{"id":724,"date":"2025-10-09T15:34:38","date_gmt":"2025-10-09T15:34:38","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=724"},"modified":"2025-11-13T15:30:48","modified_gmt":"2025-11-13T15:30:48","slug":"change-of-variables-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/change-of-variables-fresh-take\/","title":{"raw":"Change of Variables: Fresh Take","rendered":"Change of Variables: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Find how a region changes when you use a different set of variables<\/li>\r\n \t<li>Calculate the Jacobian for a transformation of variables<\/li>\r\n \t<li>Apply change of variables in double integrals<\/li>\r\n \t<li>Apply change of variables in triple integrals<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2 data-type=\"title\">Planar Transformations<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Just as [latex]u[\/latex]-substitution simplifies single-variable integrals, changing variables in multiple integrals can transform difficult integration problems into manageable ones. You've already used this idea with polar, cylindrical, and spherical coordinates. This section develops the general framework for any change of variables.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Core Idea<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">A <strong>transformation<\/strong> [latex]T[\/latex] maps points from one coordinate system to another. For example, [latex]T: G \\to R[\/latex] transforms region [latex]G[\/latex] in the [latex]uv[\/latex]-plane to region [latex]R[\/latex] in the [latex]xy[\/latex]-plane using equations [latex]x = g(u,v)[\/latex] and [latex]y = h(u,v)[\/latex]. We write [latex]T(u,v) = (x,y)[\/latex] to show that point [latex](u,v)[\/latex] maps to point [latex](x,y)[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">The goal is to convert an integral over region [latex]R[\/latex] (which might be complicated) into an integral over region [latex]G[\/latex] (which is simpler). This parallels how [latex]\\int_2^3 x(x^2-4)^5 dx[\/latex] becomes [latex]\\int_0^5 \\frac{1}{2}u^5 du[\/latex] after substituting [latex]u = x^2 - 4[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>One-to-One Transformations<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">A transformation is <strong>one-to-one<\/strong> if different input points always produce different output points. Formally, [latex]T(u_1, v_1) = T(u_2, v_2)[\/latex] implies [latex](u_1, v_1) = (u_2, v_2)[\/latex]. When [latex]T[\/latex] is one-to-one, an inverse transformation [latex]T^{-1}[\/latex] exists that maps back from [latex]R[\/latex] to [latex]G[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">For example, polar coordinates give a one-to-one transformation when [latex]0 \\leq r \\leq 1[\/latex] and [latex]0 \\leq \\theta \\leq \\frac{\\pi}{2}[\/latex]. The inverse exists: [latex]r = \\sqrt{x^2 + y^2}[\/latex] and [latex]\\theta = \\tan^{-1}(\\frac{y}{x})[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Domain and Range<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Region [latex]G[\/latex] is the <strong>domain<\/strong> of [latex]T[\/latex]\u2014the set of input points. Region [latex]R[\/latex] is the <strong>range<\/strong> or <strong>image<\/strong>\u2014the set of output points. We encounter two types of problems: finding the image [latex]R[\/latex] given [latex]G[\/latex] and [latex]T[\/latex] (forward problem), or finding the pre-image [latex]G[\/latex] given [latex]R[\/latex] and [latex]T[\/latex] (inverse problem).<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Finding Images<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">To find where a region transforms, examine how its boundaries map. For each edge or curve bounding [latex]G[\/latex], substitute the transformation equations to find the corresponding boundary in [latex]R[\/latex]. Points inside [latex]G[\/latex] map to points inside [latex]R[\/latex].<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Let a transformation [latex]T[\/latex]\u00a0 be defined as [latex]T(u, v)=(x, y)[\/latex] where [latex]x=u+v[\/latex], [latex]y=3v[\/latex]. Find the image of the rectangle [latex]{G} = {\\left \\{ {({u,v})}: {0} \\leq {u} \\leq {1,0} \\leq {v} \\leq {2}\\right \\}}[\/latex] from the [latex]uv[\/latex]-plane after the transformation into a region [latex]R[\/latex] in the [latex]xy[\/latex]-plane. Show that [latex]T[\/latex] is a one-to-one transformation and find [latex]{{T}^{-1}}{({x},{y})}[\/latex].[reveal-answer q=\"036725008\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"036725008\"][latex]T^{-1}(x,y)=(u,v)[\/latex] where\u00a0[latex]u=\\frac{3x-y}3[\/latex] and\u00a0[latex]v=\\frac{y}3[\/latex].[\/hidden-answer]<\/section>\r\n<h2 data-type=\"title\">Jacobians<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">The Jacobian measures how a transformation scales areas (or volumes in higher dimensions). When changing variables in an integral, you need to know not just the new coordinates, but also how much the transformation stretches or compresses area. The Jacobian provides this crucial scaling factor.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>What is the Jacobian?<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For a transformation [latex]T(u,v) = (x,y)[\/latex] where [latex]x = g(u,v)[\/latex] and [latex]y = h(u,v)[\/latex], the Jacobian is:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]J(u,v) = \\dfrac{\\partial(x,y)}{\\partial(u,v)} = \\begin{vmatrix} \\dfrac{\\partial x}{\\partial u} &amp; \\dfrac{\\partial x}{\\partial v} \\\\ \\dfrac{\\partial y}{\\partial u} &amp; \\dfrac{\\partial y}{\\partial v} \\end{vmatrix} = \\dfrac{\\partial x}{\\partial u} \\dfrac{\\partial y}{\\partial v} - \\dfrac{\\partial x}{\\partial v} \\dfrac{\\partial y}{\\partial u}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This [latex]2 \\times 2[\/latex] determinant captures how the transformation changes areas. A small rectangle of area [latex]\\Delta u \\Delta v[\/latex] in the [latex]uv[\/latex]-plane transforms to a region of approximate area [latex]|J(u,v)| \\Delta u \\Delta v[\/latex] in the [latex]xy[\/latex]-plane.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Why the Jacobian Matters<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">When a transformation maps region [latex]G[\/latex] to region [latex]R[\/latex], it distorts the shape. A rectangle in [latex]G[\/latex] might become a curved region in [latex]R[\/latex]. The Jacobian tells you the area scaling factor at each point. For example, polar coordinates have Jacobian [latex]J(r,\\theta) = r[\/latex], which is why [latex]dA = r \\, dr \\, d\\theta[\/latex] includes that factor of [latex]r[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Computing the Jacobian<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Set up the [latex]2 \\times 2[\/latex] matrix with partial derivatives: first row has [latex]x[\/latex] derivatives with respect to [latex]u[\/latex] and [latex]v[\/latex], second row has [latex]y[\/latex] derivatives. Compute the determinant using [latex]ad - bc[\/latex] for matrix [latex]\\begin{vmatrix} a &amp; b \\ c &amp; d \\end{vmatrix}[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Geometric Interpretation<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The transformation maps the sides of a small rectangle (length [latex]\\Delta u[\/latex], width [latex]\\Delta v[\/latex]) to tangent vectors [latex]\\Delta u \\mathbf{r}_u[\/latex] and [latex]\\Delta v \\mathbf{r}_v[\/latex] in the [latex]xy[\/latex]-plane. These vectors form a parallelogram, and its area equals [latex]|\\mathbf{r}_u \\times \\mathbf{r}_v| \\Delta u \\Delta v[\/latex]. The cross product magnitude equals [latex]|J(u,v)|[\/latex].<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">Find the Jacobian of the transformation given in the previous Try It: [latex]T(u, v)=(u+v, 3v)[\/latex].[reveal-answer q=\"203975843\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"203975843\"][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&amp;0 \\\\ 1&amp;3\\end{vmatrix}=3}[\/latex].[\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=8197114&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=51E1qqcwWzY&amp;video_target=tpm-plugin-jejxrzic-51E1qqcwWzY\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.44_transcript.html\">transcript for \u201cCP 5.44\u201d here (opens in new window).<\/a><\/center><\/section>\r\n<h2 data-type=\"title\">Change of Variables for Double Integrals<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">Changing variables in double integrals transforms a difficult integral over region [latex]R[\/latex] into a simpler integral over region [latex]S[\/latex]. The key formula replaces [latex]dA[\/latex] with [latex]|J(u,v)| , du , dv[\/latex], where the Jacobian accounts for how the transformation scales areas.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Change of Variables Formula<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For transformation [latex]T(u,v) = (x,y)[\/latex] where [latex]x = g(u,v)[\/latex] and [latex]y = h(u,v)[\/latex]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\iint} f(x,y) , dA = \\underset{S}{\\iint} f(g(u,v), h(u,v)) \\left|\\dfrac{\\partial(x,y)}{\\partial(u,v)}\\right| du \\, dv[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This means: replace [latex]x[\/latex] and [latex]y[\/latex] in the integrand with their expressions in [latex]u[\/latex] and [latex]v[\/latex], multiply by the absolute value of the Jacobian, and integrate over the transformed region [latex]S[\/latex] using [latex]du , dv[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Why This Works<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">A small rectangle in the [latex]uv[\/latex]-plane with area [latex]\\Delta u \\Delta v[\/latex] transforms into a curved region in the [latex]xy[\/latex]-plane with area approximately [latex]|J(u,v)| \\Delta u \\Delta v[\/latex]. The Jacobian captures this area scaling, ensuring the integral measures the correct total.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Choosing a Good Transformation<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Look for patterns that suggest natural substitutions. The region's boundary curves often reveal the transformation\u2014if boundaries are lines like [latex]x - y = c[\/latex] or curves like [latex]x^2 - y^2 = c[\/latex], let [latex]u = x - y[\/latex]. The integrand itself may also suggest coordinates (expressions like [latex]x^2 + y^2[\/latex] suggest polar coordinates). Choose transformations that make either the region simpler (turning curves into straight lines) or the integrand simpler (or both).<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Sketch region [latex]R[\/latex] in the [latex]xy[\/latex]-plane and identify boundary curves<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Choose transformation [latex]x = g(u,v)[\/latex], [latex]y = h(u,v)[\/latex] that simplifies the region or integrand<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Determine how boundaries transform to find the new region [latex]S[\/latex] and limits in the [latex]uv[\/latex]-plane<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Compute the Jacobian [latex]J(u,v) = \\dfrac{\\partial(x,y)}{\\partial(u,v)}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Substitute the transformation into the integrand<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Replace [latex]dA[\/latex] with [latex]|J(u,v)| , du , dv[\/latex] and evaluate over region [latex]S[\/latex]<\/li>\r\n<\/ol>\r\n<p class=\"whitespace-normal break-words\"><strong>Why the Absolute Value?<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The Jacobian can be negative if the transformation reverses orientation, but area must be positive. Using [latex]|J(u,v)|[\/latex] ensures we measure area correctly.<\/p>\r\n\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">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.[reveal-answer q=\"983458271\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"983458271\"][latex]\\displaystyle\\int_0^{\\pi\/2}\\displaystyle\\int_0^1r^3 \\ dr \\ d\\theta[\/latex][\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">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.[reveal-answer q=\"249752336\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"249752336\"][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].[\/hidden-answer]<\/section><section class=\"textbox example\" aria-label=\"Example\">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].[reveal-answer q=\"872845374\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"872845374\"][latex]\\frac12\\sin{2}-1[\/latex][\/hidden-answer]<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=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\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.47_transcript.html\">transcript for \u201cCP 5.47\u201d here (opens in new window).<\/a><\/center><\/section>\r\n<h2 data-type=\"title\">Change of Variables for Triple Integrals<\/h2>\r\n<div class=\"textbox shaded\">\r\n\r\n<strong>The Main Idea\u00a0<\/strong>\r\n<p class=\"whitespace-normal break-words\">The change of variables formula extends naturally from double integrals to triple integrals. The key difference is that the Jacobian becomes a [latex]3 \\times 3[\/latex] determinant that measures how the transformation scales volumes instead of areas. Cylindrical and spherical coordinates are just special cases of this general method.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Change of Variables Formula<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">For transformation [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]:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{D}{\\iiint} F(x,y,z) , dV = \\underset{G}{\\iiint} F(g(u,v,w), h(u,v,w), k(u,v,w)) \\left|\\dfrac{\\partial(x,y,z)}{\\partial(u,v,w)}\\right| du \\, dv \\, dw[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This replaces [latex]dV = dx \\, dy \\, dz[\/latex] with [latex]|J(u,v,w)| \\, du \\, dv \\, dw[\/latex], where the Jacobian accounts for volume scaling.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>The Three-Dimensional Jacobian<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">The Jacobian is the [latex]3 \\times 3[\/latex] determinant:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]J(u,v,w) = \\begin{vmatrix} \\dfrac{\\partial x}{\\partial u} &amp; \\dfrac{\\partial x}{\\partial v} &amp; \\dfrac{\\partial x}{\\partial w} \\\\ \\dfrac{\\partial y}{\\partial u} &amp; \\dfrac{\\partial y}{\\partial v} &amp; \\dfrac{\\partial y}{\\partial w} \\\\ \\dfrac{\\partial z}{\\partial u} &amp; \\dfrac{\\partial z}{\\partial v} &amp; \\dfrac{\\partial z}{\\partial w} \\end{vmatrix}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">You can arrange the determinant with rows for [latex]x[\/latex], [latex]y[\/latex], [latex]z[\/latex] derivatives (shown above) or with columns for [latex]u[\/latex], [latex]v[\/latex], [latex]w[\/latex] derivatives\u2014both give the same result.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Connection to Standard Coordinates<\/strong><\/p>\r\n<p class=\"whitespace-normal break-words\">Cylindrical coordinates ([latex]x = r\\cos\\theta[\/latex], [latex]y = r\\sin\\theta[\/latex], [latex]z = z[\/latex]) have Jacobian [latex]J(r,\\theta,z) = r[\/latex], giving [latex]dV = r \\, dr \\, d\\theta \\, dz[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">Spherical coordinates ([latex]x = \\rho\\sin\\varphi\\cos\\theta[\/latex], [latex]y = \\rho\\sin\\varphi\\sin\\theta[\/latex], [latex]z = \\rho\\cos\\varphi[\/latex]) have Jacobian [latex]J(\\rho,\\varphi,\\theta) = \\rho^2\\sin\\varphi[\/latex], giving [latex]dV = \\rho^2\\sin\\varphi \\, d\\rho \\, d\\varphi \\, d\\theta[\/latex].<\/p>\r\n<p class=\"whitespace-normal break-words\">These familiar volume elements come directly from the general change of variables theorem.<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Identify region [latex]D[\/latex] in [latex]xyz[\/latex]-space and its bounding surfaces<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Choose transformation equations [latex]x = g(u,v,w)[\/latex], [latex]y = h(u,v,w)[\/latex], [latex]z = k(u,v,w)[\/latex] that simplify the region or integrand<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Determine the transformed region [latex]G[\/latex] in [latex]uvw[\/latex]-space by transforming each bounding surface<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Compute the [latex]3 \\times 3[\/latex] Jacobian determinant<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Substitute the transformation into the integrand<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Replace [latex]dV[\/latex] with [latex]|J(u,v,w)| \\, du \\, dv \\, dw[\/latex] and integrate over region [latex]G[\/latex]<\/li>\r\n<\/ol>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\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<\/section><section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=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\" data-mce-fragment=\"1\"><\/iframe><\/center><center>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.48_transcript.html\">transcript for \u201cCP 5.48\u201d here (opens in new window).<\/a><\/center><\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Find how a region changes when you use a different set of variables<\/li>\n<li>Calculate the Jacobian for a transformation of variables<\/li>\n<li>Apply change of variables in double integrals<\/li>\n<li>Apply change of variables in triple integrals<\/li>\n<\/ul>\n<\/section>\n<h2 data-type=\"title\">Planar Transformations<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Just as [latex]u[\/latex]-substitution simplifies single-variable integrals, changing variables in multiple integrals can transform difficult integration problems into manageable ones. You&#8217;ve already used this idea with polar, cylindrical, and spherical coordinates. This section develops the general framework for any change of variables.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Core Idea<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">A <strong>transformation<\/strong> [latex]T[\/latex] maps points from one coordinate system to another. For example, [latex]T: G \\to R[\/latex] transforms region [latex]G[\/latex] in the [latex]uv[\/latex]-plane to region [latex]R[\/latex] in the [latex]xy[\/latex]-plane using equations [latex]x = g(u,v)[\/latex] and [latex]y = h(u,v)[\/latex]. We write [latex]T(u,v) = (x,y)[\/latex] to show that point [latex](u,v)[\/latex] maps to point [latex](x,y)[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">The goal is to convert an integral over region [latex]R[\/latex] (which might be complicated) into an integral over region [latex]G[\/latex] (which is simpler). This parallels how [latex]\\int_2^3 x(x^2-4)^5 dx[\/latex] becomes [latex]\\int_0^5 \\frac{1}{2}u^5 du[\/latex] after substituting [latex]u = x^2 - 4[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>One-to-One Transformations<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">A transformation is <strong>one-to-one<\/strong> if different input points always produce different output points. Formally, [latex]T(u_1, v_1) = T(u_2, v_2)[\/latex] implies [latex](u_1, v_1) = (u_2, v_2)[\/latex]. When [latex]T[\/latex] is one-to-one, an inverse transformation [latex]T^{-1}[\/latex] exists that maps back from [latex]R[\/latex] to [latex]G[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">For example, polar coordinates give a one-to-one transformation when [latex]0 \\leq r \\leq 1[\/latex] and [latex]0 \\leq \\theta \\leq \\frac{\\pi}{2}[\/latex]. The inverse exists: [latex]r = \\sqrt{x^2 + y^2}[\/latex] and [latex]\\theta = \\tan^{-1}(\\frac{y}{x})[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Domain and Range<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Region [latex]G[\/latex] is the <strong>domain<\/strong> of [latex]T[\/latex]\u2014the set of input points. Region [latex]R[\/latex] is the <strong>range<\/strong> or <strong>image<\/strong>\u2014the set of output points. We encounter two types of problems: finding the image [latex]R[\/latex] given [latex]G[\/latex] and [latex]T[\/latex] (forward problem), or finding the pre-image [latex]G[\/latex] given [latex]R[\/latex] and [latex]T[\/latex] (inverse problem).<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Finding Images<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">To find where a region transforms, examine how its boundaries map. For each edge or curve bounding [latex]G[\/latex], substitute the transformation equations to find the corresponding boundary in [latex]R[\/latex]. Points inside [latex]G[\/latex] map to points inside [latex]R[\/latex].<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Let a transformation [latex]T[\/latex]\u00a0 be defined as [latex]T(u, v)=(x, y)[\/latex] where [latex]x=u+v[\/latex], [latex]y=3v[\/latex]. Find the image of the rectangle [latex]{G} = {\\left \\{ {({u,v})}: {0} \\leq {u} \\leq {1,0} \\leq {v} \\leq {2}\\right \\}}[\/latex] from the [latex]uv[\/latex]-plane after the transformation into a region [latex]R[\/latex] in the [latex]xy[\/latex]-plane. Show that [latex]T[\/latex] is a one-to-one transformation and find [latex]{{T}^{-1}}{({x},{y})}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q036725008\">Show Solution<\/button><\/p>\n<div id=\"q036725008\" class=\"hidden-answer\" style=\"display: none\">[latex]T^{-1}(x,y)=(u,v)[\/latex] where\u00a0[latex]u=\\frac{3x-y}3[\/latex] and\u00a0[latex]v=\\frac{y}3[\/latex].<\/div>\n<\/div>\n<\/section>\n<h2 data-type=\"title\">Jacobians<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The Jacobian measures how a transformation scales areas (or volumes in higher dimensions). When changing variables in an integral, you need to know not just the new coordinates, but also how much the transformation stretches or compresses area. The Jacobian provides this crucial scaling factor.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>What is the Jacobian?<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For a transformation [latex]T(u,v) = (x,y)[\/latex] where [latex]x = g(u,v)[\/latex] and [latex]y = h(u,v)[\/latex], the Jacobian is:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]J(u,v) = \\dfrac{\\partial(x,y)}{\\partial(u,v)} = \\begin{vmatrix} \\dfrac{\\partial x}{\\partial u} & \\dfrac{\\partial x}{\\partial v} \\\\ \\dfrac{\\partial y}{\\partial u} & \\dfrac{\\partial y}{\\partial v} \\end{vmatrix} = \\dfrac{\\partial x}{\\partial u} \\dfrac{\\partial y}{\\partial v} - \\dfrac{\\partial x}{\\partial v} \\dfrac{\\partial y}{\\partial u}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This [latex]2 \\times 2[\/latex] determinant captures how the transformation changes areas. A small rectangle of area [latex]\\Delta u \\Delta v[\/latex] in the [latex]uv[\/latex]-plane transforms to a region of approximate area [latex]|J(u,v)| \\Delta u \\Delta v[\/latex] in the [latex]xy[\/latex]-plane.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Why the Jacobian Matters<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">When a transformation maps region [latex]G[\/latex] to region [latex]R[\/latex], it distorts the shape. A rectangle in [latex]G[\/latex] might become a curved region in [latex]R[\/latex]. The Jacobian tells you the area scaling factor at each point. For example, polar coordinates have Jacobian [latex]J(r,\\theta) = r[\/latex], which is why [latex]dA = r \\, dr \\, d\\theta[\/latex] includes that factor of [latex]r[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Computing the Jacobian<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Set up the [latex]2 \\times 2[\/latex] matrix with partial derivatives: first row has [latex]x[\/latex] derivatives with respect to [latex]u[\/latex] and [latex]v[\/latex], second row has [latex]y[\/latex] derivatives. Compute the determinant using [latex]ad - bc[\/latex] for matrix [latex]\\begin{vmatrix} a & b \\ c & d \\end{vmatrix}[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Geometric Interpretation<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The transformation maps the sides of a small rectangle (length [latex]\\Delta u[\/latex], width [latex]\\Delta v[\/latex]) to tangent vectors [latex]\\Delta u \\mathbf{r}_u[\/latex] and [latex]\\Delta v \\mathbf{r}_v[\/latex] in the [latex]xy[\/latex]-plane. These vectors form a parallelogram, and its area equals [latex]|\\mathbf{r}_u \\times \\mathbf{r}_v| \\Delta u \\Delta v[\/latex]. The cross product magnitude equals [latex]|J(u,v)|[\/latex].<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">Find the Jacobian of the transformation given in the previous Try It: [latex]T(u, v)=(u+v, 3v)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q203975843\">Show Solution<\/button><\/p>\n<div id=\"q203975843\" class=\"hidden-answer\" style=\"display: none\">[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&0 \\\\ 1&3\\end{vmatrix}=3}[\/latex].<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=8197114&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=51E1qqcwWzY&amp;video_target=tpm-plugin-jejxrzic-51E1qqcwWzY\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.44_transcript.html\">transcript for \u201cCP 5.44\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<h2 data-type=\"title\">Change of Variables for Double Integrals<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Changing variables in double integrals transforms a difficult integral over region [latex]R[\/latex] into a simpler integral over region [latex]S[\/latex]. The key formula replaces [latex]dA[\/latex] with [latex]|J(u,v)| , du , dv[\/latex], where the Jacobian accounts for how the transformation scales areas.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Change of Variables Formula<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For transformation [latex]T(u,v) = (x,y)[\/latex] where [latex]x = g(u,v)[\/latex] and [latex]y = h(u,v)[\/latex]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{R}{\\iint} f(x,y) , dA = \\underset{S}{\\iint} f(g(u,v), h(u,v)) \\left|\\dfrac{\\partial(x,y)}{\\partial(u,v)}\\right| du \\, dv[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This means: replace [latex]x[\/latex] and [latex]y[\/latex] in the integrand with their expressions in [latex]u[\/latex] and [latex]v[\/latex], multiply by the absolute value of the Jacobian, and integrate over the transformed region [latex]S[\/latex] using [latex]du , dv[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Why This Works<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">A small rectangle in the [latex]uv[\/latex]-plane with area [latex]\\Delta u \\Delta v[\/latex] transforms into a curved region in the [latex]xy[\/latex]-plane with area approximately [latex]|J(u,v)| \\Delta u \\Delta v[\/latex]. The Jacobian captures this area scaling, ensuring the integral measures the correct total.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Choosing a Good Transformation<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Look for patterns that suggest natural substitutions. The region&#8217;s boundary curves often reveal the transformation\u2014if boundaries are lines like [latex]x - y = c[\/latex] or curves like [latex]x^2 - y^2 = c[\/latex], let [latex]u = x - y[\/latex]. The integrand itself may also suggest coordinates (expressions like [latex]x^2 + y^2[\/latex] suggest polar coordinates). Choose transformations that make either the region simpler (turning curves into straight lines) or the integrand simpler (or both).<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Sketch region [latex]R[\/latex] in the [latex]xy[\/latex]-plane and identify boundary curves<\/li>\n<li class=\"whitespace-normal break-words\">Choose transformation [latex]x = g(u,v)[\/latex], [latex]y = h(u,v)[\/latex] that simplifies the region or integrand<\/li>\n<li class=\"whitespace-normal break-words\">Determine how boundaries transform to find the new region [latex]S[\/latex] and limits in the [latex]uv[\/latex]-plane<\/li>\n<li class=\"whitespace-normal break-words\">Compute the Jacobian [latex]J(u,v) = \\dfrac{\\partial(x,y)}{\\partial(u,v)}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Substitute the transformation into the integrand<\/li>\n<li class=\"whitespace-normal break-words\">Replace [latex]dA[\/latex] with [latex]|J(u,v)| , du , dv[\/latex] and evaluate over region [latex]S[\/latex]<\/li>\n<\/ol>\n<p class=\"whitespace-normal break-words\"><strong>Why the Absolute Value?<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The Jacobian can be negative if the transformation reverses orientation, but area must be positive. Using [latex]|J(u,v)|[\/latex] ensures we measure area correctly.<\/p>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q983458271\">Show Solution<\/button><\/p>\n<div id=\"q983458271\" class=\"hidden-answer\" style=\"display: none\">[latex]\\displaystyle\\int_0^{\\pi\/2}\\displaystyle\\int_0^1r^3 \\ dr \\ d\\theta[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q249752336\">Show Solution<\/button><\/p>\n<div id=\"q249752336\" class=\"hidden-answer\" style=\"display: none\">[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].<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q872845374\">Show Solution<\/button><\/p>\n<div id=\"q872845374\" class=\"hidden-answer\" style=\"display: none\">[latex]\\frac12\\sin{2}-1[\/latex]<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=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\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.47_transcript.html\">transcript for \u201cCP 5.47\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n<h2 data-type=\"title\">Change of Variables for Triple Integrals<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The change of variables formula extends naturally from double integrals to triple integrals. The key difference is that the Jacobian becomes a [latex]3 \\times 3[\/latex] determinant that measures how the transformation scales volumes instead of areas. Cylindrical and spherical coordinates are just special cases of this general method.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Change of Variables Formula<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">For transformation [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]:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\underset{D}{\\iiint} F(x,y,z) , dV = \\underset{G}{\\iiint} F(g(u,v,w), h(u,v,w), k(u,v,w)) \\left|\\dfrac{\\partial(x,y,z)}{\\partial(u,v,w)}\\right| du \\, dv \\, dw[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This replaces [latex]dV = dx \\, dy \\, dz[\/latex] with [latex]|J(u,v,w)| \\, du \\, dv \\, dw[\/latex], where the Jacobian accounts for volume scaling.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>The Three-Dimensional Jacobian<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">The Jacobian is the [latex]3 \\times 3[\/latex] determinant:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]J(u,v,w) = \\begin{vmatrix} \\dfrac{\\partial x}{\\partial u} & \\dfrac{\\partial x}{\\partial v} & \\dfrac{\\partial x}{\\partial w} \\\\ \\dfrac{\\partial y}{\\partial u} & \\dfrac{\\partial y}{\\partial v} & \\dfrac{\\partial y}{\\partial w} \\\\ \\dfrac{\\partial z}{\\partial u} & \\dfrac{\\partial z}{\\partial v} & \\dfrac{\\partial z}{\\partial w} \\end{vmatrix}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">You can arrange the determinant with rows for [latex]x[\/latex], [latex]y[\/latex], [latex]z[\/latex] derivatives (shown above) or with columns for [latex]u[\/latex], [latex]v[\/latex], [latex]w[\/latex] derivatives\u2014both give the same result.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Connection to Standard Coordinates<\/strong><\/p>\n<p class=\"whitespace-normal break-words\">Cylindrical coordinates ([latex]x = r\\cos\\theta[\/latex], [latex]y = r\\sin\\theta[\/latex], [latex]z = z[\/latex]) have Jacobian [latex]J(r,\\theta,z) = r[\/latex], giving [latex]dV = r \\, dr \\, d\\theta \\, dz[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">Spherical coordinates ([latex]x = \\rho\\sin\\varphi\\cos\\theta[\/latex], [latex]y = \\rho\\sin\\varphi\\sin\\theta[\/latex], [latex]z = \\rho\\cos\\varphi[\/latex]) have Jacobian [latex]J(\\rho,\\varphi,\\theta) = \\rho^2\\sin\\varphi[\/latex], giving [latex]dV = \\rho^2\\sin\\varphi \\, d\\rho \\, d\\varphi \\, d\\theta[\/latex].<\/p>\n<p class=\"whitespace-normal break-words\">These familiar volume elements come directly from the general change of variables theorem.<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Identify region [latex]D[\/latex] in [latex]xyz[\/latex]-space and its bounding surfaces<\/li>\n<li class=\"whitespace-normal break-words\">Choose transformation equations [latex]x = g(u,v,w)[\/latex], [latex]y = h(u,v,w)[\/latex], [latex]z = k(u,v,w)[\/latex] that simplify the region or integrand<\/li>\n<li class=\"whitespace-normal break-words\">Determine the transformed region [latex]G[\/latex] in [latex]uvw[\/latex]-space by transforming each bounding surface<\/li>\n<li class=\"whitespace-normal break-words\">Compute the [latex]3 \\times 3[\/latex] Jacobian determinant<\/li>\n<li class=\"whitespace-normal break-words\">Substitute the transformation into the integrand<\/li>\n<li class=\"whitespace-normal break-words\">Replace [latex]dV[\/latex] with [latex]|J(u,v,w)| \\, du \\, dv \\, dw[\/latex] and integrate over region [latex]G[\/latex]<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q166409830\">Show Solution<\/button><\/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<\/section>\n<section class=\"textbox watchIt\" aria-label=\"Watch It\">Watch the following video to see the worked solution to the above example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=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\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP5.48_transcript.html\">transcript for \u201cCP 5.48\u201d here (opens in new window).<\/a><\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":27,"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-724","chapter","type-chapter","status-publish","hentry"],"part":434,"module-header":"fresh_take","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/724","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":9,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/724\/revisions"}],"predecessor-version":[{"id":2216,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/724\/revisions\/2216"}],"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\/724\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=724"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=724"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=724"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=724"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}