{"id":1074,"date":"2021-11-01T19:19:03","date_gmt":"2021-11-01T19:19:03","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=1074"},"modified":"2022-11-01T04:43:05","modified_gmt":"2022-11-01T04:43:05","slug":"change-of-variables","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/change-of-variables\/","title":{"raw":"Change of Variables","rendered":"Change of Variables"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul class=\"os-abstract\">\r\n \t<li><span class=\"os-abstract-content\">Evaluate a double integral using a change of variables.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Evaluate a triple integral using a change of variables.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 data-type=\"title\">Change of Variables for Double Integrals<\/h2>\r\n<p id=\"fs-id1167794176297\">We have already seen that, under the change of variables [latex]T(u, v)=(x, y)[\/latex] where [latex]x=g(u, v)[\/latex] and [latex]y=h(u, v)[\/latex], a small region [latex]{\\Delta}{A}[\/latex] in the [latex]xy[\/latex]-plane is related to the area formed by the product [latex]{\\Delta}{u}{\\Delta}{v}[\/latex] in the [latex]uv[\/latex]-plane by the approximation<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{{\\Delta}{A}{\\approx}{J}{(u,v)}{\\Delta}{u},{\\Delta}{v}}[\/latex].<\/p>\r\nNow let\u2019s go back to the definition of double integral for a minute:\r\n<p style=\"text-align: center;\">[latex]\\large{\\underset{R}{\\displaystyle\\iint}f(x,y) \\ dA=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^n{f}(x_{ij},y_{ij})\\Delta{A}}[\/latex].<\/p>\r\nReferring to\u00a0Figure 1, observe that we divided the region [latex]S[\/latex] in the [latex]uv[\/latex]-plane into small subrectangles [latex]S_{ij}[\/latex] and we let the subrectangles [latex]R_{ij}[\/latex] in the [latex]xy[\/latex]-plane be the images of [latex]S_{ij}[\/latex] under the transformation [latex]T(u, v)=(x, y)[\/latex].\r\n\r\n[caption id=\"attachment_1422\" align=\"aligncenter\" width=\"765\"]<img class=\"size-full wp-image-1422\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28043011\/5-7-5.jpeg\" alt=\"On the left-hand side of this figure, there is a rectangle S with an inscribed red oval and a subrectangle with lower right corner point (u sub ij, v sub ij), height Delta v, and length Delta u given in the Cartesian u v-plane. Then there is an arrow from this graph to the right-hand side of the figure marked with T. On the right-hand side of this figure there is a region R with inscribed (deformed) red oval and a subrectangle R sub ij with corner point (x sub ij, y sub ij) given in the Cartesian x y-plane. The subrectangle is blown up and shown with vectors pointing along the edge from the corner point.\" width=\"765\" height=\"321\" \/> Figure 1.\u00a0The subrectangles [latex]S_{ij}[\/latex] in the\u00a0[latex]uv[\/latex]-plane transform into subrectangles [latex]R_{ij}[\/latex] in the\u00a0[latex]xy[\/latex]-plane.[\/caption]\r\n<p id=\"fs-id1167794052254\">Then the double integral becomes<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\underset{R}{\\displaystyle\\iint}f(x,y) \\ dA=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^n{f}(x_{ij},y_{ij})\\Delta{A}=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^nf(g(u_{ij},v_{ij}),h(u_{ij},v_{ij}))|J(u_{ij},v_{ij})|\\Delta{u}\\Delta{v}}[\/latex].<\/p>\r\nNotice this is exactly the double Riemann sum for the integral\r\n<p style=\"text-align: center;\">[latex]\\large{\\underset{S}{\\displaystyle\\iint}f(g(u,v),h(u,v))\\Bigg|\\frac{\\partial(x,y)}{\\partial(u,v)}\\Bigg|du \\ dv}[\/latex].<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">theorem: change of variables for double integrals<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1167793433530\">Let [latex]T(u, v)=(x, y)[\/latex] where [latex]x=g(u, v)[\/latex] and [latex]y=h(u, v)[\/latex] be a one-to-one [latex]C1[\/latex] transformation, with a nonzero Jacobian on the interior of the region [latex]S[\/latex] in the [latex]uv[\/latex]-plane; it maps [latex]S[\/latex] into the region [latex]R[\/latex] in the [latex]xy[\/latex]-plane. If [latex]f[\/latex] is continuous on [latex]R[\/latex], then<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\underset{R}{\\displaystyle\\iint}f(x,y) \\ dA=\\underset{S}{\\displaystyle\\iint}f(g(u,v),h(u,v))\\Bigg|\\frac{\\partial(x,y)}{\\partial(u,v)}\\Bigg|du \\ dv}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793976062\">With this theorem for double integrals, we can change the variables from [latex](x, y)[\/latex] to [latex](u, v)[\/latex] in a double integral simply by replacing<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{dA=dx \\ dy=\\Bigg|\\frac{\\partial(x,y)}{\\partial(u,v)}\\Bigg|du \\ dv}[\/latex]<\/p>\r\nwhen we use the substitutions [latex]x=g(u, v)[\/latex] and [latex]y=h(u, v)[\/latex] and then change the limits of integration accordingly. This change of variables often makes any computations much simpler.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: changing variables from rectangular to polar coordinates<\/h3>\r\n<p id=\"fs-id1167793926832\">Consider the integral<\/p>\r\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int_0^2\\displaystyle\\int_0^{\\sqrt{2x-x^2}}\\sqrt{x^2+y^2}dy \\ dx[\/latex].<\/p>\r\n<p id=\"fs-id1167793481509\">Use the change of variables [latex]x=r\\cos\\theta[\/latex] and [latex]y=r\\sin\\theta[\/latex], and find the resulting integral.<\/p>\r\n[reveal-answer q=\"598345190\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"598345190\"]\r\n\r\nFirst we need to find the region of integration. This region is bounded below by [latex]y=0[\/latex] and above by [latex]y=\\sqrt{2x-x^2}[\/latex] (see the following figure).\r\n\r\n[caption id=\"attachment_1423\" align=\"aligncenter\" width=\"290\"]<img class=\"size-full wp-image-1423\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28043128\/5-7-6.jpeg\" alt=\"A semicircle in the first quadrant of the xy plane with radius 1 and center (1, 0). The equation for this curve is given as y = the square root of (2x minus x squared)\" width=\"290\" height=\"234\" \/> Figure 2.\u00a0Changing a region from rectangular to polar coordinates.[\/caption]\r\n<p id=\"fs-id1167794215044\">Squaring and collecting terms, we find that the region is the upper half of the circle [latex]x^{2}+y^{2}-2x=0[\/latex] that is, [latex]y^{2}+(x-1)^{2}[\/latex]. In polar coordinates, the circle is [latex]r=2\\cos\\theta[\/latex] so the region of integration in polar coordinates is bounded by [latex]0\\leq{r}\\leq\\cos\\theta[\/latex] and [latex]0\\leq\\theta\\leq\\frac{\\pi}2[\/latex].<\/p>\r\n<p id=\"fs-id1167793618382\">The Jacobian is [latex]J(r,\\theta)=r[\/latex], as shown in\u00a0Example \"Finding the Jacobian\". Since [latex]r\\geq0[\/latex], we have [latex]|J(r,\\theta)|=r[\/latex].<\/p>\r\n<p id=\"fs-id1167793424417\">The integrand [latex]\\sqrt{x^2+y^2}[\/latex] changes to [latex]r[\/latex] in polar coordinates, so the double iterated integral is<\/p>\r\n[latex]\\displaystyle\\int_0^2\\displaystyle\\int_0^{\\sqrt{2x-x^2}}\\sqrt{x^2+y^2}dy \\ dx=\\displaystyle\\int_0^{\\pi\/2}\\displaystyle\\int_0^{2\\cos\\theta}r|J(r,\\theta)|dr \\ d\\theta=\\displaystyle\\int_0^{\\pi\/2}\\displaystyle\\int_0^{2\\cos\\theta}r^2 \\ dr \\ d\\theta[\/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\nConsidering the integral [latex]\\displaystyle\\int_0^1\\displaystyle\\int_0^{\\sqrt{1-x^2}}(x^2+y^2)dy \\ dx[\/latex], use the change of variables [latex]x=r\\cos\\theta[\/latex] and [latex]y=\\sin\\theta[\/latex], and find the resulting integral.\r\n\r\n[reveal-answer q=\"983458271\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"983458271\"]\r\n\r\n[latex]\\displaystyle\\int_0^{\\pi\/2}\\displaystyle\\int_0^1r^3 \\ dr \\ d\\theta[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNotice in the next example that the region over which we are to integrate may suggest a suitable transformation for the integration. This is a common and important situation.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: changing variables<\/h3>\r\nConsider the integral [latex]\\underset{R}{\\displaystyle\\iint}(x-y)dy \\ dx[\/latex], where [latex]R[\/latex] is the parallelogram joining the points [latex](1, 2)[\/latex], [latex](3, 4)[\/latex], [latex](4, 3)[\/latex], and [latex](6, 5)[\/latex] (Figure 3). Make appropriate changes of variables, and write the resulting integral.\r\n\r\n[caption id=\"attachment_1424\" align=\"aligncenter\" width=\"328\"]<img class=\"size-full wp-image-1424\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28043209\/5-7-7.jpeg\" alt=\"A semicircle in the first quadrant of the xy plane with radius 1 and center (1, 0). The equation for this curve is given as y = the square root of (2x minus x squared)\" width=\"328\" height=\"272\" \/> Figure 3.\u00a0The region of integration for the given integral.[\/caption]\r\n\r\n[reveal-answer q=\"134952148\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"134952148\"]\r\n<p id=\"fs-id1167794171741\">First, we need to understand the region over which we are to integrate. The sides of the parallelogram are [latex]x-y+1=0[\/latex],\u00a0[latex]x-y-1=0[\/latex],\u00a0[latex]x-3y+5=0[\/latex], and\u00a0[latex]x-3y+9=0[\/latex] (Figure 4). Another way to look at them is [latex]x-y=-1[\/latex],\u00a0[latex]x-y=1[\/latex],\u00a0[latex]x-3y=-5[\/latex], and\u00a0[latex]x-3y=-9[\/latex].<\/p>\r\n<p id=\"fs-id1167793951713\">Clearly the parallelogram is bounded by the lines\u00a0[latex]y=x+1[\/latex],\u00a0[latex]y=x-1[\/latex],\u00a0[latex]y=\\frac13(x+5)[\/latex], and\u00a0[latex]y=\\frac13(x+9)[\/latex].<\/p>\r\n<p id=\"fs-id1167793924325\">Notice that if we were to make [latex]u=x-y[\/latex] and [latex]v=x-3y[\/latex], then the limits on the integral would be [latex]-1\\leq{u}\\leq1[\/latex] and\u00a0[latex]-9\\leq{v}\\leq-5[\/latex].<\/p>\r\n<p id=\"fs-id1167793931696\">To solve for [latex]x[\/latex] and [latex]y[\/latex], we multiply the first equation by [latex]3[\/latex] and subtract the second equation, [latex]3u-v=(3x-3y)-(x-3y)=2x[\/latex]. Then we have [latex]x=\\frac{3u-v}2[\/latex]. Moreover, if we simply subtract the second equation from the first, we get [latex]u-v=(x-y)-(x-3y)=2y[\/latex] and\u00a0[latex]y=\\frac{u-v}2[\/latex].<\/p>\r\n\r\n[caption id=\"attachment_1425\" align=\"aligncenter\" width=\"652\"]<img class=\"size-full wp-image-1425\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28043237\/5-7-8.jpeg\" alt=\"A parallelogram R with corners (1, 2), (3, 4), (6, 5), and (4, 3) formed by the lines y = x + 1, y = x minus 1, y = (x + 9)\/3, and y = (x + 5)\/3.\" width=\"652\" height=\"384\" \/> Figure 4.\u00a0A parallelogram in the [latex]xy[\/latex]-plane that we want to transform by a change in variables.[\/caption]\r\n<p id=\"fs-id1167794031264\">Thus, we can choose the transformation<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{T(u,v)=\\left(\\frac{3u-v}2,\\frac{u-v}2\\right)}[\/latex]<\/p>\r\n<p id=\"fs-id1167793292020\">and compute the Jacobian [latex]J(u, v)[\/latex]. We have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{J(u,v)=\\frac{\\partial(x,y)}{\\partial(u,v)}=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{u}}&amp;\\frac{\\partial{x}}{\\partial{v}} \\\\ \\frac{\\partial{y}}{\\partial{u}}&amp;\\frac{\\partial{y}}{\\partial{v}}\\end{vmatrix}=\\begin{vmatrix}3\/2&amp;-1\/2 \\\\ 1\/2&amp;-1\/2\\end{vmatrix}=-\\frac34+\\frac14=-\\frac12}[\/latex].<\/p>\r\nTherefore, [latex]|J(u,v)|=\\frac12[\/latex]. Also, the original integrand becomes\r\n<p style=\"text-align: center;\">[latex]\\large{x-y=\\frac12[3u-v-u+v]=\\frac12[3u-u]=\\frac12[2u]=u}[\/latex].<\/p>\r\nTherefore, by the use of the transformation [latex]T[\/latex], the integral changes to\r\n<p style=\"text-align: center;\">[latex]\\large{\\underset{R}{\\displaystyle\\iint}(x-y)dy \\ dx=\\displaystyle\\int_{-9}^{-5}\\displaystyle\\int_{-1}^{1}J(u,v)u \\ du \\ dv=\\displaystyle\\int_{-9}^{-5}\\displaystyle\\int_{-1}^{1}\\left(\\frac12\\right)u \\ du \\ dv}[\/latex],<\/p>\r\n<p id=\"fs-id1167794155353\">which is much simpler to compute. In fact, it is easily found to be zero. And this is just one example of why we transform integrals like this.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nMake appropriate changes of variables in the integral [latex]\\underset{R}{\\displaystyle\\iint}\\frac4{(x-y)^2}dy \\ dx[\/latex], where [latex]R[\/latex] is the trapezoid bounded by the lines [latex]x-y=2[\/latex], [latex]x-y=4[\/latex], [latex]x=0[\/latex], and [latex]y=0[\/latex]. Write the resulting integral.\r\n\r\n[reveal-answer q=\"249752336\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"249752336\"]\r\n\r\n[latex]x=\\frac12(v+u)[\/latex] and\u00a0[latex]y=\\frac12(u-v)[\/latex] and\u00a0[latex]\\displaystyle\\int_2^4\\displaystyle\\int_{-u}^u\\frac4{u^2}\\left(\\frac12\\right)dv \\ du[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWe are ready to give a problem-solving strategy for change of variables.\r\n<div id=\"fs-id1167794333153\" class=\"problem-solving\" data-type=\"note\">\r\n<div data-type=\"title\">\r\n<div class=\"textbox examples\">\r\n<h3>Problem solving strategy: change of variables<\/h3>\r\n<ol id=\"fs-id1167793940642\" type=\"1\">\r\n \t<li>Sketch the region given by the problem in the [latex]xy[\/latex]-plane and then write the equations of the curves that form the boundary.<\/li>\r\n \t<li>Depending on the region or the integrand, choose the transformations [latex]x=g(u, v)[\/latex] and\u00a0[latex]y=h(u, v)[\/latex].<\/li>\r\n \t<li>Determine the new limits of integration in the [latex]uv[\/latex]-plane.<\/li>\r\n \t<li>Find the Jacobian [latex]J(u, v)[\/latex].<\/li>\r\n \t<li>In the integrand, replace the variables to obtain the new integrand.<\/li>\r\n \t<li>Replace [latex]dy \\ dx[\/latex] or [latex]dx \\ dy[\/latex], whichever occurs, by [latex]J(u,v)du \\ dv[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\nIn the next example, we find a substitution that makes the integrand much simpler to compute.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: evaluating an integral<\/h3>\r\n<p id=\"fs-id1167794162687\">Using the change of variables [latex]u=x-y[\/latex] and [latex]v=x+y[\/latex], evaluate the integral<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\underset{R}{\\displaystyle\\iint}(x-y)e^{x^2-y^2}dA}[\/latex],<\/p>\r\n<p id=\"fs-id1167794138969\">where [latex]R[\/latex] is the region bounded by the lines [latex]x+y=1[\/latex] and [latex]x+y=3[\/latex] and the curves [latex]x^{2}-y^{2}=-1[\/latex] and [latex]x^{2}-y^{2}=1[\/latex] (see the first region in\u00a0Figure 5).<\/p>\r\n[reveal-answer q=\"230474776\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"230474776\"]\r\n\r\nAs before, first find the region [latex]R[\/latex] and picture the transformation so it becomes easier to obtain the limits of integration after the transformations are made (Figure 5).\r\n\r\n[caption id=\"attachment_1426\" align=\"aligncenter\" width=\"796\"]<img class=\"size-full wp-image-1426\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28043332\/5-7-9.jpeg\" alt=\"A parallelogram R with corners (1, 2), (3, 4), (6, 5), and (4, 3) formed by the lines y = x + 1, y = x minus 1, y = (x + 9)\/3, and y = (x + 5)\/3.\" width=\"796\" height=\"255\" \/> Figure 5.\u00a0<span class=\"os-caption\">Transforming the region [latex]R[\/latex] into the region [latex]S[\/latex] to simplify the computation of an integral.<\/span>[\/caption]\r\n<p id=\"fs-id1167794181241\">Given [latex]u=x-y[\/latex] and [latex]v=x+y[\/latex], we have [latex]x=\\frac{u+v}2[\/latex] and [latex]y=\\frac{v-u}2[\/latex] and hence the transformation to use is [latex]T(u,v)=\\left(\\frac{u+v}2,\\frac{v-u}2\\right)[\/latex]. The lines [latex]x+y=1[\/latex] and [latex]x+y=3[\/latex] become [latex]v=1[\/latex] and [latex]v=3[\/latex], respectively. The curves [latex]x^{2}-y^{2}=1[\/latex] and [latex]x^{2}-y^{2}=-1[\/latex] become [latex]uv=1[\/latex] and [latex]uv=-1[\/latex], respectively.<\/p>\r\n<p id=\"fs-id1167794097581\">Thus we can describe the region [latex]S[\/latex] (see the second region\u00a0Figure 5) as<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{s=\\left\\{(u,v)|1\\leq{v}\\leq3,\\frac{-1}v\\leq{u}\\leq\\frac1v\\right\\}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793452878\">The Jacobian for this transformation is<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{J(u,v)=\\frac{\\partial(x,y)}{\\partial(u,v)}=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{u}}&amp;\\frac{\\partial{x}}{\\partial{v}} \\\\ \\frac{\\partial{y}}{\\partial{u}}&amp;\\frac{\\partial{y}}{\\partial{v}}\\end{vmatrix}=\\begin{vmatrix}1\/2&amp;-1\/2 \\\\ 1\/2&amp;1\/2\\end{vmatrix}=\\frac12}[\/latex].<\/p>\r\n<p id=\"fs-id1167793705384\">Therefore, by using the transformation [latex]T[\/latex], the integral changes to<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\underset{R}{\\displaystyle\\iint}(x-y)e^{x^2-y^2}dA=\\frac12\\displaystyle\\int_1^3\\displaystyle\\int_{-1\/v}^{1\/v}ue^{uv}du \\ dv}[\/latex].<\/p>\r\n<p id=\"fs-id1167793589660\">Doing the evaluation, we have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\frac12\\displaystyle\\int_1^3\\displaystyle\\int_{-1\/v}^{1\/v}ue^{uv}du \\ dv=\\frac4{3e}\\approx0.490}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nUsing the substitutions [latex]x=v[\/latex] and [latex]y=\\sqrt{u+v}[\/latex], evaluate the integral [latex]\\underset{R}{\\displaystyle\\iint}y\\sin(y^2-x)dA[\/latex] where [latex]R[\/latex] is the region bounded by the lines [latex]y=\\sqrt{x},\\ x=2[\/latex] and\u00a0[latex]y=0[\/latex].\r\n\r\n[reveal-answer q=\"872845374\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"872845374\"]\r\n\r\n[latex]\\frac12\\sin{2}-1[\/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=8197115&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=xxNKFzqcal0&amp;video_target=tpm-plugin-vljp1ebf-xxNKFzqcal0\" 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\/CP5.47_transcript.html\">transcript for \u201cCP 5.47\u201d here (opens in new window).<\/a><\/center>\r\n<h2 data-type=\"title\">Change of Variables for Triple Integrals<\/h2>\r\n<p id=\"fs-id1167793280478\">Changing variables in triple integrals works in exactly the same way. Cylindrical and spherical coordinate substitutions are special cases of this method, which we demonstrate here.<\/p>\r\nSuppose that [latex]G[\/latex] is a region in [latex]uvw[\/latex]-space and is mapped to [latex]D[\/latex] in [latex]xyz[\/latex]-space (Figure 6) by a one-to-one [latex]C1[\/latex] transformation [latex]T(u, v, w)=(x, y, z)[\/latex] where [latex]x=g(u, v, w)[\/latex],\u00a0[latex]y=h(u, v, w)[\/latex], and\u00a0[latex]z=k(u, v, w)[\/latex].\r\n\r\n[caption id=\"attachment_1427\" align=\"aligncenter\" width=\"930\"]<img class=\"size-full wp-image-1427\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28043418\/5-7-10.jpeg\" alt=\"On the left-hand side of this figure, there is a region G in u v w space. Then there is an arrow from this graph to the right-hand side of the figure marked with x = g(u, v, w), y = h(u, v, w), and z = k(u, v, w). On the right-hand side of this figure there is a region D in xyz space.\" width=\"930\" height=\"350\" \/> Figure 6. A region [latex]G[\/latex] in [latex]uvw[\/latex]-space mapped to a region\u00a0[latex]D[\/latex] in [latex]xyz[\/latex]-space.[\/caption]\r\n<p id=\"fs-id1167793950105\">Then any function [latex]F(x, y, z)[\/latex] defined on [latex]D[\/latex] can be thought of as another function [latex]H(u, v, w)[\/latex] that is defined on [latex]G[\/latex]:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{F(x,y,z)=F(g(u,v,w), \\ h(u,v,w) \\ k(u,v,w))= \\ H(u,v,w)}[\/latex].<\/p>\r\nNow we need to define the Jacobian for three variables.\r\n<div id=\"fs-id1167793957157\" class=\"ui-has-child-title\" data-type=\"note\">\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-id1167793299659\">The Jacobian determinant [latex]J(u, v, w)[\/latex] in three variables is defined as follows:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{J(u,v,w)=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{u}}&amp;\\frac{\\partial{y}}{\\partial{u}}&amp;\\frac{\\partial{z}}{\\partial{u}} \\\\ \\frac{\\partial{x}}{\\partial{v}}&amp;\\frac{\\partial{y}}{\\partial{v}}&amp;\\frac{\\partial{z}}{\\partial{v}} \\\\ \\frac{\\partial{x}}{\\partial{w}}&amp;\\frac{\\partial{y}}{\\partial{w}}&amp;\\frac{\\partial{z}}{\\partial{w}}\\end{vmatrix}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793384577\">This is also the same as<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{J(u,v,w)=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{u}}&amp;\\frac{\\partial{x}}{\\partial{v}}&amp;\\frac{\\partial{x}}{\\partial{w}} \\\\ \\frac{\\partial{y}}{\\partial{u}}&amp;\\frac{\\partial{y}}{\\partial{v}}&amp;\\frac{\\partial{y}}{\\partial{w}} \\\\ \\frac{\\partial{z}}{\\partial{u}}&amp;\\frac{\\partial{z}}{\\partial{v}}&amp;\\frac{\\partial{z}}{\\partial{w}}\\end{vmatrix}}[\/latex].<\/p>\r\n<p id=\"fs-id1167793299607\">The Jacobian can also be simply denoted as [latex]\\frac{\\partial(x,y,z)}{\\partial(u,v,w)}[\/latex].<\/p>\r\n\r\n<\/div>\r\nWith the transformations and the Jacobian for three variables, we are ready to establish the theorem that describes change of variables for triple integrals.\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">theorem: change of variables for triple integrals<\/h3>\r\n\r\n<hr \/>\r\n\r\nLet [latex]T(u, v, w)=(x, y, z)[\/latex] where [latex]x=g(u, v, w)[\/latex], [latex]y=h(u, v, w)[\/latex], and [latex]z=k(u, v, w)[\/latex], be a one-to-one [latex]C1[\/latex] transformation, with a nonzero Jacobian, that maps the region [latex]G[\/latex] in the [latex]uvw[\/latex]-plane into the region [latex]D[\/latex] in the [latex]xyz[\/latex]-plane. As in the two-dimensional case, if [latex]F[\/latex] is continuous on [latex]D[\/latex], then\r\n\r\n[latex]\\hspace{3cm}\\large{\\begin{align}\r\n\r\n\\underset{R}{\\displaystyle\\iiint}F(x,y,z)dV&amp;=\\underset{G}{\\displaystyle\\iiint}F(g(u,v,w), \\ h(u,v,w), \\ k(u,v,w))\\Bigg|\\frac{\\partial(x,y,z)}{\\partial(u,v,w)}\\Bigg|du \\ dv \\ dw \\\\\r\n\r\n&amp;=\\underset{G}{\\displaystyle\\iiint}H(u,v,w)|J(u,v,w)|du \\ dv \\ dw\r\n\r\n\\end{align}}[\/latex].\r\n\r\n<\/div>\r\nLet us now see how changes in triple integrals for cylindrical and spherical coordinates are affected by this theorem. We expect to obtain the same formulas as in\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/introduction-to-triple-integrals-in-cylindrical-and-spherical-coordinates\/\" data-page-slug=\"5-5-triple-integrals-in-cylindrical-and-spherical-coordinates\" data-page-uuid=\"ebd07590-c9ca-4701-95cf-6fa9131d20db\" data-page-fragment=\"page_ebd07590-c9ca-4701-95cf-6fa9131d20db\">Triple Integrals in Cylindrical and Spherical Coordinates<\/a>.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: obtaining formulas in triple integrals for cylindrical and spherical coordinates<\/h3>\r\n<p id=\"fs-id1167793293911\">Derive the formula in triple integrals for<\/p>\r\n\r\n<ol id=\"fs-id1167793293914\" type=\"a\">\r\n \t<li>cylindrical and<\/li>\r\n \t<li>spherical coordinates.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"827454432\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"827454432\"]\r\n<ol id=\"fs-id1167793498790\" type=\"a\">\r\n \t<li>For cylindrical coordinates, the transformation is [latex]T(r,\\theta,z)=(x,y,z)[\/latex] from the Cartesian [latex]r\\theta{z}[\/latex]--plane\u00a0\u00a0to the Cartesian [latex]xyz[\/latex]-plane\u00a0(Figure 7). Here [latex]x=r\\cos\\theta[\/latex], [latex]y=r\\sin\\theta[\/latex], and [latex]z=z[\/latex]. The Jacobian for the transformation is<span data-type=\"newline\">\r\n<\/span>\r\n[latex]\\hspace{1.5cm}\\large{\\begin{align}J(r,\\theta,z)&amp;=\\frac{\\partial(x,y,z)}{\\partial(r,\\theta,z)}=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{r}}&amp;\\frac{\\partial{x}}{\\partial{\\theta}}&amp;\\frac{\\partial{x}}{\\partial{z}} \\\\ \\frac{\\partial{y}}{\\partial{r}}&amp;\\frac{\\partial{y}}{\\partial{\\theta}}&amp;\\frac{\\partial{y}}{\\partial{z}} \\\\ \\frac{\\partial{z}}{\\partial{r}}&amp;\\frac{\\partial{z}}{\\partial{\\theta}}&amp;\\frac{\\partial{z}}{\\partial{z}}\\end{vmatrix} \\\\&amp;=\\begin{vmatrix}\\cos\\theta&amp;-r\\sin\\theta&amp;0 \\\\ \\sin\\theta&amp;r\\cos\\theta&amp;0 \\\\ 0&amp;0&amp;1\\end{vmatrix}=r\\cos^2\\theta+r\\sin^2\\theta=r(\\cos^2\\theta+\\sin^s\\theta)=r \\end{align}}[\/latex].\r\n<span data-type=\"newline\">\r\n<\/span>We know that [latex]r\\geq0[\/latex], so [latex]|J(r,\\theta,z)|=r[\/latex]. Then the triple integral is<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167793640526\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\large{\\underset{D}{\\displaystyle\\iiint}f(x,y,z)dV=\\underset{G}{\\displaystyle\\iiint}f(r\\cos\\theta,r\\sin\\theta,z)r \\ dr \\ d\\theta \\ dz}[\/latex].<\/div>\r\n[caption id=\"attachment_1428\" align=\"aligncenter\" width=\"898\"]<img class=\"size-full wp-image-1428\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28043458\/5-7-11.jpeg\" alt=\"On the left-hand side of this figure, there is a cube G with sides parallel to the coordinate axes in cylindrical coordinate space. Then there is an arrow from this graph to the right-hand side of the figure marked with x = r cos theta, y = r sin theta, and z = z. On the right-hand side of this figure there is a region D in x y z space that is a thick annulus. The top is labeled z = constant, the flat vertical side is labeled theta = constant, and the outermost side is labeled r = constant.\" width=\"898\" height=\"391\" \/> Figure 7. The transformation from rectangular coordinates to cylindrical coordinates can be treated as a change of variables from region [latex]G[\/latex] in [latex]r\\theta{z}[\/latex]-space to region\u00a0[latex]D[\/latex] in [latex]xyz[\/latex]-space.[\/caption]<\/li>\r\n \t<li>For spherical coordinates, the transformation is [latex]T(\\rho,\\theta,\\varphi)=(x,y,z)[\/latex] from the Cartesian [latex]p\\theta\\varphi[\/latex]--plane\u00a0\u00a0to the Cartesian [latex]xyz[\/latex]--plane\u00a0\u00a0(Figure 8). Here [latex]x=\\rho\\sin\\varphi\\cos\\theta[\/latex], [latex]y=\\rho\\sin\\varphi\\sin\\theta[\/latex], and [latex]z=\\rho\\cos\\varphi[\/latex]. The Jacobian for the transformation is<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1167793373656\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\large{J(\\rho,\\theta,\\varphi)=\\frac{\\partial{(x,y,z)}}{\\partial{(\\rho,\\theta\\varphi)}}=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{\\rho}}&amp;\\frac{\\partial{x}}{\\partial{\\theta}}&amp;\\frac{\\partial{x}}{\\partial{\\varphi}} \\\\ \\frac{\\partial{y}}{\\partial{\\rho}}&amp;\\frac{\\partial{y}}{\\partial{\\theta}}&amp;\\frac{\\partial{y}}{\\partial{\\varphi}} \\\\ \\frac{\\partial{z}}{\\partial{\\rho}}&amp;\\frac{\\partial{z}}{\\partial{\\theta}}&amp;\\frac{\\partial{z}}{\\partial{\\varphi}}\\end{vmatrix}=\\begin{vmatrix}\\sin\\varphi\\cos\\theta&amp;-\\rho\\sin\\varphi\\sin\\theta&amp;\\rho\\cos\\varphi\\cos\\theta \\\\ \\sin\\varphi\\sin\\theta&amp;-\\rho\\sin\\varphi\\cos\\theta&amp;\\rho\\cos\\varphi\\sin\\theta \\\\ \\cos\\theta&amp;0&amp;-\\rho\\sin\\varphi\\end{vmatrix}}[\/latex]<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>Expanding the determinant with respect to the third row:<span data-type=\"newline\">\r\n<\/span>\r\n[latex]\\hspace{2cm}\\begin{align}\r\n\r\n&amp;=\\cos\\varphi\\begin{vmatrix}-\\rho\\sin\\varphi\\sin\\theta&amp;\\rho\\cos\\varphi\\cos\\theta \\\\ \\rho\\sin\\varphi\\sin\\theta&amp;\\rho\\cos\\varphi\\sin\\theta\\end{vmatrix}-\\rho\\sin\\varphi\\begin{vmatrix}\\sin\\varphi\\cos\\theta&amp;-\\rho\\sin\\varphi\\sin\\theta \\\\ \\sin\\varphi\\sin\\theta&amp;\\rho\\sin\\varphi\\cos\\theta\\end{vmatrix} \\\\\r\n\r\n&amp;=\\cos\\varphi(-\\rho^2\\sin\\varphi\\cos\\varphi\\sin^2\\theta-\\rho^2\\sin\\varphi\\cos\\varphi\\cos^2\\theta)-\\rho\\sin\\varphi(\\rho\\sin^2\\varphi\\cos^2\\theta+\\rho\\sin^2\\varphi\\sin^2\\theta) \\\\\r\n\r\n&amp;=-\\rho^2\\sin\\varphi\\cos^2\\varphi(\\sin^2\\theta+\\cos^2\\theta)-\\rho^2\\sin\\varphi\\sin^2\\varphi(\\sin^2\\theta+\\cos^2\\theta) \\\\\r\n\r\n&amp;=-\\rho^2\\sin\\varphi\\cos^2\\varphi-\\rho^2\\sin\\varphi\\sin^2\\varphi \\\\\r\n\r\n&amp;=-\\rho^2\\sin\\varphi(\\cos^2\\varphi+\\sin^2\\varphi)=-\\rho^2\\sin\\varphi\\end{align}[\/latex]\r\n<span data-type=\"newline\">\r\n<\/span>Since [latex]0\\leq\\varphi\\leq\\pi[\/latex], we must have [latex]\\sin\\varphi\\geq0[\/latex]. Thus [latex]|J(\\rho,\\theta,\\varphi)|=|-\\rho^2\\sin\\varphi|=\\rho^2\\sin\\varphi[\/latex].\r\n<div id=\"CNX_Calc_Figure_15_07_012\" class=\"os-figure\">\r\n<div class=\"os-caption-container\">[caption id=\"attachment_1429\" align=\"aligncenter\" width=\"895\"]<img class=\"size-full wp-image-1429\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28043542\/5-7-12.jpeg\" alt=\"On the left-hand side of this figure, there is a cube G with sides parallel to the coordinate axes in rho phi theta space. Then there is an arrow from this graph to the right-hand side of the figure marked with x = rho sin phi cos theta, y = rho sin phi sin theta, and z = rho cos phi. On the right-hand side of this figure there is a region D in xyz space that is a thick annulus and has the point (x, y, z) shown as being equal to (rho, phi, theta). The top is labeled phi = constant, the flat vertical side is labeled theta = constant, and the outermost side is labeled rho = constant.\" width=\"895\" height=\"391\" \/> Figure 8. The transformation from rectangular coordinates to cylindrical coordinates can be treated as a change of variables from region [latex]G[\/latex] in [latex]\\rho\\theta\\varphi[\/latex]-space to region\u00a0[latex]D[\/latex] in [latex]xyz[\/latex]-space.[\/caption]<\/div>\r\n<\/div>\r\n<span data-type=\"newline\">\r\n<\/span>Then the triple integral becomes<span data-type=\"newline\">\r\n<\/span>\r\n[latex]\\underset{D}{\\displaystyle\\iiint}f(x,y,z)dV=\\underset{G}{\\displaystyle\\iiint}f(\\rho\\sin\\varphi\\cos\\theta,\\rho\\sin\\varphi\\sin\\theta,\\rho\\cos\\varphi)\\rho^2\\sin\\varphi \\ d\\rho \\ d\\varphi \\ d\\theta[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nLet\u2019s try another example with a different substitution.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: evaluating a triple integral with a change of variables<\/h3>\r\n<p id=\"fs-id1167793376275\">Evaluate the triple integral<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_0^3\\displaystyle\\int_0^4\\displaystyle\\int_{y\/2}^{(y\/2)+1}\\left(x+\\frac{z}3\\right)dx \\ dy \\ dz}[\/latex]<\/p>\r\n<p id=\"fs-id1167793638297\">in [latex]xyz[\/latex]-space by using the transformation<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{u=(2x-y)\/2, \\ v=y\/2,\\text{ and }w=z\/3}[\/latex].<\/p>\r\n<p id=\"fs-id1167793930128\">Then integrate over an appropriate region in [latex]uvw[\/latex]-space<\/p>\r\n[reveal-answer q=\"458273487\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"458273487\"]\r\n<p id=\"fs-id1167793930147\">As before, some kind of sketch of the region [latex]G[\/latex] in [latex]xyz[\/latex]-space over which we have to perform the integration can help identify the region [latex]D[\/latex] in [latex]uvw[\/latex]-space (Figure 9). Clearly [latex]G[\/latex] in [latex]xyz[\/latex]-space is bounded by the planes [latex]x=y\/2[\/latex], [latex]x=(y\/2)+1[\/latex], [latex]y=0[\/latex],[latex]y=4[\/latex], [latex]z=0[\/latex], and [latex]z=4[\/latex]. We also know that we have to use [latex]u=(2x-y)\/2[\/latex], [latex]v=y\/2[\/latex] and [latex]w=z\/3[\/latex] for the transformations. We need to solve for [latex]x[\/latex], [latex]y[\/latex], and [latex]z[\/latex]. Here we find that\u00a0[latex]x=u+v[\/latex],\u00a0[latex]y=2v[\/latex], and\u00a0[latex]z=3w[\/latex].<\/p>\r\n<p id=\"fs-id1167794223120\">Using elementary algebra, we can find the corresponding surfaces for the region [latex]G[\/latex] and the limits of integration in [latex]uvw[\/latex]-space. It is convenient to list these equations in a table.<\/p>\r\n\r\n<div id=\"fs-id1167794223141\" class=\"os-table\">\r\n<table class=\"unnumbered\" style=\"height: 166px; width: 622px;\" data-id=\"fs-id1167794223141\" data-label=\"\">\r\n<thead>\r\n<tr style=\"height: 24px;\" valign=\"top\">\r\n<th style=\"height: 24px; width: 259.734px;\" scope=\"col\" data-valign=\"top\" data-align=\"left\">Equations in [latex]xyz[\/latex] for the region [latex]D[\/latex]<\/th>\r\n<th style=\"height: 24px; width: 204.016px;\" scope=\"col\" data-valign=\"top\" data-align=\"left\">Corresponding equations in [latex]uvw[\/latex] for the region [latex]G[\/latex]<\/th>\r\n<th style=\"height: 24px; width: 123.219px;\" scope=\"col\" data-valign=\"top\" data-align=\"left\">Limits for the integration in [latex]uvw[\/latex]<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr style=\"height: 24px;\" valign=\"top\">\r\n<td style=\"height: 22px; width: 259.734px;\" data-valign=\"top\" data-align=\"left\">[latex]x=y\/2[\/latex]<\/td>\r\n<td style=\"height: 22px; width: 204.016px;\" data-valign=\"top\" data-align=\"left\">[latex]u+v=2v\/2=v[\/latex]<\/td>\r\n<td style=\"height: 22px; width: 123.219px;\" data-valign=\"top\" data-align=\"left\">[latex]u=0[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 24px;\" valign=\"top\">\r\n<td style=\"height: 24px; width: 259.734px;\" data-valign=\"top\" data-align=\"left\">[latex]x=y\/2[\/latex]<\/td>\r\n<td style=\"height: 24px; width: 204.016px;\" data-valign=\"top\" data-align=\"left\">[latex]u+v=(2v\/2)+1=v+1[\/latex]<\/td>\r\n<td style=\"height: 24px; width: 123.219px;\" data-valign=\"top\" data-align=\"left\">[latex]u=1[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 24px;\" valign=\"top\">\r\n<td style=\"height: 24px; width: 259.734px;\" data-valign=\"top\" data-align=\"left\">[latex]y=0[\/latex]<\/td>\r\n<td style=\"height: 24px; width: 204.016px;\" data-valign=\"top\" data-align=\"left\">[latex]2v=0[\/latex]<\/td>\r\n<td style=\"height: 24px; width: 123.219px;\" data-valign=\"top\" data-align=\"left\">[latex]v=0[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 24px;\" valign=\"top\">\r\n<td style=\"height: 24px; width: 259.734px;\" data-valign=\"top\" data-align=\"left\">[latex]y=4[\/latex]<\/td>\r\n<td style=\"height: 24px; width: 204.016px;\" data-valign=\"top\" data-align=\"left\">[latex]2v=4[\/latex]<\/td>\r\n<td style=\"height: 24px; width: 123.219px;\" data-valign=\"top\" data-align=\"left\">[latex]v=2[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 24px;\" valign=\"top\">\r\n<td style=\"height: 24px; width: 259.734px;\" data-valign=\"top\" data-align=\"left\">[latex]z=0[\/latex]<\/td>\r\n<td style=\"height: 24px; width: 204.016px;\" data-valign=\"top\" data-align=\"left\">[latex]3w=0[\/latex]<\/td>\r\n<td style=\"height: 24px; width: 123.219px;\" data-valign=\"top\" data-align=\"left\">[latex]w=0[\/latex]<\/td>\r\n<\/tr>\r\n<tr style=\"height: 24px;\" valign=\"top\">\r\n<td style=\"height: 24px; width: 259.734px;\" data-valign=\"top\" data-align=\"left\">[latex]z=3[\/latex]<\/td>\r\n<td style=\"height: 24px; width: 204.016px;\" data-valign=\"top\" data-align=\"left\">[latex]3w=3[\/latex]<\/td>\r\n<td style=\"height: 24px; width: 123.219px;\" data-valign=\"top\" data-align=\"left\">[latex]w=1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div id=\"CNX_Calc_Figure_15_07_013\" class=\"os-figure\">\r\n<div class=\"os-caption-container\">[caption id=\"attachment_1430\" align=\"aligncenter\" width=\"894\"]<img class=\"size-full wp-image-1430\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28043626\/5-7-13.jpeg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20210823.155019\/resources\/9ac91afe22764bb4840cdcc0310bae65f7605c69&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;On the left-hand side of this figure, there is a box G with sides 1, 2, and 1 along the u, v, and w axes, respectively. Then there is an arrow from this graph to the right-hand side of the figure marked with x = u + v, y = 2v, and z = 3w. On the right-hand side of this figure there is a region D in xyz space that is a rotated box with sides 1, 4, and 3 along the x, y, and z axes. The rear plane is marked x = y\/2 or y = 2x. The front plane is marked x = y\/2 + 1 or y = 2x minus 2.&quot; id=&quot;34&quot;&gt;\" width=\"894\" height=\"435\" \/> Figure 9. The region [latex]G[\/latex] in [latex]uvw[\/latex]-space is transformed to region [latex]D[\/latex] in [latex]xyz[\/latex]-space[\/caption]<\/div>\r\n<div><\/div>\r\n<\/div>\r\n<p id=\"fs-id1167793607338\">Now we can calculate the Jacobian for the transformation:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{J(u,v,w)=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{u}}&amp;\\frac{\\partial{x}}{\\partial{v}}&amp;\\frac{\\partial{x}}{\\partial{w}} \\\\ \\frac{\\partial{y}}{\\partial{u}}&amp;\\frac{\\partial{y}}{\\partial{v}}&amp;\\frac{\\partial{y}}{\\partial{w}} \\\\ \\frac{\\partial{z}}{\\partial{u}}&amp;\\frac{\\partial{z}}{\\partial{v}}&amp;\\frac{\\partial{z}}{\\partial{w}}\\end{vmatrix}=\\begin{vmatrix}1&amp;1&amp;0 \\\\ 0&amp;2&amp;0 \\\\ 0&amp;0&amp;3\\end{vmatrix}=6}[\/latex].<\/p>\r\n<p id=\"fs-id1167794296460\">The function to be integrated becomes<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{f(x,y,z)=x+\\frac{z}3=u+v+\\frac{3w}3=u+v+w}[\/latex].<\/p>\r\n<p id=\"fs-id1167793420356\">We are now ready to put everything together and complete the problem.<\/p>\r\n[latex]\\hspace{3cm}\\large{\\begin{align}\r\n\r\n&amp;\\hspace{1cm}\\displaystyle\\int_0^3\\displaystyle\\int_0^4\\displaystyle\\int_{y\/2}^{(y\/2)+1}\\left(x+\\frac{z}3\\right)dx \\ dy \\ dz \\\\\r\n\r\n&amp;=\\displaystyle\\int_0^1\\displaystyle\\int_0^2\\displaystyle\\int_0^1(u+v+w)|J(u,v,w)|du \\ dv \\ dw=\\displaystyle\\int_0^1\\displaystyle\\int_0^2\\displaystyle\\int_0^1(u+v+w)|6|du \\ dv \\ dw \\\\\r\n\r\n&amp;=6\\displaystyle\\int_0^1\\displaystyle\\int_0^2\\displaystyle\\int_0^1(u+v+w)du \\ dv \\ dw=6\\displaystyle\\int_0^1\\displaystyle\\int_0^2\\left[\\frac{u^2}2+vu+wu\\right]_0^1dv \\ dw \\\\\r\n\r\n&amp;=6\\displaystyle\\int_0^1\\displaystyle\\int_0^2\\left(\\frac12+v+w\\right)dv \\ dw=6\\displaystyle\\int_0^1\\left[\\frac12v+\\frac{v^2}2+wv\\right]_0^2dw \\\\\r\n\r\n&amp;=6\\displaystyle\\int_0^1(3+2w)dw=6\\left[3w+w^2\\right]_0^1=24\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\n<p id=\"fs-id1167794212114\">Let [latex]D[\/latex] be the region in [latex]xyz[\/latex]-space defined by [latex]1\\leq{x}\\leq2, \\ 0\\leq{xy}\\leq2, \\text{and} \\ 0\\leq{z}\\leq1[\/latex].<\/p>\r\n<p id=\"fs-id1167793420423\">Evaluate [latex]\\underset{D}{\\displaystyle\\iiint}(x^2y+3xyz)dx \\ dy \\ dz[\/latex] by using the transformation [latex]u=x[\/latex],\u00a0[latex]v=xy[\/latex], and\u00a0[latex]w=3z[\/latex].<\/p>\r\n[reveal-answer q=\"166409830\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"166409830\"]\r\n\r\n[latex]\\displaystyle\\int_0^3\\displaystyle\\int_0^2\\displaystyle\\int_1^2\\left(\\frac{v}3+\\frac{vw}{3u}\\right)dy \\ dv \\ dw=2+\\ln8 = 2+ 3\\ln2[\/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=8197116&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=TChzX4WNbC4&amp;video_target=tpm-plugin-bxj0q4mi-TChzX4WNbC4\" 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\/CP5.48_transcript.html\">transcript for \u201cCP 5.48\u201d here (opens in new window).<\/a><\/center><\/div>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul class=\"os-abstract\">\n<li><span class=\"os-abstract-content\">Evaluate a double integral using a change of variables.<\/span><\/li>\n<li><span class=\"os-abstract-content\">Evaluate a triple integral using a change of variables.<\/span><\/li>\n<\/ul>\n<\/div>\n<h2 data-type=\"title\">Change of Variables for Double Integrals<\/h2>\n<p id=\"fs-id1167794176297\">We have already seen that, under the change of variables [latex]T(u, v)=(x, y)[\/latex] where [latex]x=g(u, v)[\/latex] and [latex]y=h(u, v)[\/latex], a small region [latex]{\\Delta}{A}[\/latex] in the [latex]xy[\/latex]-plane is related to the area formed by the product [latex]{\\Delta}{u}{\\Delta}{v}[\/latex] in the [latex]uv[\/latex]-plane by the approximation<\/p>\n<p style=\"text-align: center;\">[latex]\\large{{\\Delta}{A}{\\approx}{J}{(u,v)}{\\Delta}{u},{\\Delta}{v}}[\/latex].<\/p>\n<p>Now let\u2019s go back to the definition of double integral for a minute:<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\underset{R}{\\displaystyle\\iint}f(x,y) \\ dA=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^n{f}(x_{ij},y_{ij})\\Delta{A}}[\/latex].<\/p>\n<p>Referring to\u00a0Figure 1, observe that we divided the region [latex]S[\/latex] in the [latex]uv[\/latex]-plane into small subrectangles [latex]S_{ij}[\/latex] and we let the subrectangles [latex]R_{ij}[\/latex] in the [latex]xy[\/latex]-plane be the images of [latex]S_{ij}[\/latex] under the transformation [latex]T(u, v)=(x, y)[\/latex].<\/p>\n<div id=\"attachment_1422\" style=\"width: 775px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1422\" class=\"size-full wp-image-1422\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28043011\/5-7-5.jpeg\" alt=\"On the left-hand side of this figure, there is a rectangle S with an inscribed red oval and a subrectangle with lower right corner point (u sub ij, v sub ij), height Delta v, and length Delta u given in the Cartesian u v-plane. Then there is an arrow from this graph to the right-hand side of the figure marked with T. On the right-hand side of this figure there is a region R with inscribed (deformed) red oval and a subrectangle R sub ij with corner point (x sub ij, y sub ij) given in the Cartesian x y-plane. The subrectangle is blown up and shown with vectors pointing along the edge from the corner point.\" width=\"765\" height=\"321\" \/><\/p>\n<p id=\"caption-attachment-1422\" class=\"wp-caption-text\">Figure 1.\u00a0The subrectangles [latex]S_{ij}[\/latex] in the\u00a0[latex]uv[\/latex]-plane transform into subrectangles [latex]R_{ij}[\/latex] in the\u00a0[latex]xy[\/latex]-plane.<\/p>\n<\/div>\n<p id=\"fs-id1167794052254\">Then the double integral becomes<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\underset{R}{\\displaystyle\\iint}f(x,y) \\ dA=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^n{f}(x_{ij},y_{ij})\\Delta{A}=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^nf(g(u_{ij},v_{ij}),h(u_{ij},v_{ij}))|J(u_{ij},v_{ij})|\\Delta{u}\\Delta{v}}[\/latex].<\/p>\n<p>Notice this is exactly the double Riemann sum for the integral<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\underset{S}{\\displaystyle\\iint}f(g(u,v),h(u,v))\\Bigg|\\frac{\\partial(x,y)}{\\partial(u,v)}\\Bigg|du \\ dv}[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">theorem: change of variables for double integrals<\/h3>\n<hr \/>\n<p id=\"fs-id1167793433530\">Let [latex]T(u, v)=(x, y)[\/latex] where [latex]x=g(u, v)[\/latex] and [latex]y=h(u, v)[\/latex] be a one-to-one [latex]C1[\/latex] transformation, with a nonzero Jacobian on the interior of the region [latex]S[\/latex] in the [latex]uv[\/latex]-plane; it maps [latex]S[\/latex] into the region [latex]R[\/latex] in the [latex]xy[\/latex]-plane. If [latex]f[\/latex] is continuous on [latex]R[\/latex], then<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\underset{R}{\\displaystyle\\iint}f(x,y) \\ dA=\\underset{S}{\\displaystyle\\iint}f(g(u,v),h(u,v))\\Bigg|\\frac{\\partial(x,y)}{\\partial(u,v)}\\Bigg|du \\ dv}[\/latex].<\/p>\n<\/div>\n<p id=\"fs-id1167793976062\">With this theorem for double integrals, we can change the variables from [latex](x, y)[\/latex] to [latex](u, v)[\/latex] in a double integral simply by replacing<\/p>\n<p style=\"text-align: center;\">[latex]\\large{dA=dx \\ dy=\\Bigg|\\frac{\\partial(x,y)}{\\partial(u,v)}\\Bigg|du \\ dv}[\/latex]<\/p>\n<p>when we use the substitutions [latex]x=g(u, v)[\/latex] and [latex]y=h(u, v)[\/latex] and then change the limits of integration accordingly. This change of variables often makes any computations much simpler.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: changing variables from rectangular to polar coordinates<\/h3>\n<p id=\"fs-id1167793926832\">Consider the integral<\/p>\n<p style=\"text-align: center;\">[latex]\\displaystyle\\int_0^2\\displaystyle\\int_0^{\\sqrt{2x-x^2}}\\sqrt{x^2+y^2}dy \\ dx[\/latex].<\/p>\n<p id=\"fs-id1167793481509\">Use the change of variables [latex]x=r\\cos\\theta[\/latex] and [latex]y=r\\sin\\theta[\/latex], and find the resulting integral.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q598345190\">Show Solution<\/span><\/p>\n<div id=\"q598345190\" class=\"hidden-answer\" style=\"display: none\">\n<p>First we need to find the region of integration. This region is bounded below by [latex]y=0[\/latex] and above by [latex]y=\\sqrt{2x-x^2}[\/latex] (see the following figure).<\/p>\n<div id=\"attachment_1423\" style=\"width: 300px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1423\" class=\"size-full wp-image-1423\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28043128\/5-7-6.jpeg\" alt=\"A semicircle in the first quadrant of the xy plane with radius 1 and center (1, 0). The equation for this curve is given as y = the square root of (2x minus x squared)\" width=\"290\" height=\"234\" \/><\/p>\n<p id=\"caption-attachment-1423\" class=\"wp-caption-text\">Figure 2.\u00a0Changing a region from rectangular to polar coordinates.<\/p>\n<\/div>\n<p id=\"fs-id1167794215044\">Squaring and collecting terms, we find that the region is the upper half of the circle [latex]x^{2}+y^{2}-2x=0[\/latex] that is, [latex]y^{2}+(x-1)^{2}[\/latex]. In polar coordinates, the circle is [latex]r=2\\cos\\theta[\/latex] so the region of integration in polar coordinates is bounded by [latex]0\\leq{r}\\leq\\cos\\theta[\/latex] and [latex]0\\leq\\theta\\leq\\frac{\\pi}2[\/latex].<\/p>\n<p id=\"fs-id1167793618382\">The Jacobian is [latex]J(r,\\theta)=r[\/latex], as shown in\u00a0Example &#8220;Finding the Jacobian&#8221;. Since [latex]r\\geq0[\/latex], we have [latex]|J(r,\\theta)|=r[\/latex].<\/p>\n<p id=\"fs-id1167793424417\">The integrand [latex]\\sqrt{x^2+y^2}[\/latex] changes to [latex]r[\/latex] in polar coordinates, so the double iterated integral is<\/p>\n<p>[latex]\\displaystyle\\int_0^2\\displaystyle\\int_0^{\\sqrt{2x-x^2}}\\sqrt{x^2+y^2}dy \\ dx=\\displaystyle\\int_0^{\\pi\/2}\\displaystyle\\int_0^{2\\cos\\theta}r|J(r,\\theta)|dr \\ d\\theta=\\displaystyle\\int_0^{\\pi\/2}\\displaystyle\\int_0^{2\\cos\\theta}r^2 \\ dr \\ d\\theta[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Considering the integral [latex]\\displaystyle\\int_0^1\\displaystyle\\int_0^{\\sqrt{1-x^2}}(x^2+y^2)dy \\ dx[\/latex], use the change of variables [latex]x=r\\cos\\theta[\/latex] and [latex]y=\\sin\\theta[\/latex], and find the resulting integral.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q983458271\">Show Solution<\/span><\/p>\n<div id=\"q983458271\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\displaystyle\\int_0^{\\pi\/2}\\displaystyle\\int_0^1r^3 \\ dr \\ d\\theta[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Notice in the next example that the region over which we are to integrate may suggest a suitable transformation for the integration. This is a common and important situation.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: changing variables<\/h3>\n<p>Consider the integral [latex]\\underset{R}{\\displaystyle\\iint}(x-y)dy \\ dx[\/latex], where [latex]R[\/latex] is the parallelogram joining the points [latex](1, 2)[\/latex], [latex](3, 4)[\/latex], [latex](4, 3)[\/latex], and [latex](6, 5)[\/latex] (Figure 3). Make appropriate changes of variables, and write the resulting integral.<\/p>\n<div id=\"attachment_1424\" style=\"width: 338px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1424\" class=\"size-full wp-image-1424\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28043209\/5-7-7.jpeg\" alt=\"A semicircle in the first quadrant of the xy plane with radius 1 and center (1, 0). The equation for this curve is given as y = the square root of (2x minus x squared)\" width=\"328\" height=\"272\" \/><\/p>\n<p id=\"caption-attachment-1424\" class=\"wp-caption-text\">Figure 3.\u00a0The region of integration for the given integral.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q134952148\">Show Solution<\/span><\/p>\n<div id=\"q134952148\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794171741\">First, we need to understand the region over which we are to integrate. The sides of the parallelogram are [latex]x-y+1=0[\/latex],\u00a0[latex]x-y-1=0[\/latex],\u00a0[latex]x-3y+5=0[\/latex], and\u00a0[latex]x-3y+9=0[\/latex] (Figure 4). Another way to look at them is [latex]x-y=-1[\/latex],\u00a0[latex]x-y=1[\/latex],\u00a0[latex]x-3y=-5[\/latex], and\u00a0[latex]x-3y=-9[\/latex].<\/p>\n<p id=\"fs-id1167793951713\">Clearly the parallelogram is bounded by the lines\u00a0[latex]y=x+1[\/latex],\u00a0[latex]y=x-1[\/latex],\u00a0[latex]y=\\frac13(x+5)[\/latex], and\u00a0[latex]y=\\frac13(x+9)[\/latex].<\/p>\n<p id=\"fs-id1167793924325\">Notice that if we were to make [latex]u=x-y[\/latex] and [latex]v=x-3y[\/latex], then the limits on the integral would be [latex]-1\\leq{u}\\leq1[\/latex] and\u00a0[latex]-9\\leq{v}\\leq-5[\/latex].<\/p>\n<p id=\"fs-id1167793931696\">To solve for [latex]x[\/latex] and [latex]y[\/latex], we multiply the first equation by [latex]3[\/latex] and subtract the second equation, [latex]3u-v=(3x-3y)-(x-3y)=2x[\/latex]. Then we have [latex]x=\\frac{3u-v}2[\/latex]. Moreover, if we simply subtract the second equation from the first, we get [latex]u-v=(x-y)-(x-3y)=2y[\/latex] and\u00a0[latex]y=\\frac{u-v}2[\/latex].<\/p>\n<div id=\"attachment_1425\" style=\"width: 662px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1425\" class=\"size-full wp-image-1425\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28043237\/5-7-8.jpeg\" alt=\"A parallelogram R with corners (1, 2), (3, 4), (6, 5), and (4, 3) formed by the lines y = x + 1, y = x minus 1, y = (x + 9)\/3, and y = (x + 5)\/3.\" width=\"652\" height=\"384\" \/><\/p>\n<p id=\"caption-attachment-1425\" class=\"wp-caption-text\">Figure 4.\u00a0A parallelogram in the [latex]xy[\/latex]-plane that we want to transform by a change in variables.<\/p>\n<\/div>\n<p id=\"fs-id1167794031264\">Thus, we can choose the transformation<\/p>\n<p style=\"text-align: center;\">[latex]\\large{T(u,v)=\\left(\\frac{3u-v}2,\\frac{u-v}2\\right)}[\/latex]<\/p>\n<p id=\"fs-id1167793292020\">and compute the Jacobian [latex]J(u, v)[\/latex]. We have<\/p>\n<p style=\"text-align: center;\">[latex]\\large{J(u,v)=\\frac{\\partial(x,y)}{\\partial(u,v)}=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{u}}&\\frac{\\partial{x}}{\\partial{v}} \\\\ \\frac{\\partial{y}}{\\partial{u}}&\\frac{\\partial{y}}{\\partial{v}}\\end{vmatrix}=\\begin{vmatrix}3\/2&-1\/2 \\\\ 1\/2&-1\/2\\end{vmatrix}=-\\frac34+\\frac14=-\\frac12}[\/latex].<\/p>\n<p>Therefore, [latex]|J(u,v)|=\\frac12[\/latex]. Also, the original integrand becomes<\/p>\n<p style=\"text-align: center;\">[latex]\\large{x-y=\\frac12[3u-v-u+v]=\\frac12[3u-u]=\\frac12[2u]=u}[\/latex].<\/p>\n<p>Therefore, by the use of the transformation [latex]T[\/latex], the integral changes to<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\underset{R}{\\displaystyle\\iint}(x-y)dy \\ dx=\\displaystyle\\int_{-9}^{-5}\\displaystyle\\int_{-1}^{1}J(u,v)u \\ du \\ dv=\\displaystyle\\int_{-9}^{-5}\\displaystyle\\int_{-1}^{1}\\left(\\frac12\\right)u \\ du \\ dv}[\/latex],<\/p>\n<p id=\"fs-id1167794155353\">which is much simpler to compute. In fact, it is easily found to be zero. And this is just one example of why we transform integrals like this.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Make appropriate changes of variables in the integral [latex]\\underset{R}{\\displaystyle\\iint}\\frac4{(x-y)^2}dy \\ dx[\/latex], where [latex]R[\/latex] is the trapezoid bounded by the lines [latex]x-y=2[\/latex], [latex]x-y=4[\/latex], [latex]x=0[\/latex], and [latex]y=0[\/latex]. Write the resulting integral.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q249752336\">Show Solution<\/span><\/p>\n<div id=\"q249752336\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=\\frac12(v+u)[\/latex] and\u00a0[latex]y=\\frac12(u-v)[\/latex] and\u00a0[latex]\\displaystyle\\int_2^4\\displaystyle\\int_{-u}^u\\frac4{u^2}\\left(\\frac12\\right)dv \\ du[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>We are ready to give a problem-solving strategy for change of variables.<\/p>\n<div id=\"fs-id1167794333153\" class=\"problem-solving\" data-type=\"note\">\n<div data-type=\"title\">\n<div class=\"textbox examples\">\n<h3>Problem solving strategy: change of variables<\/h3>\n<ol id=\"fs-id1167793940642\" type=\"1\">\n<li>Sketch the region given by the problem in the [latex]xy[\/latex]-plane and then write the equations of the curves that form the boundary.<\/li>\n<li>Depending on the region or the integrand, choose the transformations [latex]x=g(u, v)[\/latex] and\u00a0[latex]y=h(u, v)[\/latex].<\/li>\n<li>Determine the new limits of integration in the [latex]uv[\/latex]-plane.<\/li>\n<li>Find the Jacobian [latex]J(u, v)[\/latex].<\/li>\n<li>In the integrand, replace the variables to obtain the new integrand.<\/li>\n<li>Replace [latex]dy \\ dx[\/latex] or [latex]dx \\ dy[\/latex], whichever occurs, by [latex]J(u,v)du \\ dv[\/latex].<\/li>\n<\/ol>\n<\/div>\n<p>In the next example, we find a substitution that makes the integrand much simpler to compute.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: evaluating an integral<\/h3>\n<p id=\"fs-id1167794162687\">Using the change of variables [latex]u=x-y[\/latex] and [latex]v=x+y[\/latex], evaluate the integral<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\underset{R}{\\displaystyle\\iint}(x-y)e^{x^2-y^2}dA}[\/latex],<\/p>\n<p id=\"fs-id1167794138969\">where [latex]R[\/latex] is the region bounded by the lines [latex]x+y=1[\/latex] and [latex]x+y=3[\/latex] and the curves [latex]x^{2}-y^{2}=-1[\/latex] and [latex]x^{2}-y^{2}=1[\/latex] (see the first region in\u00a0Figure 5).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q230474776\">Show Solution<\/span><\/p>\n<div id=\"q230474776\" class=\"hidden-answer\" style=\"display: none\">\n<p>As before, first find the region [latex]R[\/latex] and picture the transformation so it becomes easier to obtain the limits of integration after the transformations are made (Figure 5).<\/p>\n<div id=\"attachment_1426\" style=\"width: 806px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1426\" class=\"size-full wp-image-1426\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28043332\/5-7-9.jpeg\" alt=\"A parallelogram R with corners (1, 2), (3, 4), (6, 5), and (4, 3) formed by the lines y = x + 1, y = x minus 1, y = (x + 9)\/3, and y = (x + 5)\/3.\" width=\"796\" height=\"255\" \/><\/p>\n<p id=\"caption-attachment-1426\" class=\"wp-caption-text\">Figure 5.\u00a0<span class=\"os-caption\">Transforming the region [latex]R[\/latex] into the region [latex]S[\/latex] to simplify the computation of an integral.<\/span><\/p>\n<\/div>\n<p id=\"fs-id1167794181241\">Given [latex]u=x-y[\/latex] and [latex]v=x+y[\/latex], we have [latex]x=\\frac{u+v}2[\/latex] and [latex]y=\\frac{v-u}2[\/latex] and hence the transformation to use is [latex]T(u,v)=\\left(\\frac{u+v}2,\\frac{v-u}2\\right)[\/latex]. The lines [latex]x+y=1[\/latex] and [latex]x+y=3[\/latex] become [latex]v=1[\/latex] and [latex]v=3[\/latex], respectively. The curves [latex]x^{2}-y^{2}=1[\/latex] and [latex]x^{2}-y^{2}=-1[\/latex] become [latex]uv=1[\/latex] and [latex]uv=-1[\/latex], respectively.<\/p>\n<p id=\"fs-id1167794097581\">Thus we can describe the region [latex]S[\/latex] (see the second region\u00a0Figure 5) as<\/p>\n<p style=\"text-align: center;\">[latex]\\large{s=\\left\\{(u,v)|1\\leq{v}\\leq3,\\frac{-1}v\\leq{u}\\leq\\frac1v\\right\\}}[\/latex].<\/p>\n<p id=\"fs-id1167793452878\">The Jacobian for this transformation is<\/p>\n<p style=\"text-align: center;\">[latex]\\large{J(u,v)=\\frac{\\partial(x,y)}{\\partial(u,v)}=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{u}}&\\frac{\\partial{x}}{\\partial{v}} \\\\ \\frac{\\partial{y}}{\\partial{u}}&\\frac{\\partial{y}}{\\partial{v}}\\end{vmatrix}=\\begin{vmatrix}1\/2&-1\/2 \\\\ 1\/2&1\/2\\end{vmatrix}=\\frac12}[\/latex].<\/p>\n<p id=\"fs-id1167793705384\">Therefore, by using the transformation [latex]T[\/latex], the integral changes to<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\underset{R}{\\displaystyle\\iint}(x-y)e^{x^2-y^2}dA=\\frac12\\displaystyle\\int_1^3\\displaystyle\\int_{-1\/v}^{1\/v}ue^{uv}du \\ dv}[\/latex].<\/p>\n<p id=\"fs-id1167793589660\">Doing the evaluation, we have<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\frac12\\displaystyle\\int_1^3\\displaystyle\\int_{-1\/v}^{1\/v}ue^{uv}du \\ dv=\\frac4{3e}\\approx0.490}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Using the substitutions [latex]x=v[\/latex] and [latex]y=\\sqrt{u+v}[\/latex], evaluate the integral [latex]\\underset{R}{\\displaystyle\\iint}y\\sin(y^2-x)dA[\/latex] where [latex]R[\/latex] is the region bounded by the lines [latex]y=\\sqrt{x},\\ x=2[\/latex] and\u00a0[latex]y=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q872845374\">Show Solution<\/span><\/p>\n<div id=\"q872845374\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\frac12\\sin{2}-1[\/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=8197115&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=xxNKFzqcal0&amp;video_target=tpm-plugin-vljp1ebf-xxNKFzqcal0\" 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\/CP5.47_transcript.html\">transcript for \u201cCP 5.47\u201d here (opens in new window).<\/a><\/div>\n<h2 data-type=\"title\">Change of Variables for Triple Integrals<\/h2>\n<p id=\"fs-id1167793280478\">Changing variables in triple integrals works in exactly the same way. Cylindrical and spherical coordinate substitutions are special cases of this method, which we demonstrate here.<\/p>\n<p>Suppose that [latex]G[\/latex] is a region in [latex]uvw[\/latex]-space and is mapped to [latex]D[\/latex] in [latex]xyz[\/latex]-space (Figure 6) by a one-to-one [latex]C1[\/latex] transformation [latex]T(u, v, w)=(x, y, z)[\/latex] where [latex]x=g(u, v, w)[\/latex],\u00a0[latex]y=h(u, v, w)[\/latex], and\u00a0[latex]z=k(u, v, w)[\/latex].<\/p>\n<div id=\"attachment_1427\" style=\"width: 940px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1427\" class=\"size-full wp-image-1427\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28043418\/5-7-10.jpeg\" alt=\"On the left-hand side of this figure, there is a region G in u v w space. Then there is an arrow from this graph to the right-hand side of the figure marked with x = g(u, v, w), y = h(u, v, w), and z = k(u, v, w). On the right-hand side of this figure there is a region D in xyz space.\" width=\"930\" height=\"350\" \/><\/p>\n<p id=\"caption-attachment-1427\" class=\"wp-caption-text\">Figure 6. A region [latex]G[\/latex] in [latex]uvw[\/latex]-space mapped to a region\u00a0[latex]D[\/latex] in [latex]xyz[\/latex]-space.<\/p>\n<\/div>\n<p id=\"fs-id1167793950105\">Then any function [latex]F(x, y, z)[\/latex] defined on [latex]D[\/latex] can be thought of as another function [latex]H(u, v, w)[\/latex] that is defined on [latex]G[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\large{F(x,y,z)=F(g(u,v,w), \\ h(u,v,w) \\ k(u,v,w))= \\ H(u,v,w)}[\/latex].<\/p>\n<p>Now we need to define the Jacobian for three variables.<\/p>\n<div id=\"fs-id1167793957157\" class=\"ui-has-child-title\" data-type=\"note\">\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">definition<\/h3>\n<hr \/>\n<p id=\"fs-id1167793299659\">The Jacobian determinant [latex]J(u, v, w)[\/latex] in three variables is defined as follows:<\/p>\n<p style=\"text-align: center;\">[latex]\\large{J(u,v,w)=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{u}}&\\frac{\\partial{y}}{\\partial{u}}&\\frac{\\partial{z}}{\\partial{u}} \\\\ \\frac{\\partial{x}}{\\partial{v}}&\\frac{\\partial{y}}{\\partial{v}}&\\frac{\\partial{z}}{\\partial{v}} \\\\ \\frac{\\partial{x}}{\\partial{w}}&\\frac{\\partial{y}}{\\partial{w}}&\\frac{\\partial{z}}{\\partial{w}}\\end{vmatrix}}[\/latex].<\/p>\n<p id=\"fs-id1167793384577\">This is also the same as<\/p>\n<p style=\"text-align: center;\">[latex]\\large{J(u,v,w)=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{u}}&\\frac{\\partial{x}}{\\partial{v}}&\\frac{\\partial{x}}{\\partial{w}} \\\\ \\frac{\\partial{y}}{\\partial{u}}&\\frac{\\partial{y}}{\\partial{v}}&\\frac{\\partial{y}}{\\partial{w}} \\\\ \\frac{\\partial{z}}{\\partial{u}}&\\frac{\\partial{z}}{\\partial{v}}&\\frac{\\partial{z}}{\\partial{w}}\\end{vmatrix}}[\/latex].<\/p>\n<p id=\"fs-id1167793299607\">The Jacobian can also be simply denoted as [latex]\\frac{\\partial(x,y,z)}{\\partial(u,v,w)}[\/latex].<\/p>\n<\/div>\n<p>With the transformations and the Jacobian for three variables, we are ready to establish the theorem that describes change of variables for triple integrals.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">theorem: change of variables for triple integrals<\/h3>\n<hr \/>\n<p>Let [latex]T(u, v, w)=(x, y, z)[\/latex] where [latex]x=g(u, v, w)[\/latex], [latex]y=h(u, v, w)[\/latex], and [latex]z=k(u, v, w)[\/latex], be a one-to-one [latex]C1[\/latex] transformation, with a nonzero Jacobian, that maps the region [latex]G[\/latex] in the [latex]uvw[\/latex]-plane into the region [latex]D[\/latex] in the [latex]xyz[\/latex]-plane. As in the two-dimensional case, if [latex]F[\/latex] is continuous on [latex]D[\/latex], then<\/p>\n<p>[latex]\\hspace{3cm}\\large{\\begin{align}    \\underset{R}{\\displaystyle\\iiint}F(x,y,z)dV&=\\underset{G}{\\displaystyle\\iiint}F(g(u,v,w), \\ h(u,v,w), \\ k(u,v,w))\\Bigg|\\frac{\\partial(x,y,z)}{\\partial(u,v,w)}\\Bigg|du \\ dv \\ dw \\\\    &=\\underset{G}{\\displaystyle\\iiint}H(u,v,w)|J(u,v,w)|du \\ dv \\ dw    \\end{align}}[\/latex].<\/p>\n<\/div>\n<p>Let us now see how changes in triple integrals for cylindrical and spherical coordinates are affected by this theorem. We expect to obtain the same formulas as in\u00a0<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/introduction-to-triple-integrals-in-cylindrical-and-spherical-coordinates\/\" data-page-slug=\"5-5-triple-integrals-in-cylindrical-and-spherical-coordinates\" data-page-uuid=\"ebd07590-c9ca-4701-95cf-6fa9131d20db\" data-page-fragment=\"page_ebd07590-c9ca-4701-95cf-6fa9131d20db\">Triple Integrals in Cylindrical and Spherical Coordinates<\/a>.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: obtaining formulas in triple integrals for cylindrical and spherical coordinates<\/h3>\n<p id=\"fs-id1167793293911\">Derive the formula in triple integrals for<\/p>\n<ol id=\"fs-id1167793293914\" type=\"a\">\n<li>cylindrical and<\/li>\n<li>spherical coordinates.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q827454432\">Show Solution<\/span><\/p>\n<div id=\"q827454432\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1167793498790\" type=\"a\">\n<li>For cylindrical coordinates, the transformation is [latex]T(r,\\theta,z)=(x,y,z)[\/latex] from the Cartesian [latex]r\\theta{z}[\/latex]&#8211;plane\u00a0\u00a0to the Cartesian [latex]xyz[\/latex]-plane\u00a0(Figure 7). Here [latex]x=r\\cos\\theta[\/latex], [latex]y=r\\sin\\theta[\/latex], and [latex]z=z[\/latex]. The Jacobian for the transformation is<span data-type=\"newline\"><br \/>\n<\/span><br \/>\n[latex]\\hspace{1.5cm}\\large{\\begin{align}J(r,\\theta,z)&=\\frac{\\partial(x,y,z)}{\\partial(r,\\theta,z)}=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{r}}&\\frac{\\partial{x}}{\\partial{\\theta}}&\\frac{\\partial{x}}{\\partial{z}} \\\\ \\frac{\\partial{y}}{\\partial{r}}&\\frac{\\partial{y}}{\\partial{\\theta}}&\\frac{\\partial{y}}{\\partial{z}} \\\\ \\frac{\\partial{z}}{\\partial{r}}&\\frac{\\partial{z}}{\\partial{\\theta}}&\\frac{\\partial{z}}{\\partial{z}}\\end{vmatrix} \\\\&=\\begin{vmatrix}\\cos\\theta&-r\\sin\\theta&0 \\\\ \\sin\\theta&r\\cos\\theta&0 \\\\ 0&0&1\\end{vmatrix}=r\\cos^2\\theta+r\\sin^2\\theta=r(\\cos^2\\theta+\\sin^s\\theta)=r \\end{align}}[\/latex].<br \/>\n<span data-type=\"newline\"><br \/>\n<\/span>We know that [latex]r\\geq0[\/latex], so [latex]|J(r,\\theta,z)|=r[\/latex]. Then the triple integral is<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167793640526\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\large{\\underset{D}{\\displaystyle\\iiint}f(x,y,z)dV=\\underset{G}{\\displaystyle\\iiint}f(r\\cos\\theta,r\\sin\\theta,z)r \\ dr \\ d\\theta \\ dz}[\/latex].<\/div>\n<div id=\"attachment_1428\" style=\"width: 908px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1428\" class=\"size-full wp-image-1428\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28043458\/5-7-11.jpeg\" alt=\"On the left-hand side of this figure, there is a cube G with sides parallel to the coordinate axes in cylindrical coordinate space. Then there is an arrow from this graph to the right-hand side of the figure marked with x = r cos theta, y = r sin theta, and z = z. On the right-hand side of this figure there is a region D in x y z space that is a thick annulus. The top is labeled z = constant, the flat vertical side is labeled theta = constant, and the outermost side is labeled r = constant.\" width=\"898\" height=\"391\" \/><\/p>\n<p id=\"caption-attachment-1428\" class=\"wp-caption-text\">Figure 7. The transformation from rectangular coordinates to cylindrical coordinates can be treated as a change of variables from region [latex]G[\/latex] in [latex]r\\theta{z}[\/latex]-space to region\u00a0[latex]D[\/latex] in [latex]xyz[\/latex]-space.<\/p>\n<\/div>\n<\/li>\n<li>For spherical coordinates, the transformation is [latex]T(\\rho,\\theta,\\varphi)=(x,y,z)[\/latex] from the Cartesian [latex]p\\theta\\varphi[\/latex]&#8211;plane\u00a0\u00a0to the Cartesian [latex]xyz[\/latex]&#8211;plane\u00a0\u00a0(Figure 8). Here [latex]x=\\rho\\sin\\varphi\\cos\\theta[\/latex], [latex]y=\\rho\\sin\\varphi\\sin\\theta[\/latex], and [latex]z=\\rho\\cos\\varphi[\/latex]. The Jacobian for the transformation is<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1167793373656\" class=\"unnumbered\" style=\"text-align: center;\" data-type=\"equation\" data-label=\"\">[latex]\\large{J(\\rho,\\theta,\\varphi)=\\frac{\\partial{(x,y,z)}}{\\partial{(\\rho,\\theta\\varphi)}}=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{\\rho}}&\\frac{\\partial{x}}{\\partial{\\theta}}&\\frac{\\partial{x}}{\\partial{\\varphi}} \\\\ \\frac{\\partial{y}}{\\partial{\\rho}}&\\frac{\\partial{y}}{\\partial{\\theta}}&\\frac{\\partial{y}}{\\partial{\\varphi}} \\\\ \\frac{\\partial{z}}{\\partial{\\rho}}&\\frac{\\partial{z}}{\\partial{\\theta}}&\\frac{\\partial{z}}{\\partial{\\varphi}}\\end{vmatrix}=\\begin{vmatrix}\\sin\\varphi\\cos\\theta&-\\rho\\sin\\varphi\\sin\\theta&\\rho\\cos\\varphi\\cos\\theta \\\\ \\sin\\varphi\\sin\\theta&-\\rho\\sin\\varphi\\cos\\theta&\\rho\\cos\\varphi\\sin\\theta \\\\ \\cos\\theta&0&-\\rho\\sin\\varphi\\end{vmatrix}}[\/latex]<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Expanding the determinant with respect to the third row:<span data-type=\"newline\"><br \/>\n<\/span><br \/>\n[latex]\\hspace{2cm}\\begin{align}    &=\\cos\\varphi\\begin{vmatrix}-\\rho\\sin\\varphi\\sin\\theta&\\rho\\cos\\varphi\\cos\\theta \\\\ \\rho\\sin\\varphi\\sin\\theta&\\rho\\cos\\varphi\\sin\\theta\\end{vmatrix}-\\rho\\sin\\varphi\\begin{vmatrix}\\sin\\varphi\\cos\\theta&-\\rho\\sin\\varphi\\sin\\theta \\\\ \\sin\\varphi\\sin\\theta&\\rho\\sin\\varphi\\cos\\theta\\end{vmatrix} \\\\    &=\\cos\\varphi(-\\rho^2\\sin\\varphi\\cos\\varphi\\sin^2\\theta-\\rho^2\\sin\\varphi\\cos\\varphi\\cos^2\\theta)-\\rho\\sin\\varphi(\\rho\\sin^2\\varphi\\cos^2\\theta+\\rho\\sin^2\\varphi\\sin^2\\theta) \\\\    &=-\\rho^2\\sin\\varphi\\cos^2\\varphi(\\sin^2\\theta+\\cos^2\\theta)-\\rho^2\\sin\\varphi\\sin^2\\varphi(\\sin^2\\theta+\\cos^2\\theta) \\\\    &=-\\rho^2\\sin\\varphi\\cos^2\\varphi-\\rho^2\\sin\\varphi\\sin^2\\varphi \\\\    &=-\\rho^2\\sin\\varphi(\\cos^2\\varphi+\\sin^2\\varphi)=-\\rho^2\\sin\\varphi\\end{align}[\/latex]<br \/>\n<span data-type=\"newline\"><br \/>\n<\/span>Since [latex]0\\leq\\varphi\\leq\\pi[\/latex], we must have [latex]\\sin\\varphi\\geq0[\/latex]. Thus [latex]|J(\\rho,\\theta,\\varphi)|=|-\\rho^2\\sin\\varphi|=\\rho^2\\sin\\varphi[\/latex].<\/p>\n<div id=\"CNX_Calc_Figure_15_07_012\" class=\"os-figure\">\n<div class=\"os-caption-container\">\n<div id=\"attachment_1429\" style=\"width: 905px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1429\" class=\"size-full wp-image-1429\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28043542\/5-7-12.jpeg\" alt=\"On the left-hand side of this figure, there is a cube G with sides parallel to the coordinate axes in rho phi theta space. Then there is an arrow from this graph to the right-hand side of the figure marked with x = rho sin phi cos theta, y = rho sin phi sin theta, and z = rho cos phi. On the right-hand side of this figure there is a region D in xyz space that is a thick annulus and has the point (x, y, z) shown as being equal to (rho, phi, theta). The top is labeled phi = constant, the flat vertical side is labeled theta = constant, and the outermost side is labeled rho = constant.\" width=\"895\" height=\"391\" \/><\/p>\n<p id=\"caption-attachment-1429\" class=\"wp-caption-text\">Figure 8. The transformation from rectangular coordinates to cylindrical coordinates can be treated as a change of variables from region [latex]G[\/latex] in [latex]\\rho\\theta\\varphi[\/latex]-space to region\u00a0[latex]D[\/latex] in [latex]xyz[\/latex]-space.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><span data-type=\"newline\"><br \/>\n<\/span>Then the triple integral becomes<span data-type=\"newline\"><br \/>\n<\/span><br \/>\n[latex]\\underset{D}{\\displaystyle\\iiint}f(x,y,z)dV=\\underset{G}{\\displaystyle\\iiint}f(\\rho\\sin\\varphi\\cos\\theta,\\rho\\sin\\varphi\\sin\\theta,\\rho\\cos\\varphi)\\rho^2\\sin\\varphi \\ d\\rho \\ d\\varphi \\ d\\theta[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>Let\u2019s try another example with a different substitution.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: evaluating a triple integral with a change of variables<\/h3>\n<p id=\"fs-id1167793376275\">Evaluate the triple integral<\/p>\n<p style=\"text-align: center;\">[latex]\\large{\\displaystyle\\int_0^3\\displaystyle\\int_0^4\\displaystyle\\int_{y\/2}^{(y\/2)+1}\\left(x+\\frac{z}3\\right)dx \\ dy \\ dz}[\/latex]<\/p>\n<p id=\"fs-id1167793638297\">in [latex]xyz[\/latex]-space by using the transformation<\/p>\n<p style=\"text-align: center;\">[latex]\\large{u=(2x-y)\/2, \\ v=y\/2,\\text{ and }w=z\/3}[\/latex].<\/p>\n<p id=\"fs-id1167793930128\">Then integrate over an appropriate region in [latex]uvw[\/latex]-space<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q458273487\">Show Solution<\/span><\/p>\n<div id=\"q458273487\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793930147\">As before, some kind of sketch of the region [latex]G[\/latex] in [latex]xyz[\/latex]-space over which we have to perform the integration can help identify the region [latex]D[\/latex] in [latex]uvw[\/latex]-space (Figure 9). Clearly [latex]G[\/latex] in [latex]xyz[\/latex]-space is bounded by the planes [latex]x=y\/2[\/latex], [latex]x=(y\/2)+1[\/latex], [latex]y=0[\/latex],[latex]y=4[\/latex], [latex]z=0[\/latex], and [latex]z=4[\/latex]. We also know that we have to use [latex]u=(2x-y)\/2[\/latex], [latex]v=y\/2[\/latex] and [latex]w=z\/3[\/latex] for the transformations. We need to solve for [latex]x[\/latex], [latex]y[\/latex], and [latex]z[\/latex]. Here we find that\u00a0[latex]x=u+v[\/latex],\u00a0[latex]y=2v[\/latex], and\u00a0[latex]z=3w[\/latex].<\/p>\n<p id=\"fs-id1167794223120\">Using elementary algebra, we can find the corresponding surfaces for the region [latex]G[\/latex] and the limits of integration in [latex]uvw[\/latex]-space. It is convenient to list these equations in a table.<\/p>\n<div id=\"fs-id1167794223141\" class=\"os-table\">\n<table class=\"unnumbered\" style=\"height: 166px; width: 622px;\" data-id=\"fs-id1167794223141\" data-label=\"\">\n<thead>\n<tr style=\"height: 24px;\" valign=\"top\">\n<th style=\"height: 24px; width: 259.734px;\" scope=\"col\" data-valign=\"top\" data-align=\"left\">Equations in [latex]xyz[\/latex] for the region [latex]D[\/latex]<\/th>\n<th style=\"height: 24px; width: 204.016px;\" scope=\"col\" data-valign=\"top\" data-align=\"left\">Corresponding equations in [latex]uvw[\/latex] for the region [latex]G[\/latex]<\/th>\n<th style=\"height: 24px; width: 123.219px;\" scope=\"col\" data-valign=\"top\" data-align=\"left\">Limits for the integration in [latex]uvw[\/latex]<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr style=\"height: 24px;\" valign=\"top\">\n<td style=\"height: 22px; width: 259.734px;\" data-valign=\"top\" data-align=\"left\">[latex]x=y\/2[\/latex]<\/td>\n<td style=\"height: 22px; width: 204.016px;\" data-valign=\"top\" data-align=\"left\">[latex]u+v=2v\/2=v[\/latex]<\/td>\n<td style=\"height: 22px; width: 123.219px;\" data-valign=\"top\" data-align=\"left\">[latex]u=0[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 24px;\" valign=\"top\">\n<td style=\"height: 24px; width: 259.734px;\" data-valign=\"top\" data-align=\"left\">[latex]x=y\/2[\/latex]<\/td>\n<td style=\"height: 24px; width: 204.016px;\" data-valign=\"top\" data-align=\"left\">[latex]u+v=(2v\/2)+1=v+1[\/latex]<\/td>\n<td style=\"height: 24px; width: 123.219px;\" data-valign=\"top\" data-align=\"left\">[latex]u=1[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 24px;\" valign=\"top\">\n<td style=\"height: 24px; width: 259.734px;\" data-valign=\"top\" data-align=\"left\">[latex]y=0[\/latex]<\/td>\n<td style=\"height: 24px; width: 204.016px;\" data-valign=\"top\" data-align=\"left\">[latex]2v=0[\/latex]<\/td>\n<td style=\"height: 24px; width: 123.219px;\" data-valign=\"top\" data-align=\"left\">[latex]v=0[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 24px;\" valign=\"top\">\n<td style=\"height: 24px; width: 259.734px;\" data-valign=\"top\" data-align=\"left\">[latex]y=4[\/latex]<\/td>\n<td style=\"height: 24px; width: 204.016px;\" data-valign=\"top\" data-align=\"left\">[latex]2v=4[\/latex]<\/td>\n<td style=\"height: 24px; width: 123.219px;\" data-valign=\"top\" data-align=\"left\">[latex]v=2[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 24px;\" valign=\"top\">\n<td style=\"height: 24px; width: 259.734px;\" data-valign=\"top\" data-align=\"left\">[latex]z=0[\/latex]<\/td>\n<td style=\"height: 24px; width: 204.016px;\" data-valign=\"top\" data-align=\"left\">[latex]3w=0[\/latex]<\/td>\n<td style=\"height: 24px; width: 123.219px;\" data-valign=\"top\" data-align=\"left\">[latex]w=0[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 24px;\" valign=\"top\">\n<td style=\"height: 24px; width: 259.734px;\" data-valign=\"top\" data-align=\"left\">[latex]z=3[\/latex]<\/td>\n<td style=\"height: 24px; width: 204.016px;\" data-valign=\"top\" data-align=\"left\">[latex]3w=3[\/latex]<\/td>\n<td style=\"height: 24px; width: 123.219px;\" data-valign=\"top\" data-align=\"left\">[latex]w=1[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div id=\"CNX_Calc_Figure_15_07_013\" class=\"os-figure\">\n<div class=\"os-caption-container\">\n<div id=\"attachment_1430\" style=\"width: 904px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1430\" class=\"size-full wp-image-1430\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/11\/28043626\/5-7-13.jpeg\" alt=\"&lt;img src=&quot;\/apps\/archive\/20210823.155019\/resources\/9ac91afe22764bb4840cdcc0310bae65f7605c69&quot; data-media-type=&quot;image\/jpeg&quot; alt=&quot;On the left-hand side of this figure, there is a box G with sides 1, 2, and 1 along the u, v, and w axes, respectively. Then there is an arrow from this graph to the right-hand side of the figure marked with x = u + v, y = 2v, and z = 3w. On the right-hand side of this figure there is a region D in xyz space that is a rotated box with sides 1, 4, and 3 along the x, y, and z axes. The rear plane is marked x = y\/2 or y = 2x. The front plane is marked x = y\/2 + 1 or y = 2x minus 2.&quot; id=&quot;34&quot;&gt;\" width=\"894\" height=\"435\" \/><\/p>\n<p id=\"caption-attachment-1430\" class=\"wp-caption-text\">Figure 9. The region [latex]G[\/latex] in [latex]uvw[\/latex]-space is transformed to region [latex]D[\/latex] in [latex]xyz[\/latex]-space<\/p>\n<\/div>\n<\/div>\n<div><\/div>\n<\/div>\n<p id=\"fs-id1167793607338\">Now we can calculate the Jacobian for the transformation:<\/p>\n<p style=\"text-align: center;\">[latex]\\large{J(u,v,w)=\\begin{vmatrix}\\frac{\\partial{x}}{\\partial{u}}&\\frac{\\partial{x}}{\\partial{v}}&\\frac{\\partial{x}}{\\partial{w}} \\\\ \\frac{\\partial{y}}{\\partial{u}}&\\frac{\\partial{y}}{\\partial{v}}&\\frac{\\partial{y}}{\\partial{w}} \\\\ \\frac{\\partial{z}}{\\partial{u}}&\\frac{\\partial{z}}{\\partial{v}}&\\frac{\\partial{z}}{\\partial{w}}\\end{vmatrix}=\\begin{vmatrix}1&1&0 \\\\ 0&2&0 \\\\ 0&0&3\\end{vmatrix}=6}[\/latex].<\/p>\n<p id=\"fs-id1167794296460\">The function to be integrated becomes<\/p>\n<p style=\"text-align: center;\">[latex]\\large{f(x,y,z)=x+\\frac{z}3=u+v+\\frac{3w}3=u+v+w}[\/latex].<\/p>\n<p id=\"fs-id1167793420356\">We are now ready to put everything together and complete the problem.<\/p>\n<p>[latex]\\hspace{3cm}\\large{\\begin{align}    &\\hspace{1cm}\\displaystyle\\int_0^3\\displaystyle\\int_0^4\\displaystyle\\int_{y\/2}^{(y\/2)+1}\\left(x+\\frac{z}3\\right)dx \\ dy \\ dz \\\\    &=\\displaystyle\\int_0^1\\displaystyle\\int_0^2\\displaystyle\\int_0^1(u+v+w)|J(u,v,w)|du \\ dv \\ dw=\\displaystyle\\int_0^1\\displaystyle\\int_0^2\\displaystyle\\int_0^1(u+v+w)|6|du \\ dv \\ dw \\\\    &=6\\displaystyle\\int_0^1\\displaystyle\\int_0^2\\displaystyle\\int_0^1(u+v+w)du \\ dv \\ dw=6\\displaystyle\\int_0^1\\displaystyle\\int_0^2\\left[\\frac{u^2}2+vu+wu\\right]_0^1dv \\ dw \\\\    &=6\\displaystyle\\int_0^1\\displaystyle\\int_0^2\\left(\\frac12+v+w\\right)dv \\ dw=6\\displaystyle\\int_0^1\\left[\\frac12v+\\frac{v^2}2+wv\\right]_0^2dw \\\\    &=6\\displaystyle\\int_0^1(3+2w)dw=6\\left[3w+w^2\\right]_0^1=24    \\end{align}}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p id=\"fs-id1167794212114\">Let [latex]D[\/latex] be the region in [latex]xyz[\/latex]-space defined by [latex]1\\leq{x}\\leq2, \\ 0\\leq{xy}\\leq2, \\text{and} \\ 0\\leq{z}\\leq1[\/latex].<\/p>\n<p id=\"fs-id1167793420423\">Evaluate [latex]\\underset{D}{\\displaystyle\\iiint}(x^2y+3xyz)dx \\ dy \\ dz[\/latex] by using the transformation [latex]u=x[\/latex],\u00a0[latex]v=xy[\/latex], and\u00a0[latex]w=3z[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q166409830\">Show Solution<\/span><\/p>\n<div id=\"q166409830\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\displaystyle\\int_0^3\\displaystyle\\int_0^2\\displaystyle\\int_1^2\\left(\\frac{v}3+\\frac{vw}{3u}\\right)dy \\ dv \\ dw=2+\\ln8 = 2+ 3\\ln2[\/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=8197116&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=TChzX4WNbC4&amp;video_target=tpm-plugin-bxj0q4mi-TChzX4WNbC4\" 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\/CP5.48_transcript.html\">transcript for \u201cCP 5.48\u201d here (opens in new window).<\/a><\/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-1074\">\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 5.47. <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 5.48. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":428269,"menu_order":30,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"CP 5.47\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"CP 5.48\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1074","chapter","type-chapter","status-publish","hentry"],"part":23,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1074","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/428269"}],"version-history":[{"count":140,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1074\/revisions"}],"predecessor-version":[{"id":6399,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1074\/revisions\/6399"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/23"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/1074\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=1074"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=1074"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=1074"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=1074"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}