# Jacobian (Change of Variables in Multiple Integrals)

• By David A. Smith, Founder & CEO, Direct Knowledge
• David Smith has a B.S. and M.S. in Mathematics and has enjoyed teaching calculus, linear algebra, and number theory at both Tarrant County College and the University of Texas at Arlington. David is the Founder and current CEO of Direct Knowledge.

## 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{align*} \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{align*}

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{align*} 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{align*}

More generally for

\begin{align*} & x_1=x_1\left(u_1,\ldots,u_n\right), \\ & x_2 = x_2\left(u_1,\ldots,u_n\right), \\ & \ldots \\ & x_n = x_n \left(u_1,\ldots ,u_n\right)\end{align*}

the Jacobian of $x_1,\ldots , x_n$ with respect to $u_1,\ldots,u_n$ is

\begin{align*} 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{align*}

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{align*} 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{align*}

we find

\begin{align*} J(u,v)=\frac{1}{J(x,y)}=\frac{-1}{\left(u^2+v^2\right)^2}. \end{align*}

## 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{align*} & \iint_R f(x,y)dx\, dy \\ & =\iint_D f(x(u,v),y(u,v))| J(u,v)| du dv. \end{align*}

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{align*}{r}(u,v)=x(u,v){i}+y(u,v){j} \end{align*}

and also $a = {r}(u_0+\Delta u,v_0)-{r}(u_0,v_0)$ and $b = {r}(u_0,v_0+\Delta v)-{r}(u_0,v_0).$

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{align*} {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{align*}

which implies

\begin{align*} |{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{align*}

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 $y$

to find the Jacobian is

\begin{align*} J(u,v) =\left| \begin{array}{cc} 1/2 & 1/2 \\ 1/2 & -1/2 \end{array} \right|=-\frac{1}{2}.\end{align*}

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{align*}D=\{(u,v)|1\leq v\leq 2, -v\leq u\leq v\}. \end{align*}

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

\begin{align*}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}.\end{align*}

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{align*} 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{align*}

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{align*} \iint_R y^3(2x-y)\cos (2x-y) \, dy \, dx. \end{align*}

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{align*} J(u,v)=\left| \begin{array}{cc} 1/2 & 1/2 \\ 0 & 1 \end{array} \right|=\frac{1}{2}, \end{align*}

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 \\ & =\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 $$\iint_R \left(x^4-y^4\right)e^{x y} \, dA$$ 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{align*} \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{align*}

Notice $v^2=\left(x^2-y^2\right)^2=x^4-2x^2y^2+y^4$ we have

\begin{align*} \left(x^2+y^2\right)^2 & =x^4-2x^2y^2+y^4+4x^2y^2 \\ & =v^2+4u^2.\end{align*}

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{align*} J(u,v) & =\left| \begin{array}{cc} 1/2 & 1/2 \\ -1/2 & 1/2 \end{array} \right| \\ & =\frac{1}{2}.\end{align*}

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{align*} – \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{align*}

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{align*} \iint_D f(x,y)dx\, dy=\iint_R f(x(u,v),y(u,v)) |J(u,v)| \, du \, dv \end{align*}

so we compute the Jacobian,

\begin{align*} 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{align*}

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{align*} \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{align*}

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 $x=u \cos \theta -v \sin \theta$ and $y=u \sin \theta +v \cos \theta.$ 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

$$x=\frac{\sqrt{2}}{2}(u-v)$$ and $$y=\frac{\sqrt{2}}{2}(u+v).$$

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 $$\iint_R\frac{1}{\sqrt{x^2+y^2}} \, dy \, dx.$$

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 $$\underset{E}{\int \int }y dy dx.$$ Evaluate the transformed integral.