{"id":1304,"date":"2025-10-17T13:23:52","date_gmt":"2025-10-17T13:23:52","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=1304"},"modified":"2025-11-04T14:20:08","modified_gmt":"2025-11-04T14:20:08","slug":"change-of-variables-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/change-of-variables-learn-it-3\/","title":{"raw":"Change of Variables: Learn It 3","rendered":"Change of Variables: Learn It 3"},"content":{"raw":"

Change of Variables for Double Integrals<\/h2>\r\n

We've established 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 [latex]\\Delta u \\Delta v[\/latex] in the [latex]uv[\/latex]-plane by:<\/p>\r\n

[latex]\\Delta A \\approx J(u,v) \\Delta u \\Delta v[\/latex]<\/p>\r\n

Let's revisit the definition of the double integral:<\/p>\r\n

[latex]\\underset{R}{\\displaystyle\\iint}f(x,y) \\ dA=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^nf(x_{ij}, y_{ij}) \\Delta A[\/latex]<\/p>\r\n

Looking at Figure 1, we divide region [latex]S[\/latex] in the [latex]uv[\/latex]-plane into small subrectangles [latex]S_{ij}[\/latex]. The subrectangles [latex]R_{ij}[\/latex] in the [latex]xy[\/latex]-plane are the images of [latex]S_{ij}[\/latex] under transformation [latex]T(u,v) = (x,y)[\/latex].<\/p>\r\n\r\n[caption id=\"attachment_1422\" align=\"aligncenter\" width=\"765\"]\"On 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

The double integral becomes:<\/p>\r\n

[latex]\\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\n

This is exactly the double Riemann sum for the integral:<\/p>\r\n

[latex]\\underset{S}{\\displaystyle\\iint}f(g(u,v),h(u,v))\\Bigg|\\dfrac{\\partial(x,y)}{\\partial(u,v)}\\Bigg|du \\ dv[\/latex].<\/p>\r\n\r\n

\r\n

change of variables for double integrals<\/h3>\r\n

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]C^1[\/latex] transformation with a nonzero Jacobian on the interior of region [latex]S[\/latex] in the [latex]uv[\/latex]-plane. If [latex]T[\/latex] maps [latex]S[\/latex] into region [latex]R[\/latex] in the [latex]xy[\/latex]-plane and [latex]f[\/latex] is continuous on [latex]R[\/latex], then:<\/p>\r\n

[latex]\\underset{R}{\\displaystyle\\iint}f(x,y) \\ dA=\\underset{S}{\\displaystyle\\iint}f(g(u,v),h(u,v))\\Bigg|\\dfrac{\\partial(x,y)}{\\partial(u,v)}\\Bigg|du \\ dv[\/latex].<\/p>\r\n\r\n<\/section>\r\n

With this theorem, 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

[latex]dA=dx \\ dy=\\Bigg|\\dfrac{\\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\r\n

\r\n

Why the Absolute Value?\u00a0<\/strong><\/p>\r\n

We use [latex]|J(u,v)|[\/latex] in the integral because the Jacobian can be negative, but area must be positive. The absolute value ensures we're measuring area correctly regardless of whether the transformation reverses orientation.<\/p>\r\n\r\n<\/section>

\r\n

Choosing a Good Transformation<\/strong><\/p>\r\n

Look for patterns in:<\/p>\r\n\r\n

    \r\n \t
  • The boundary curves<\/strong> of the region (can they be simplified?)<\/li>\r\n \t
  • The integrand<\/strong> itself (does it suggest a coordinate system?)<\/li>\r\n \t
  • Symmetry<\/strong> in the problem (circular, elliptical, etc.)<\/li>\r\n<\/ul>\r\n

    Often the region's shape will suggest natural substitutions that make the limits constants.<\/p>\r\n\r\n<\/section>

    \r\n

    How to: Apply Change of Variables to Double Integrals<\/strong><\/p>\r\n\r\n

      \r\n \t
    1. Sketch the region<\/strong> in the [latex]xy[\/latex]-plane and write the equations of the curves that form the boundary.<\/li>\r\n \t
    2. Choose the transformation<\/strong> [latex]x = g(u,v)[\/latex] and [latex]y = h(u,v)[\/latex] based on the region or integrand structure.<\/li>\r\n \t
    3. Determine the new limits<\/strong> of integration in the [latex]uv[\/latex]-plane.<\/li>\r\n \t
    4. Find the Jacobian<\/strong> [latex]J(u,v) = \\dfrac{\\partial(x,y)}{\\partial(u,v)}[\/latex].<\/li>\r\n \t
    5. Replace the variables<\/strong> in the integrand to obtain the new integrand.<\/li>\r\n \t
    6. Replace [latex]dA[\/latex]<\/strong> (whether [latex]dy , dx[\/latex] or [latex]dx , dy[\/latex]) by [latex]|J(u,v)| , du , dv[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>
      \r\n

      Consider the integral<\/p>\r\n

      [latex]\\displaystyle\\int_0^2\\displaystyle\\int_0^{\\sqrt{2x-x^2}}\\sqrt{x^2+y^2}dy \\ dx[\/latex].<\/p>\r\n

      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\"]\"A Figure 2.\u00a0Changing a region from rectangular to polar coordinates.[\/caption]\r\n

      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

      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

      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<\/section>\r\n

      Notice in the next example that the region of integration often suggests a suitable transformation. This is a common and important situation.<\/p>\r\n\r\n

      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.\r\n\r\n[caption id=\"attachment_1424\" align=\"aligncenter\" width=\"328\"]\"A 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

      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

      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

      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

      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\"]\"A Figure 4.\u00a0A parallelogram in the [latex]xy[\/latex]-plane that we want to transform by a change in variables.[\/caption]\r\n

      Thus, we can choose the transformation<\/p>\r\n

      [latex]\\large{T(u,v)=\\left(\\frac{3u-v}2,\\frac{u-v}2\\right)}[\/latex]<\/p>\r\n

      and compute the Jacobian [latex]J(u, v)[\/latex]. We have<\/p>\r\n

      [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>\r\nTherefore, [latex]|J(u,v)|=\\frac12[\/latex]. Also, the original integrand becomes\r\n

      [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

      [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

      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<\/section>

      \r\n

      Using the change of variables [latex]u=x-y[\/latex] and [latex]v=x+y[\/latex], evaluate the integral<\/p>\r\n

      [latex]\\large{\\underset{R}{\\displaystyle\\iint}(x-y)e^{x^2-y^2}dA}[\/latex],<\/p>\r\n

      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\"]\"A Figure 5.\u00a0Transforming the region [latex]R[\/latex] into the region [latex]S[\/latex] to simplify the computation of an integral.<\/span>[\/caption]\r\n

      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

      Thus we can describe the region [latex]S[\/latex] (see the second region\u00a0Figure 5) as<\/p>\r\n

      [latex]\\large{s=\\left\\{(u,v)|1\\leq{v}\\leq3,\\frac{-1}v\\leq{u}\\leq\\frac1v\\right\\}}[\/latex].<\/p>\r\n

      The Jacobian for this transformation is<\/p>\r\n

      [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>\r\n

      Therefore, by using the transformation [latex]T[\/latex], the integral changes to<\/p>\r\n

      [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

      Doing the evaluation, we have<\/p>\r\n

      [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<\/section>","rendered":"

      Change of Variables for Double Integrals<\/h2>\n

      We’ve established 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 [latex]\\Delta u \\Delta v[\/latex] in the [latex]uv[\/latex]-plane by:<\/p>\n

      [latex]\\Delta A \\approx J(u,v) \\Delta u \\Delta v[\/latex]<\/p>\n

      Let’s revisit the definition of the double integral:<\/p>\n

      [latex]\\underset{R}{\\displaystyle\\iint}f(x,y) \\ dA=\\displaystyle\\lim_{m,n\\to\\infty}\\displaystyle\\sum_{i=1}^m\\displaystyle\\sum_{j=1}^nf(x_{ij}, y_{ij}) \\Delta A[\/latex]<\/p>\n

      Looking at Figure 1, we divide region [latex]S[\/latex] in the [latex]uv[\/latex]-plane into small subrectangles [latex]S_{ij}[\/latex]. The subrectangles [latex]R_{ij}[\/latex] in the [latex]xy[\/latex]-plane are the images of [latex]S_{ij}[\/latex] under transformation [latex]T(u,v) = (x,y)[\/latex].<\/p>\n

      \"On
      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.<\/figcaption><\/figure>\n

      The double integral becomes:<\/p>\n

      [latex]\\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

      This is exactly the double Riemann sum for the integral:<\/p>\n

      [latex]\\underset{S}{\\displaystyle\\iint}f(g(u,v),h(u,v))\\Bigg|\\dfrac{\\partial(x,y)}{\\partial(u,v)}\\Bigg|du \\ dv[\/latex].<\/p>\n

      \n

      change of variables for double integrals<\/h3>\n

      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]C^1[\/latex] transformation with a nonzero Jacobian on the interior of region [latex]S[\/latex] in the [latex]uv[\/latex]-plane. If [latex]T[\/latex] maps [latex]S[\/latex] into region [latex]R[\/latex] in the [latex]xy[\/latex]-plane and [latex]f[\/latex] is continuous on [latex]R[\/latex], then:<\/p>\n

      [latex]\\underset{R}{\\displaystyle\\iint}f(x,y) \\ dA=\\underset{S}{\\displaystyle\\iint}f(g(u,v),h(u,v))\\Bigg|\\dfrac{\\partial(x,y)}{\\partial(u,v)}\\Bigg|du \\ dv[\/latex].<\/p>\n<\/section>\n

      With this theorem, we can change the variables from [latex](x, y)[\/latex] to [latex](u, v)[\/latex] in a double integral simply by replacing:<\/p>\n

      [latex]dA=dx \\ dy=\\Bigg|\\dfrac{\\partial(x,y)}{\\partial(u,v)}\\Bigg|du \\ dv[\/latex]<\/p>\n

      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

      \n

      Why the Absolute Value?\u00a0<\/strong><\/p>\n

      We use [latex]|J(u,v)|[\/latex] in the integral because the Jacobian can be negative, but area must be positive. The absolute value ensures we’re measuring area correctly regardless of whether the transformation reverses orientation.<\/p>\n<\/section>\n

      \n

      Choosing a Good Transformation<\/strong><\/p>\n

      Look for patterns in:<\/p>\n

        \n
      • The boundary curves<\/strong> of the region (can they be simplified?)<\/li>\n
      • The integrand<\/strong> itself (does it suggest a coordinate system?)<\/li>\n
      • Symmetry<\/strong> in the problem (circular, elliptical, etc.)<\/li>\n<\/ul>\n

        Often the region’s shape will suggest natural substitutions that make the limits constants.<\/p>\n<\/section>\n

        \n

        How to: Apply Change of Variables to Double Integrals<\/strong><\/p>\n

          \n
        1. Sketch the region<\/strong> in the [latex]xy[\/latex]-plane and write the equations of the curves that form the boundary.<\/li>\n
        2. Choose the transformation<\/strong> [latex]x = g(u,v)[\/latex] and [latex]y = h(u,v)[\/latex] based on the region or integrand structure.<\/li>\n
        3. Determine the new limits<\/strong> of integration in the [latex]uv[\/latex]-plane.<\/li>\n
        4. Find the Jacobian<\/strong> [latex]J(u,v) = \\dfrac{\\partial(x,y)}{\\partial(u,v)}[\/latex].<\/li>\n
        5. Replace the variables<\/strong> in the integrand to obtain the new integrand.<\/li>\n
        6. Replace [latex]dA[\/latex]<\/strong> (whether [latex]dy , dx[\/latex] or [latex]dx , dy[\/latex]) by [latex]|J(u,v)| , du , dv[\/latex].<\/li>\n<\/ol>\n<\/section>\n
          \n

          Consider the integral<\/p>\n

          [latex]\\displaystyle\\int_0^2\\displaystyle\\int_0^{\\sqrt{2x-x^2}}\\sqrt{x^2+y^2}dy \\ dx[\/latex].<\/p>\n

          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