{"id":1248,"date":"2025-10-15T15:50:38","date_gmt":"2025-10-15T15:50:38","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=1248"},"modified":"2025-10-31T15:25:25","modified_gmt":"2025-10-31T15:25:25","slug":"triple-integrals-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/triple-integrals-learn-it-3\/","title":{"raw":"Triple Integrals: Learn It 3","rendered":"Triple Integrals: Learn It 3"},"content":{"raw":"<h2 data-type=\"title\">Changing the Order of Integration<\/h2>\r\n<p class=\"whitespace-normal break-words\">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.<\/p>\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\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].<\/p>\r\nExpress 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 [latex]f(x, y, z)=xyz[\/latex].\r\n\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}\\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} \\\\&amp;=\\displaystyle\\int^{x=1}_{x=0}\\displaystyle\\int^{y=x^2}_{y=0}\\left(x\\frac{y^5}2\\right)dy \\ dx \\\\&amp;=\\displaystyle\\int^{x=1}_{x=0}\\left[x\\frac{y^6}{12}\\bigg|^{y=x^2}_{y=0}\\right]dx \\\\&amp; =\\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} &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\\\\&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 \\\\&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 \\\\&amp;=\\displaystyle\\int^{y=1}_{y=0}\\left(\\frac{y^5}4-\\frac{y^6}4\\right)dy=\\frac1{168}.\\end{align}[\/latex]<\/center>\r\n<p id=\"fs-id1167793617989\">The answers match.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">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].\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<\/section>","rendered":"<h2 data-type=\"title\">Changing the Order of Integration<\/h2>\n<p class=\"whitespace-normal break-words\">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.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\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].<\/p>\n<p>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 [latex]f(x, y, z)=xyz[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q567723190\">Show Solution<\/button><\/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<figure id=\"attachment_1380\" aria-describedby=\"caption-attachment-1380\" style=\"width: 899px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" 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\" \/><figcaption id=\"caption-attachment-1380\" class=\"wp-caption-text\">Figure 8.\u00a0The three cross sections of\u00a0[latex]E[\/latex] on the three coordinate planes.<\/figcaption><\/figure>\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}\\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]<\/div>\n<p id=\"fs-id1167793617989\">The answers match.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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<figure id=\"attachment_1382\" aria-describedby=\"caption-attachment-1382\" style=\"width: 577px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" 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\" \/><figcaption id=\"caption-attachment-1382\" class=\"wp-caption-text\">Figure 9.\u00a0Integrating a triple integral over a paraboloid.<\/figcaption><\/figure>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q843152286\">Show Solution<\/button><\/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<figure id=\"attachment_1383\" aria-describedby=\"caption-attachment-1383\" style=\"width: 417px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" 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\" \/><figcaption id=\"caption-attachment-1383\" class=\"wp-caption-text\">Figure 10.\u00a0Cross section in the [latex]xy[\/latex]-plane of the paraboloid in Figure 9.<\/figcaption><\/figure>\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<\/section>\n","protected":false},"author":15,"menu_order":7,"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-1248","chapter","type-chapter","status-publish","hentry"],"part":434,"module-header":"- Select Header -","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\/1248","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":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1248\/revisions"}],"predecessor-version":[{"id":1912,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1248\/revisions\/1912"}],"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\/1248\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=1248"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=1248"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=1248"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=1248"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}