{"id":708,"date":"2025-10-09T15:33:18","date_gmt":"2025-10-09T15:33:18","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=708"},"modified":"2025-11-13T15:09:11","modified_gmt":"2025-11-13T15:09:11","slug":"triple-integrals-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/triple-integrals-fresh-take\/","title":{"raw":"Triple Integrals: Fresh Take","rendered":"Triple Integrals: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Evaluate triple integrals over box-shaped regions<\/li>\r\n \t<li>Calculate triple integrals over more complex 3D regions<\/li>\r\n \t<li>Simplify a calculation by changing the order of integration of a triple integral<\/li>\r\n \t<li>Find the average value of a function of three variables<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">Integrable Functions of Three Variables<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Triple integrals extend the concept of double integrals into three dimensions. Just as a double integral sums up values over a flat region, a triple integral sums up values throughout a three-dimensional rectangular box. Think of it as filling a room with tiny cubes and adding up the function's value at each cube\u2014then shrinking those cubes infinitely small.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Setup<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">A rectangular box [latex]B[\/latex] in 3D space is defined by intervals for [latex]x[\/latex], [latex]y[\/latex], and [latex]z[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]B = {(x,y,z) \\mid a \\leq x \\leq b, c \\leq y \\leq d, e \\leq z \\leq f}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">We slice this box into smaller subboxes, each with volume [latex]\\Delta V = \\Delta x \\Delta y \\Delta z[\/latex]. The triple integral is the limit of the Riemann sum as these subboxes get infinitely small:<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\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}}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Computing Triple Integrals<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Fubini's theorem is your computational workhorse here. For continuous functions over rectangular boxes, you can evaluate the triple integral as an iterated integral in any order:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{B}{\\iiint} f(x,y,z) , dV = \\int_{e}^{f} \\int_{c}^{d} \\int_{a}^{b} f(x,y,z) , dx , dy , dz[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">There are six possible orderings (integrate [latex]x[\/latex], [latex]y[\/latex], [latex]z[\/latex] in any sequence), and for rectangular boxes, all produce the same result.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Work from the inside out\u2014integrate with respect to the innermost variable first, treating the other variables as constants. Then move outward, evaluating at the bounds after each integration step. For rectangular regions, the order doesn't affect difficulty, so choose whichever feels most natural.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Key Condition<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The function must be continuous on [latex]B[\/latex] for the triple integral to exist, though continuity everywhere except possibly on the boundary is sufficient.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Evaluate the triple integral [latex]\\underset{B}{\\displaystyle\\iiint}{z} \\ {\\sin \\ x} \\ {\\cos \\ y} \\ {dV}[\/latex] where[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].[reveal-answer q=\"159035786\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"159035786\"][latex]\\underset{B}{\\displaystyle\\iiint}z\\sin{x}\\cos{y} \\ dV=8[\/latex][\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<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\" data-mce-fragment=\"1\"><\/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><\/section>\r\n<h2 data-type=\"title\">Triple Integrals over a General Bounded Region<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Moving beyond rectangular boxes, triple integrals can handle more complex three-dimensional regions by using projections and bounding surfaces. The key is to visualize the solid region as sitting between two surfaces, with its footprint (projection) on one of the coordinate planes.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Core Approach<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For a general region [latex]E[\/latex], identify which coordinate plane works best for projection. If projecting onto the [latex]xy[\/latex]-plane, describe the region as:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]E = {(x,y,z) \\mid (x,y) \\in D, u_1(x,y) \\leq z \\leq u_2(x,y)}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">Here, [latex]D[\/latex] is the shadow of [latex]E[\/latex] on the [latex]xy[\/latex]-plane, and [latex]z[\/latex] varies between two surfaces [latex]z = u_1(x,y)[\/latex] (bottom) and [latex]z = u_2(x,y)[\/latex] (top). The triple integral becomes:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{E}{\\iiint} f(x,y,z) , dV = \\underset{D}{\\iint} \\left[\\int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z) , dz\\right] dA[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">You can project onto any coordinate plane\u2014just adjust which variable is bounded by surfaces and which two form the projection region.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Setting Up the Integral<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The projection region [latex]D[\/latex] can be Type I or Type II. For Type I regions in the [latex]xy[\/latex]-plane, [latex]x[\/latex] has constant bounds and [latex]y[\/latex] depends on [latex]x[\/latex]. For Type II regions, [latex]y[\/latex] has constant bounds and [latex]x[\/latex] depends on [latex]y[\/latex]. Choose the type that simplifies your bounds.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Sketch the region and identify natural bounding surfaces<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Determine which projection plane makes the bounds clearest<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Write out the region description carefully, noting which variables have constant bounds versus function bounds<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Integrate from innermost to outermost, working through one variable at a time<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Finding Volume<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">To find the volume of a solid region [latex]E[\/latex], evaluate [latex]\\underset{E}{\\iiint} 1 , dV[\/latex]. The integrand is simply 1, so you're adding up all the tiny volume elements in the region.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">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.[reveal-answer q=\"178836902\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"178836902\"][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][\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<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\" data-mce-fragment=\"1\"><\/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><\/section>\r\n<h2 data-type=\"title\">Changing the Order of Integration<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">For triple integrals over general regions, the order of integration can dramatically affect how difficult the computation becomes. Unlike rectangular boxes where order doesn't matter much, choosing the right order for bounded regions can mean the difference between a manageable integral and an impossible one.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Why Change the Order?<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Sometimes the given order leads to complicated integrands or difficult bounds. By changing the order, you might simplify the function you're integrating or get cleaner limits. Occasionally, switching to polar coordinates in one of the coordinate planes can also make the calculation much easier.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>How to Change the Order<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The key tool is visualization. Sketch the region [latex]E[\/latex] and all three of its projections onto the [latex]xy[\/latex]-, [latex]xz[\/latex]-, and [latex]yz[\/latex]-planes. Each projection gives you a different way to describe the same region. From the original integral's bounds, determine what the region [latex]E[\/latex] looks like. Then rewrite those bounds to reflect your new desired order.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Sketch the region [latex]E[\/latex] in three dimensions if possible<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Draw all three projections onto the coordinate planes<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Write out the region description in set notation from the original bounds<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Determine which projection plane makes the most sense for your new order<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Rewrite the bounds based on that projection, ensuring each variable's limits match the geometry<\/li>\r\n \t<li class=\"whitespace-normal break-words\">If you see circular or radial symmetry in a projection, consider using polar coordinates in that plane<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>When to Consider Polar Coordinates<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">If a projection onto the [latex]xz[\/latex]- or [latex]yz[\/latex]-plane shows circular symmetry (like [latex]x^2 + z^2 \\leq r^2[\/latex]), use polar substitution in that plane just as you would in the [latex]xy[\/latex]-plane.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">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].[reveal-answer q=\"297541139\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"297541139\"](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],(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],(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,(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],(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<\/section>\r\n<h2>Average Value of a Function of Three Variables<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Just as you can find the average value of a function over a region in the plane, you can find the average value of a function throughout a three-dimensional solid. The idea is intuitive: integrate the function over the entire solid, then divide by the solid's volume to get the \"typical\" value.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Formula<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For a function [latex]f(x,y,z)[\/latex] that's integrable over a solid region [latex]E[\/latex] with positive volume [latex]V(E)[\/latex], the average value is:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]f_{\\text{ave}} = \\frac{1}{V(E)} \\underset{E}{\\iiint} f(x,y,z) \\, dV[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This formula has two parts: the numerator is the triple integral of the function over [latex]E[\/latex], and the denominator is the volume of [latex]E[\/latex], which you find by evaluating [latex]V(E) = \\underset{E}{\\iiint} 1 \\, dV[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Set up the region [latex]E[\/latex] with appropriate bounds based on the geometry<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Evaluate the triple integral of the function: [latex]\\underset{E}{\\iiint} f(x,y,z) \\, dV[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Evaluate the volume integral: [latex]V(E) = \\underset{E}{\\iiint} 1 \\, dV[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Divide the function integral by the volume to get the average value<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Why This Makes Sense<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Think of it like finding the average temperature in a room. You're summing up all the temperature values at every point in the space, then dividing by the total volume to get a single representative temperature. The triple integral accumulates all the values, and dividing by volume normalizes it to a meaningful average.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Find the average value of the function [latex]f(x, y, z)=xyz[\/latex] over the cube with sides of length 4 units in the first octant with one vertex at the origin and edges parallel to the coordinate axes.[reveal-answer q=\"151360489\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"151360489\"][latex]f_{\\text{ave}}=8[\/latex][\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8197107&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=6U0zv9wquVo&amp;video_target=tpm-plugin-7r1ilu60-6U0zv9wquVo\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.26_transcript.html\">transcript for \u201cCP 5.26\u201d here (opens in new window).<\/a><\/center><\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Evaluate triple integrals over box-shaped regions<\/li>\n<li>Calculate triple integrals over more complex 3D regions<\/li>\n<li>Simplify a calculation by changing the order of integration of a triple integral<\/li>\n<li>Find the average value of a function of three variables<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Integrable Functions of Three Variables<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Triple integrals extend the concept of double integrals into three dimensions. Just as a double integral sums up values over a flat region, a triple integral sums up values throughout a three-dimensional rectangular box. Think of it as filling a room with tiny cubes and adding up the function&#8217;s value at each cube\u2014then shrinking those cubes infinitely small.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Setup<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">A rectangular box [latex]B[\/latex] in 3D space is defined by intervals for [latex]x[\/latex], [latex]y[\/latex], and [latex]z[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]B = {(x,y,z) \\mid a \\leq x \\leq b, c \\leq y \\leq d, e \\leq z \\leq f}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">We slice this box into smaller subboxes, each with volume [latex]\\Delta V = \\Delta x \\Delta y \\Delta z[\/latex]. The triple integral is the limit of the Riemann sum as these subboxes get infinitely small:<\/p>\n<p style=\"text-align: center;\">[latex]{\\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}}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Computing Triple Integrals<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Fubini&#8217;s theorem is your computational workhorse here. For continuous functions over rectangular boxes, you can evaluate the triple integral as an iterated integral in any order:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{B}{\\iiint} f(x,y,z) , dV = \\int_{e}^{f} \\int_{c}^{d} \\int_{a}^{b} f(x,y,z) , dx , dy , dz[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">There are six possible orderings (integrate [latex]x[\/latex], [latex]y[\/latex], [latex]z[\/latex] in any sequence), and for rectangular boxes, all produce the same result.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Work from the inside out\u2014integrate with respect to the innermost variable first, treating the other variables as constants. Then move outward, evaluating at the bounds after each integration step. For rectangular regions, the order doesn&#8217;t affect difficulty, so choose whichever feels most natural.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Key Condition<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The function must be continuous on [latex]B[\/latex] for the triple integral to exist, though continuity everywhere except possibly on the boundary is sufficient.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Evaluate the triple integral [latex]\\underset{B}{\\displaystyle\\iiint}{z} \\ {\\sin \\ x} \\ {\\cos \\ y} \\ {dV}[\/latex] where[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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q159035786\">Show Solution<\/button><\/p>\n<div id=\"q159035786\" class=\"hidden-answer\" style=\"display: none\">[latex]\\underset{B}{\\displaystyle\\iiint}z\\sin{x}\\cos{y} \\ dV=8[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/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\" data-mce-fragment=\"1\"><\/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<\/section>\n<h2 data-type=\"title\">Triple Integrals over a General Bounded Region<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Moving beyond rectangular boxes, triple integrals can handle more complex three-dimensional regions by using projections and bounding surfaces. The key is to visualize the solid region as sitting between two surfaces, with its footprint (projection) on one of the coordinate planes.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Core Approach<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For a general region [latex]E[\/latex], identify which coordinate plane works best for projection. If projecting onto the [latex]xy[\/latex]-plane, describe the region as:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]E = {(x,y,z) \\mid (x,y) \\in D, u_1(x,y) \\leq z \\leq u_2(x,y)}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">Here, [latex]D[\/latex] is the shadow of [latex]E[\/latex] on the [latex]xy[\/latex]-plane, and [latex]z[\/latex] varies between two surfaces [latex]z = u_1(x,y)[\/latex] (bottom) and [latex]z = u_2(x,y)[\/latex] (top). The triple integral becomes:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{E}{\\iiint} f(x,y,z) , dV = \\underset{D}{\\iint} \\left[\\int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z) , dz\\right] dA[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">You can project onto any coordinate plane\u2014just adjust which variable is bounded by surfaces and which two form the projection region.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Setting Up the Integral<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The projection region [latex]D[\/latex] can be Type I or Type II. For Type I regions in the [latex]xy[\/latex]-plane, [latex]x[\/latex] has constant bounds and [latex]y[\/latex] depends on [latex]x[\/latex]. For Type II regions, [latex]y[\/latex] has constant bounds and [latex]x[\/latex] depends on [latex]y[\/latex]. Choose the type that simplifies your bounds.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Sketch the region and identify natural bounding surfaces<\/li>\n<li class=\"whitespace-normal break-words\">Determine which projection plane makes the bounds clearest<\/li>\n<li class=\"whitespace-normal break-words\">Write out the region description carefully, noting which variables have constant bounds versus function bounds<\/li>\n<li class=\"whitespace-normal break-words\">Integrate from innermost to outermost, working through one variable at a time<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Finding Volume<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">To find the volume of a solid region [latex]E[\/latex], evaluate [latex]\\underset{E}{\\iiint} 1 , dV[\/latex]. The integrand is simply 1, so you&#8217;re adding up all the tiny volume elements in the region.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q178836902\">Show Solution<\/button><\/p>\n<div id=\"q178836902\" class=\"hidden-answer\" style=\"display: none\">[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]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/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\" data-mce-fragment=\"1\"><\/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<\/section>\n<h2 data-type=\"title\">Changing the Order of Integration<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For triple integrals over general regions, the order of integration can dramatically affect how difficult the computation becomes. Unlike rectangular boxes where order doesn&#8217;t matter much, choosing the right order for bounded regions can mean the difference between a manageable integral and an impossible one.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Why Change the Order?<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Sometimes the given order leads to complicated integrands or difficult bounds. By changing the order, you might simplify the function you&#8217;re integrating or get cleaner limits. Occasionally, switching to polar coordinates in one of the coordinate planes can also make the calculation much easier.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>How to Change the Order<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The key tool is visualization. Sketch the region [latex]E[\/latex] and all three of its projections onto the [latex]xy[\/latex]-, [latex]xz[\/latex]-, and [latex]yz[\/latex]-planes. Each projection gives you a different way to describe the same region. From the original integral&#8217;s bounds, determine what the region [latex]E[\/latex] looks like. Then rewrite those bounds to reflect your new desired order.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Sketch the region [latex]E[\/latex] in three dimensions if possible<\/li>\n<li class=\"whitespace-normal break-words\">Draw all three projections onto the coordinate planes<\/li>\n<li class=\"whitespace-normal break-words\">Write out the region description in set notation from the original bounds<\/li>\n<li class=\"whitespace-normal break-words\">Determine which projection plane makes the most sense for your new order<\/li>\n<li class=\"whitespace-normal break-words\">Rewrite the bounds based on that projection, ensuring each variable&#8217;s limits match the geometry<\/li>\n<li class=\"whitespace-normal break-words\">If you see circular or radial symmetry in a projection, consider using polar coordinates in that plane<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>When to Consider Polar Coordinates<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">If a projection onto the [latex]xz[\/latex]&#8211; or [latex]yz[\/latex]-plane shows circular symmetry (like [latex]x^2 + z^2 \\leq r^2[\/latex]), use polar substitution in that plane just as you would in the [latex]xy[\/latex]-plane.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q297541139\">Show Solution<\/button><\/p>\n<div id=\"q297541139\" class=\"hidden-answer\" style=\"display: none\">(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],(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],(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,(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],(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<\/section>\n<h2>Average Value of a Function of Three Variables<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Just as you can find the average value of a function over a region in the plane, you can find the average value of a function throughout a three-dimensional solid. The idea is intuitive: integrate the function over the entire solid, then divide by the solid&#8217;s volume to get the &#8220;typical&#8221; value.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Formula<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For a function [latex]f(x,y,z)[\/latex] that&#8217;s integrable over a solid region [latex]E[\/latex] with positive volume [latex]V(E)[\/latex], the average value is:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]f_{\\text{ave}} = \\frac{1}{V(E)} \\underset{E}{\\iiint} f(x,y,z) \\, dV[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This formula has two parts: the numerator is the triple integral of the function over [latex]E[\/latex], and the denominator is the volume of [latex]E[\/latex], which you find by evaluating [latex]V(E) = \\underset{E}{\\iiint} 1 \\, dV[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Set up the region [latex]E[\/latex] with appropriate bounds based on the geometry<\/li>\n<li class=\"whitespace-normal break-words\">Evaluate the triple integral of the function: [latex]\\underset{E}{\\iiint} f(x,y,z) \\, dV[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Evaluate the volume integral: [latex]V(E) = \\underset{E}{\\iiint} 1 \\, dV[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Divide the function integral by the volume to get the average value<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Why This Makes Sense<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Think of it like finding the average temperature in a room. You&#8217;re summing up all the temperature values at every point in the space, then dividing by the total volume to get a single representative temperature. The triple integral accumulates all the values, and dividing by volume normalizes it to a meaningful average.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Find the average value of the function [latex]f(x, y, z)=xyz[\/latex] over the cube with sides of length 4 units in the first octant with one vertex at the origin and edges parallel to the coordinate axes.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q151360489\">Show Solution<\/button><\/p>\n<div id=\"q151360489\" class=\"hidden-answer\" style=\"display: none\">[latex]f_{\\text{ave}}=8[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8197107&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=6U0zv9wquVo&amp;video_target=tpm-plugin-7r1ilu60-6U0zv9wquVo\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/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\/CP5.26_transcript.html\">transcript for \u201cCP 5.26\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":10,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-708","chapter","type-chapter","status-publish","hentry"],"part":434,"module-header":"fresh_take","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/708","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":10,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/708\/revisions"}],"predecessor-version":[{"id":2199,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/708\/revisions\/2199"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/434"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/708\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=708"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=708"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=708"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=708"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}