jacobian

Jacobian (Change of Variables in Multiple Integrals)


Master of Science in Mathematics
Lecture Notes. Accessed on: 2019-10-21 18:36:53

Jacobians

If $x=x(u,v)$ and $y=y(u,v)$ then the Jacobian of $x$ and $y$ with respect to $u$ and $v$ is \begin{equation} \frac{\partial (x,y)}{\partial (u,v)} =J(u,v)=\left| \begin{array}{cc} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array} \right|=\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial y}{\partial u}\frac{\partial x}{\partial v}. \end{equation}

Example. Determine the Jacobian for the transformation from the rectangular plane to the polar plane.

Solution. The conversion formulas are $x=r \cos \theta $ and $y=r \sin \theta .$ So the Jacobian is \begin{equation} J(r,\theta )=\left| \begin{array}{cc} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta } \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta } \end{array} \right|=\left| \begin{array}{cc} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{array} \right| =r \cos ^2 \theta -\left(-r \sin ^2 \theta \right)=r. \end{equation}

More generally for \begin{equation} x_1=x_1\left(u_1,\ldots,u_n\right), \quad x_2 = x_2\left(u_1,\ldots,u_n\right), \quad\ldots \quad x_n = x_n \left(u_1,\ldots ,u_n\right)\end{equation} the Jacobian of $x_1,\ldots , x_n$ with respect to $u_1,\ldots,u_n$ is \begin{equation} J\left(u_1, \ldots ,u_n\right)=\left| \begin{array}{ccc} \frac{\partial x_1}{\partial u_1} & \ldots & \frac{\partial x_1}{\partial u_n} \\ \ldots & \ldots & \ldots \\ \frac{\partial x_n}{\partial u_1} &\ldots & \frac{\partial x_n}{\partial u_n} \end{array} \right|. \end{equation}

Example. Determine the Jacobian for the transformation from rectangular coordinates to cylindrical coordinates.

Solution. The conversion formulas are $x=r \cos \theta ,$ $y=r \sin \theta ,$ and $z=z.$ So the Jacobian is \begin{align} J(r,\theta ,z) =\left| \begin{array}{ccc} \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{array} \right| & =\left| \begin{array}{ccc} \cos \theta & -r \sin \theta & 0 \\ \sin \theta & r \cos \theta & 0 \\ 0 & 0 & 1 \end{array} \right| \\ & =r \cos ^2 \theta -\left(-r \sin ^2 \theta \right)=r. \end{align}

Example. Determine the Jacobian for the transformation from rectangular coordinates to spherical coordinates.

Solution. The conversion formulas are $x=\rho \sin \phi \cos \theta ,$ $y=\rho \sin \phi \sin \theta $ and $z=\rho \cos \phi .$ So the Jacobian $J(\rho ,\theta ,\phi )$ is \begin{align} \left| \begin{array}{ccc} \frac{\partial x}{\partial \rho } & \frac{\partial x}{\partial \theta } & \frac{\partial x}{\partial \phi } \\ \frac{\partial y}{\partial \rho } & \frac{\partial y}{\partial \theta } & \frac{\partial y}{\partial \phi } \\ \frac{\partial z}{\partial \rho } & \frac{\partial z}{\partial \theta } & \frac{\partial z}{\partial \phi } \end{array} \right| & =\left| \begin{array}{ccc} \sin \phi \cos \theta & -\text{$\rho $sin} \phi \sin \theta & \rho \cos \phi \cos \theta \\ \sin \phi \sin \theta & \rho \sin \phi \cos \theta & \rho \cos \phi \sin \theta \\ \cos \phi & 0 & -\rho \sin \phi \end{array} \right| \\ & =-\rho ^2 \sin \phi . \end{align}

Example. Find the Jacobian $\frac{\partial (x,y)}{\partial (u,v)}$ given $$ u=\frac{x}{x^2+y^2} \quad \text{and} \quad v=\frac{y}{x^2+y^2}. $$

Solution. The Jacobian is \begin{align} J(x,y) & = \left| \begin{array}{cc} \partial _x \left(\frac{x}{x^2+y^2}\right) & \partial _y \left(\frac{x}{x^2+y^2}\right) \\ \partial _x \left(\frac{y}{x^2+y^2}\right) & \partial _y \left(\frac{y}{x^2+y^2}\right) \end{array} \right| =\left| \begin{array}{cc} \frac{y^2-x^2}{\left(x^2+y^2\right)^2} & -\frac{2 x y}{\left(x^2+y^2\right)^2} \\ -\frac{2 x y}{\left(x^2+y^2\right)^2} & \frac{x^2-y^2}{\left(x^2+y^2\right)^2} \end{array} \right| \\ & =\frac{y^2-x^2}{\left(x^2+y^2\right)^2}\frac{x^2-y^2}{\left(x^2+y^2\right)^2}-\frac{2 x y}{\left(x^2+y^2\right)^2}\frac{2 x y}{\left(x^2+y^2\right)^2} =-\frac{1}{\left(x^2+y^2\right)^2}. \end{align} Since
\begin{equation} u^2+v^2=\left(\frac{x}{x^2+y^2}\right)^2+\left(\frac{y}{x^2+y^2}\right)^2=\frac{1}{x^2+y^2} \end{equation} we find \begin{equation} J(u,v)=\frac{1}{J(x,y)}=\frac{-1}{\left(u^2+v^2\right)^2}. \end{equation}

Change of Variable in a Double Integral

Theorem. (Change of Variable in a Double Integral) Let $z=f(x,y)$ be a continuous function on a region $R$ in the $x y$-plane, and let $T$ be a one-to-one transformation that maps the region $D$ in the $u v$-plane onto $R$ under the change of variables $x=x(u,v)$ and $y=y(u,v),$ where $x$ and $y$ are continuously differentiable functions on $R.$ If $J(u,v)\neq 0,$ then \begin{equation} \iint_R f(x,y)dx\, dy=\iint_D f(x(u,v),y(u,v))| J(u,v)| du dv. \end{equation}

Proof. Let $T$ be the transformation from the $u v$-plane to the $x y$-plane given by $x=x(u,v)$ and $y=y(u,v).$ We will consider the effect that $T$ has on the area of a small rectangular region $D$ in the $uv$-plane with vertices $(u_0,v_0)$, $(u_0+\Delta u,v_0)$, $(u_0,v_0+\Delta v)$, and $(u_0+\Delta u,v_0+\Delta v)$ where $(u_0,v_0)$ is a given point and both $\Delta u$ and $\Delta v$ are increments of $u$ and $v$ respectively. Let $L_1$ be the line segment between the vertices $(u_0,v_0)$ and $(u_0+\Delta u,v_0)$ and also let $L_2$ be the line segment between the vertices $(u_0,v_0)$ and $(u_0,v_0+\Delta v).$

Define the vector function ${r}$ by \begin{equation}{r}(u,v)=x(u,v){i}+y(u,v){j} \end{equation} and also \begin{equation} {a} = {r}(u_0+\Delta u,v_0)-{r}(u_0,v_0) \quad \text{and} \quad {b} = {r}(u_0,v_0+\Delta v)-{r}(u_0,v_0). \end{equation} Notice that $T(D)=R$ is a region in the $xy$-plane whose area can be approximated by $|{a}\times {b}|$ because we are assuming $\Delta u$ and $\Delta v$ are small increments in $u$ and $v$ and $T$ is a continuous one-to-one onto transformation. Let $\Delta A$ denote the area of $R$, it follows \begin{align} \Delta A \approx |{a}\times {b}| & = \left| \frac{{r}(u_0+\Delta u,v_0)-{r}(u_0,v_0) \Delta u}{ \Delta u} \times \frac{{r}(u_0,v_0+\Delta v)-{r}(u_0,v_0) \Delta v}{ \Delta v} \right | \\ & \approx \left| {r}_u \Delta u \times {r}_v \Delta v \right| = \left| {r}_u \times {r}_v \right| \Delta u \Delta v \end{align} If we consider the factor $\left| {r}_u \times {r}_v \right|$ in terms of the functions which define the transformation $T$ we are lead to \begin{equation} {r}_u \times {r}_v =\left| \begin{array}{ccc} {i} & {j} & {k} \\ x_u & y_u & 0 \\ x_v & y_v & 0 \end{array} \right| = 0{i}-0{j}+(x_u y_v-x_v y_u) {k}\end{equation} which implies \begin{equation} |{r}_u \times {r}_v | =x_u y_v-x_v y_u = \left| \begin{array}{cc} x_u & y_u \\ x_v & y_v \end{array} \right| = \left| \begin{array}{cc} x_u & x_v \\ y_u & y_v \end{array} \right| = \left| \frac{\partial (x,y)}{\partial (u,v)} \right|. \end{equation}

Example. Use a change of variables to evaluate the integral $$ \iint_R e^{(x+y)/(x-y)} \, dA, $$ where $R$ is the trapezoidal region with vertices $(1,0),$ $(2,0),$ $(0,-2),$ and $(0,-1).$

Solution. Since it is not easy to integrate $f(x,y)=e^{(x+y)/(x-y)},$ we make a change of variables suggested by the form of $f$ namely: $u=x+y$ and $v=x-y.$ Solving for $x$ and $y$ \begin{equation} x =\frac{1}{2}(u+v)\qquad \text{and} \qquad y=\frac{1}{2}(u-v) \end{equation} to find the Jacobian is \begin{equation} J(u,v)=\left| \begin{array}{cc} 1/2 & 1/2 \\ 1/2 & -1/2 \end{array} \right|=-\frac{1}{2}.\end{equation} To find the region $D$ in the $u v$-plane corresponding to $R$ we note that the sides of $R$ lie on the lines $y=0,$ $x-y=2,$ $x=0,$ $x-y=1$ and using the rules for the transformation, $x=(u+v)/2 $ and $y=(u-v)/2$ the images of the lines in the $u v$-plane are $u=v,$ $v=2,$ $u=-v ,$ and $v=1.$ Thus the region $D$ is the trapezoidal region with vertices $(1,1),$ $(2,2),$ $(-2,2)$ and $(-1,1);$ that is \begin{equation}D=\{(u,v)|1\leq v\leq 2, -v\leq u\leq v\}. \end{equation} Therefore we can evaluate the integral as follows \begin{align} \iint_R e^{(x+y)/(x-y)} \, dA & =\iint_D e^{u/v} \left(\frac{1}{2}\right) \, dA \\ &=\int_1^2\int_{-v}^ve^{u/v}\left(\frac{1}{2}\right) \, du \, dv \\ & =\frac{1}{2}\int_1^2\left(e-e^{-1}\right)v dv = \frac{3}{4}\left(e-e^{-1}\right). \end{align}

Example. Use a change of variables to evaluate the integral $$ \iint_R 3 x y \, dA $$ where $R$ is the region bounded by the lines $x-2y=0,$ $x-2y=-4,$ $x+y=4,$ and $x+y=1.$

Solution. Let $u=x+y$ and $v=x-2y.$ Then solving for $x$ and $y$ produces $ x=\frac{1}{3}(2u+v)$ and $y=\frac{1}{3}(u-v).$ The Jacobian of $u$ and $v$ is $$ J(u,v)=\left|\begin{array}{cc} \frac{2}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{-1}{3} \end{array} \right|=\left(\frac{2}{3}\right)\left(\frac{-1}{3}\right)-\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)=-\frac{1}{3}.$$ The bounds under the transformation determined by the following equations \begin{align} & x+y =1 \Longrightarrow u=1 & x+y =4 \Longrightarrow u=4 \\ & x-2y =0 \Longrightarrow v =0 & x-2y =-4 \Longrightarrow v=-4\end{align} We consider the region $D$ in the $u v$-plane as vertically simple. Then it follows \begin{align} \iint_R 3 x y \, dA & =\iint_D 3 \frac{1}{3}(2u+v) \frac{1}{3}(u-v)\, |J(u,v)| \, dv \, du \\ & =\int_1^4\int_{-4}^0\frac{1}{3}\left(2u^2-u v-v^2\right)\left|-\frac{1}{3}\right| \, dv \, du \\ & =\int_1^4\int_{-4}^0\frac{1}{9}\left(2u^2-u v-v^2\right) dv du \\ & = \int_1^4 \frac{1}{9} \left(8 u^2+8 u-\frac{64}{3}\right) \, du =\frac{164}{9}. \end{align}

Example. Use a change of variables to evaluate the integral $$ \iint_R (x+y)^2\sin ^2(x-y)\, dA, $$ where $R$ is the region bounded by the square with vertices $(0,1),$ $(1,2),$ $(2,1),$ and $(1,0).$

Solution. The region $R$ is bounded by the lines $x-y=1,$ $x-y=-1,$ $x+y=1,$ and $x+y=3.$ Let $u=x+y$ and $v=x-y,$ then solving for $x$ and $y$ produces $x=\frac{1}{2}(u+v)$ and $y=\frac{1}{2}(u-v).$ The Jacobian of $u$ and $v$ is \begin{equation} J(u,v)=\left|\begin{array}{cc} 1/2 & 1/2 \\ 1/2 & -1/2 \end{array} \right|=\left(\frac{1}{2}\right)\left(\frac{-1}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=-\frac{1}{2}.\end{equation} The bounds under the transformation determined by the following equations \begin{align} & x+y=1 \Longrightarrow u=1 & x+y=3 \Longrightarrow u=3 \\ & x-y=-1 \Longrightarrow v=-1 & x-y=1 \Longrightarrow v=1 \end{align} Let’s consider the region $D$ in the $u v$-plane as horizontally simple. It follows \begin{align} \iint_R (x+y)^2\sin ^2(x-y) \, dA & =\iint_D u^2 \sin ^2v |J(u,v)| \, du \, dv \\ &=\frac{1}{2}\int_{-1}^1\int_1^3u^2 \sin ^2(v) \, du \, dv \\ & =\frac{1}{2}\int{-1}^1 \frac{26 \sin ^2(v)}{3} \, dv =\frac{13}{6} (2-\sin 2). \end{align}

Example. Use a change of variables to evaluate the integral \begin{equation} \iint_R y^3(2x-y)\cos (2x-y) \, dy \, dx. \end{equation} where $R$ is the region bounded by the parallelogram with vertices $(0,0),$ $(2,0),$ $(3,2),$ and $(1,2).$

Solution. The boundary lines of the parallelogram are $y=0,$ $y=2,$ $y=2x,$ and $y=2x-4.$ Let $u=2x-y$ and $v=y$ with boundary lines $u=0,$ $u=4,$ $v=0,$ and $v=2.$ Solving for $x$ and $y$ produces $x=\frac{1}{2}(u+v)$ and $y=v.$ Since the Jacobian is \begin{equation} J(u,v)=\left| \begin{array}{cc} 1/2 & 1/2 \\ 0 & 1 \end{array} \right|=\frac{1}{2}, \end{equation} we evaluate the integral as follows
\begin{align} &\iint_D y^3(2x-y)\cos (2x-y) \, dy \, dx =\int_0^4\int_0^2v^3u \cos (u)\frac{1}{2} \, dv \, du \\ &\quad =\int_0^4 2 u \cos (u) \, du =-2+2 \cos (4)+8 \sin (4). \end{align}

Example. Use a change of variables to evaluate the integral \begin{equation} \iint_R \left(x^4-y^4\right)e^{x y} \, dA \end{equation} where $R$ is the region bounded by the hyperbolas $x y=1,$ $x y=2,$ $x^2-y^2=1,$ and $x^2-y^2=4.$

Solution. Let $u=x y$ and $v=x^2-y^2.$ Then \begin{equation} \frac{\partial (u,v)}{\partial (x,y)}=\left| \begin{array}{cc} y & x \\ 2x & -2y \end{array} \right|=-2y^2-2x^2=-2(x^2+y^2).\end{equation} Notice $v^2=\left(x^2-y^2\right)^2=x^4-2x^2y^2+y^4$ we have \begin{equation}
\left(x^2+y^2\right)^2 =x^4-2x^2y^2+y^4+4x^2y^2 =v^2+4u^2.\end{equation} Thus $x^2+y^2=\sqrt{v^2+4u^2}$ and so $J(u,v)=\frac{-1}{2\sqrt{v^2+4u^2}}.$ We will make use of $$\frac{\partial (u,v)}{\partial (x,y)}\frac{\partial (x,y)}{\partial (u,v)}=1$$ because it is easier not to solve for $x$ and $y$ in terms of $u$ and $v.$ So we evaluate the integral as follows \begin{align} \iint_R\left(x^4-y^4\right)e^{x y} \, dA & =\int_1^4\int_1^2e^uv\sqrt{v^2+4u^2}\left|\frac{-1}{2\sqrt{v^2+4u^2}} \right|dudv \\ & =\frac{1}{2}\int_1^4\int_1^2v e^u \, du \, dv \\ & =\frac{1}{2}\int_1^4 \left(-e+e^2\right) v \, dv =\frac{15}{4}e(e-1). \end{align}

Example. Use a change of variables to evaluate the integral $$ I=\iint_R \ln \left(\frac{x-y}{x+y}\right) \, dy \, dx $$ where $R$ is the triangular region bounded by the vertices $(1,0),$ $(4,-3),$ $(4,1)$

Solution. Let $u=x-y$ and $v=x+y$ so that $x=\frac{1}{2}(u+v)$ and $y=\frac{-1}{2}u+\frac{1}{2}.$ Then the Jacobian is \begin{equation} J(u,v)=\left| \begin{array}{cc} 1/2 & 1/2 \\ -1/2 & 1/2 \end{array} \right|=\frac{1}{2}.\end{equation} The given the region $R$ is bounded by the lines $x-3y=1,$ $x+y=1,$ and $x=4$ which transform into $2u-v=1,$ $v=1,$ and $u+v=8.$ Therefore we evaluate the integral as follows,
\begin{align} I& =\frac{1}{2}\int_1^5\int_{(v+1)/2}^{8-v}\ln \left(\frac{u}{v}\right)du dv \\ &=\frac{1}{4}\int_1^5 \left[-(v+1)\ln \left(\frac{v+1}{2v}\right)+2(8-v)\ln \left(\frac{8-v}{v}\right)+3(v-5)\right] \, dv \\ &=\frac{1}{4}\left(49\ln 7-\frac{75}{2}\ln 5-27\ln 3+6\right). \end{align}

Example. Use a change of variables to evaluate the integral $$ I=\iint_D\exp \left(-\frac{x^2}{a^2}-\frac{y^2}{b^2}\right) \, dy \, dx, $$ where $D$ is the region bounded by the quarter ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ in the first octant.

Solution. Let’s try the change of variables $ x=a r \cos \theta $ and $y=b r \sin \theta $ with $0\leq r<\infty $ and $0\leq \theta <2\pi ;$ to see if we can simply the region of integration and the integrand. Notice \begin{equation} – \frac{x^2}{a^2} – \frac{y^2}{b^2}=-\left[\frac{x^2}{a^2}+\frac{y^2}{b^2}\right]=-\left[\frac{(a r \cos \theta )^2}{a^2}+\frac{(b r \sin \theta )^2}{b^2}\right]=-r^2 \end{equation} using $\sin ^2\theta +\cos ^2\theta =1.$ Further the region of becomes $r^2=1$ or $r=1$ which is the circle of radius 1. We wish to use \begin{equation} \iint_D f(x,y)dx\, dy=\iint_R f(x(u,v),y(u,v)) |J(u,v)| \, du \, dv \end{equation} so we compute the Jacobian, \begin{equation} J(r,\theta )=\left| \begin{array}{cc} a \cos \theta & -a \text{rsin} \theta \\ b \sin \theta & b r \cos \theta \end{array} \right|=a b r \cos \theta +a b r \sin ^2\theta =a b r. \end{equation} Now then we have, \begin{align} I& =\int_0^{\pi /2}\int_0^1e^{-r^2}|a b r| \, dr \, d\theta =\int_0^{\pi /2}\int_0^1e^{-r^2}a b r \, dr \, d\theta \\ & =\int_0^{\pi /2} a b \left(\frac{1}{2}-\frac{1}{2 e}\right) \, d\theta =\frac{a b \pi }{4}\left(1-e^{-1}\right). \end{align}

Example. Use a change of variables to evaluate the integral $$ \iint_D \left(\frac{x-y}{x+y}\right)^4 \, dy \, dx, $$ where $D$ be the region in the $x y$-plane that is bounded by the coordinate axes and the line $x+y=1$

Solution. The region is transformed into $0\leq v\leq 1$ and $-v\leq u\leq v$ and so \begin{equation} \iint_R \left(\frac{x-y}{x+y}\right)^4 \, dy \, dx =\int_0^1\int_{-v}^v\frac{u^4}{v^4}\left(\frac{1}{2}\right) \, du \, dv =\int_0^1 \frac{v}{5} \, dv =\frac{1}{10}. \end{equation}

Example. Use integration and a change of variables determine the area of an ellipse.

Solution. Assume the ellipse is given in standard form by $$ \left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=1. $$ Let $u=\frac{x}{a}$ and $v=\frac{y}{b},$ then the ellipse in the $x y$-plane corresponds to the unit circle $u^2+v^2=1$ in the $u v$-plane. Since $x=a u$ and $y=b v,$ the Jacobian is $\left| \begin{array}{cc} a & 0 \\ 0 & b \end{array}\right|=$ $a b $ and so the area of an ellipse is given by \begin{align*} A & =4\int_0^a\int_0^{b \sqrt{1-x^2/a^2}} dy \, dx \\ & =4\int_0^1\int_0^{ \sqrt{1-u^2}}a b dv du \\ & =ab 4 \int_0^1 \int_0^{ \sqrt{1-u^2}} dv du \\ & =4\int_0^1 a b \sqrt{1-u^2} \, du \\ & = a b \pi\end{align*} since $\pi$ is the area of the unit circle.

Example. A rotation of the $x y$-plane through the fixed angle $\theta $ is given by \begin{equation} x=u \cos \theta -v \sin \theta \qquad \text{and} \qquad y=u \sin \theta +v \cos \theta \end{equation} Compute the Jacobian $\frac{\partial (x,y)}{\partial (u,v)}.$ Let $E$ denote the region bounded by the ellipse $x^2+x y+y^2=3.$ Use a rotation of $\pi /4$ to obtain an integral that is equivalent to $$
\iint_E y \, dy \, dx. $$ Evaluate the transformed integral.

Solution. We find $J(u,v)=1$ and so $dx\, dy=du\, dv.$ A rotation of $\frac{\pi }{4}$ eliminates the $u v$-term so we use the transformation \begin{equation} x=\frac{\sqrt{2}}{2}(u-v) \qquad \text{and}\qquad y=\frac{\sqrt{2}}{2}(u+v)\end{equation} Then $x^2+x y+y^2$ becomes $\left(\frac{u}{\sqrt{2}}\right)^2+\left(\frac{v}{\sqrt{6}}\right)^2=1.$ Therefore, \begin{align*} \iint_E y \, dy \, dx & =4\int_0^{\sqrt{2}}\int_0^{\sqrt{6-3u^2}}\frac{\sqrt{2}}{2}(u+v)(1)\, dv \, du \\ & =4\int_0^{\sqrt{2}} \frac{-\frac{3 u^2}{2}+\sqrt{6-3 u^2} u+3}{\sqrt{2}} \, du \\ & =8+\frac{8}{\sqrt{3}} \\ & =\frac{8\sqrt{3}+8}{\sqrt{3}} \\ & =\frac{24+8\sqrt{3}}{3} \\ & =\frac{8}{3}\left(3+\sqrt{3}\right) \\ & \approx 12.6188. \end{align*}

Change of Variable in a Triple Integral

Theorem. (Change of Variable in a Triple Integral) Let $f$ be a continuous function on a region $R$ in the $x y z$-space, and let $T$ be a one-to-one transformation that maps the region $D$ in the $u v w$-space onto $R$ under the change of variables $x=x(u,v,w),$ $y=y(u,v,w),$ and $z=z(u,v,w)$ where functions $x,$ $y,$ and $z$ are continuously differentiable functions on $D.$ If $J(u,v,w)\neq 0,$ then \begin{align} & \iiint_Rf(x,y,z) \, dx \, dy \, dz \\ & = \iiint_D f(x(u,v,w),y(u,v,w),z(u,v,w)) \left|J(u,v,w)\right| \, du \, dv \, dw. \end{align}

Example. Use integration and a change of variables the volume of an ellipsoid.

Solution. Assume the ellipsoid is given in standard form by $$ \left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2+\left(\frac{z}{c}\right)^2=1. $$ Let $u=\frac{x}{a},$ $v=\frac{y}{b},$ and $w=\frac{z}{c}$ then the ellipsoid corresponds to the unit sphere $u^2+v^2+w^2=1.$ Since $x=a u,$ $y=b v,$ $z=c w,$ the Jacobian is $a b c$ $$J(u,v,w)=\left| \begin{array}{ccc} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{array} \right|=a b c $$ and so the volume of an ellipsoid is given by \begin{align} V & = 8\int_0^a\int_0^{ \sqrt{1-x^2/a^2}}\int_0^{ \sqrt{1-x^2/a^2-y^2/b^2}} \, dz \, dy \, dx \\ & =8\int_0^1\int_0^{ \sqrt{1-u^2}}\int_0^{ \sqrt{1-u^2-v^2}} a b c \, dw \, dv \, du \\ & = \frac{4}{3}a b c \pi . \end{align}

Exercises on Jacobians and Change of Variables in Multiple Integrals

Exercise. For each of the following find the Jacobian of the change of variables.

$(1) \quad x=u^2$ and $y=u+v$

$(2) \quad x=u \cos v$ and $y=u \sin v$

$(3) \quad x=\frac{u}{v},$ $y=\frac{v}{w},$ and $z=\frac{w}{u}$

$(4) \quad u=y e^{-x}$ and $v=e^x$

$(5) \quad u=\frac{x}{x^2+y^2}$ and $v=\frac{y}{x^2+y^2}$

$(6) \quad x=u+v-w,$ $y=2u-v+3w,$ and $z=-u+2v-w$

Exercise. Use a change of variables to compute the following integrals.

$(1) \quad \displaystyle \underset{D}{\int \int }\left(\frac{x-y}{x+y}\right)^5 \, dy \, dx$ where $D$ is the region in the $x y$-plane bounded by the coordinate axes and the line $x+y=1$

$(2) \quad \displaystyle \underset{D}{\int \int }(x-y)e^{x^2+y^2} \, dy \, dx$ where $D$ is the region in the $x y$-plane bounded by the coordinate axes and the line $x+y=1$

$(3) \quad \displaystyle \underset{D}{\int \int }\left(\frac{2x+y}{x-2y+5}\right)^2 \, dy \, dx$ where $D$ is the square in the $x y$-plane with vertices $(0,0),$ $(1,-2),$ $(3,-1),$ and $(2,1)$

$(4) \quad \displaystyle \underset{D}{\int \int }\sqrt{(2x+y)(x-2y)} \, dy \, dx$ where $D$ is the square in the $x y$-plane with vertices $(0,0),$ $(1,-2),$ $(3,-1),$ and $(2,1)$

$(5) \quad \displaystyle \underset{D}{\int \int }e^{(2y-x)(y+2x)} \, dA$ where $D$ is the trapezoid with vertices $(0,2),$ $(1,0),$ $(4,0),$ and $(0,8)$

$(6) \quad \displaystyle \iint_Ry^3(2x-y)\cos (2x-y)\, dy \, dx$ where $D$ is the region bounded by the parallelogram with vertices $(0,0),$ $(2,0),$ $(3,2),$ and $(1,2)$

Exercise. Under the change of variables $x=s^2-t^2,$ $y=2s t,$ the quarter circle region in the $s t$-plane given by $s^2+t^2\leq 1,$ $s\geq 0,$ $t\geq 0$ is mapped onto a certain region $D$ of the $x y$-plane. Evaluate \begin{equation} \iint_R\frac{1}{\sqrt{x^2+y^2}} \, dy \, dx. \end{equation}

Exercise. A rotation of the $x y$-plane through the fixed angle $\theta $ is given by $x = u \cos \theta -v \sin \theta$ and $y=u \sin \theta +v \cos \theta.$ Determine the Jacobian $\frac{\partial (x,y)}{\partial (u,v)}.$ Let $E$ denote the region bounded by the ellipse $x^2+x +y^2=3.$ Use a rotation of $\pi /4$ to obtain an integral that is equivalent to \begin{equation} \underset{E}{\int \int }y dy dx. \end{equation} Evaluate the transformed integral.

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