\ dx \\ dy[\/latex]\u00a0,<\/p>\n<p>(iv)\u00a0[latex]\\displaystyle\\int_{x=0}^{x=2}\\displaystyle\\int_{y=x^2}^{y=4}\\displaystyle\\int_{z=0}^{z=4-y}f(x,y,z)dz \\ dy \\ dx[\/latex],<\/p>\n<p>(v)\u00a0[latex]\\displaystyle\\int_{x=0}^{x=2}\\displaystyle\\int_{z=0}^{z=4-x^2}\\displaystyle\\int_{y=x^2}^{y=4-y}f(x,y,z)dy \\ dz \\ dx[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: changing integration order and coordinate systems<\/h3>\n<p>Evaluate the triple integral [latex]\\underset{E}{\\displaystyle\\iiint}{\\sqrt{{x^2}+{z^2}}}{dV}[\/latex], where [latex]E[\/latex] is the region bounded by the paraboloid [latex]y=x^{2}+z^{2}[\/latex] (Figure 9) and the plane [latex]y=4[\/latex].<\/p>\n<div id=\"attachment_1382\" style=\"width: 587px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1382\" class=\"size-full wp-image-1382\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25144936\/5-4-9.jpeg\" alt=\"The paraboloid y = x squared + z squared is shown opening up along the y axis to y = 4.\" width=\"577\" height=\"327\" \/><\/p>\n<p id=\"caption-attachment-1382\" class=\"wp-caption-text\">Figure 9.\u00a0Integrating a triple integral over a paraboloid.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q843152286\">Show Solution<\/span><\/p>\n<div id=\"q843152286\" class=\"hidden-answer\" style=\"display: none\">\n<p>The projection of the solid region [latex]E[\/latex] onto the [latex]xy[\/latex]-plane is the region bounded above by [latex]y=4[\/latex] and below by the parabola [latex]y=x^{2}[\/latex] as shown.<\/p>\n<div id=\"attachment_1383\" style=\"width: 427px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1383\" class=\"size-full wp-image-1383\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/25145013\/5-4-10.jpeg\" alt=\"In the x y plane, the graph of y = x squared is shown with the line y = 4 intersecting the graph at (negative 2, 4) and (2, 4).\" width=\"417\" height=\"272\" \/><\/p>\n<p id=\"caption-attachment-1383\" class=\"wp-caption-text\">Figure 10.\u00a0Cross section in the [latex]xy[\/latex]-plane of the paraboloid in Figure 9.<\/p>\n<\/div>\n<p id=\"fs-id1167793285294\">Thus, we have<\/p>\n<p style=\"text-align: center;\">[latex]{E} = {\\left \\{ {(x,y,z)}{\\mid}{-2} \\ {\\leq} \\ {x} \\ {\\leq} \\ {2,{x^2}} \\ {\\leq} \\ {y} \\ {\\leq} \\ {{4},-{\\sqrt{y-x^2}}} \\ {\\leq} \\ {z} \\ {\\leq} \\ {\\sqrt{y-x^2}} \\right \\}}[\/latex].<\/p>\n<p id=\"fs-id1167794200516\">The triple integral becomes<\/p>\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{\\sqrt{x^2+z^2}dV} = {\\displaystyle\\int^{x=2}_{x=-2}} \\ {\\displaystyle\\int^{y=4}_{y=x^2}} \\ {\\displaystyle\\int^{z={\\sqrt{y-x^2}}}_{z=-{\\sqrt{y-x^2}}}}{\\sqrt{{x^2}+{z^2}}}{dz}{dy}{dx}[\/latex].<\/p>\n<p id=\"fs-id1167793607349\">This expression is difficult to compute, so consider the projection of [latex]E[\/latex] onto the [latex]xz[\/latex]-plane. This is a circular disc [latex]{x^2} + {z^2} \\ {\\leq} \\ 4[\/latex]. So we obtain<\/p>\n<p style=\"text-align: center;\">[latex]\\underset{E}{\\displaystyle\\iiint}{\\sqrt{x^2+z^2}dV} = {\\displaystyle\\int^{x=2}_{x=-2}} \\ {\\displaystyle\\int^{y=4}_{y=x^2}} \\ {\\displaystyle\\int^{z={\\sqrt{y-x^2}}}_{z=-{\\sqrt{y-x^2}}}}{\\sqrt{{x^2}+{z^2}}}{dz}{dy}{dx} = {\\displaystyle\\int^{x=2}_{x=-2}} \\ {\\displaystyle\\int^{z={\\sqrt{4-x^2}}}_{z=-{\\sqrt{4-x^2}}}} \\ {\\displaystyle\\int^{y=4}_{y=x^2+z^2}}{\\sqrt{{x^2}+{z^2}}}{dy}{dz}{dx}[\/latex].<\/p>\n<p id=\"fs-id1167794246456\">Here the order of integration changes from being first with respect to [latex]z[\/latex], then [latex]y[\/latex] and then [latex]x[\/latex] to being first with respect to [latex]y[\/latex], then to [latex]z[\/latex], and then to [latex]x[\/latex]. It will soon be clear how this change can be beneficial for computation. We have<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{x=2}_{x=-2}} \\ {\\displaystyle\\int^{z={\\sqrt{4-x^2}}}_{z=-{\\sqrt{4-x^2}}}} \\ {\\displaystyle\\int^{y=4}_{y=x^2+z^2}}{\\sqrt{{x^2}+{z^2}}}{dy}{dz}{dx} = {\\displaystyle\\int^{x=2}_{x=-2}} \\ {\\displaystyle\\int^{z={\\sqrt{4-x^2}}}_{z=-{\\sqrt{4-x^2}}}}{(4-x^2-z^2)}{\\sqrt{{x^2}+{z^2}}}{dz}{dx}[\/latex].<\/p>\n<p id=\"fs-id1167793485217\">Now use the polar substitution [latex]{{x} = {r}{\\cos}{\\theta}},{{z} = {r}{\\sin}{\\theta}}[\/latex], and [latex]{dz}{dx} = {r}{dr}{{d}{\\theta}}[\/latex] in the [latex]xz[\/latex]-plane. This is essentially the same thing as when we used polar coordinates in the [latex]xy[\/latex]-plane, except we are replacing [latex]y[\/latex] by [latex]z[\/latex]. Consequently the limits of integration change and we have, by using [latex]r^{2}=x^{2}+z^{2}[\/latex],<\/p>\n<p style=\"text-align: center;\">[latex]{\\displaystyle\\int^{x=2}_{x=-2}} \\ {\\displaystyle\\int^{z={\\sqrt{4-x^2}}}_{z=-{\\sqrt{4-x^2}}}}{(4-x^2-z^2)}{\\sqrt{{x^2}+{z^2}}}{dz}{dx} = {\\displaystyle\\int^{{\\theta} = 2{\\pi}}_{{\\theta} = 0}} \\ {\\displaystyle\\int^{r=2}_{r=0}}{(4-{r^2})}{rr}{dr}{{d}{\\theta}}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]= {\\displaystyle\\int^{2{\\pi}}_{0}} \\ {\\left [ {\\frac{4r^3}{3}}-{\\frac{r^5}{5}}{{\\bigg\\vert}^2_0} \\right ]} \\ {{d}{\\theta}} = {\\displaystyle\\int^{2{\\pi}}_{0}} \\ {\\frac{64}{15}}{{d}{\\theta}} = {\\frac{{128}{\\pi}}{15}}[\/latex].<\/p>\n<\/div>\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-3986\">\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 5.23. <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 5.24. <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":17,"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 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