### 2 The Jacobian

Given an integral of the form $\phantom{\rule{1em}{0ex}}\int \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}{\int }_{A}f\left(x,y\right)\phantom{\rule{0.3em}{0ex}}dxdy$

Assume we have a change of variables of the form $x=x\left(u,v\right)$ and $y=y\left(u,v\right)$ then the Jacobian of the transformation is defined as

$\phantom{\rule{2em}{0ex}}J\left(u,v\right)=\left|\begin{array}{cc}\hfill \frac{\partial x}{\partial u}\hfill & \hfill \frac{\partial x}{\partial v}\hfill \\ \hfill \hfill & \hfill \hfill \\ \hfill \frac{\partial y}{\partial u}\hfill & \hfill \frac{\partial y}{\partial v}\hfill \end{array}\right|$

##### Key Point 10

Jacobian in Two Variables

For given transformations $x=x\left(u,v\right)$ and $y=y\left(u,v\right)$ the Jacobian is

$J\left(u,v\right)=\left|\begin{array}{cc}\hfill \frac{\partial x}{\partial u}\hfill & \hfill \frac{\partial x}{\partial v}\hfill \\ \hfill \hfill & \hfill \hfill \\ \hfill \frac{\partial y}{\partial u}\hfill & \hfill \frac{\partial y}{\partial v}\hfill \end{array}\right|$

Notice the pattern occurring in the $x$ , $y$ , $u$ and $v$ . Across a row of the determinant the numerators are the same and down a column the denominators are the same.

Notation

Different textbooks use different notation for the Jacobian. The following are equivalent.

$\phantom{\rule{2em}{0ex}}J\left(u,v\right)=J\left(x,y;u,v\right)=J\left(\frac{x,y}{u,v}\right)=\left|\frac{\partial \left(x,y\right)}{\partial \left(u,v\right)}\right|$

The Jacobian correctly describes how area elements change under such a transformation. The required relationship is

$\phantom{\rule{2em}{0ex}}dxdy\to \left|J\left(u,v\right)\right|\phantom{\rule{0.3em}{0ex}}dudv$

that is, $\left|J\left(u,v\right)\right|\phantom{\rule{0.3em}{0ex}}dudv$ plays the role of $dxdy$ .

##### Key Point 11

Jacobian for Transforming Areas

When transforming area elements employing the Jacobian it is the modulus of the Jacobian that must be used.

##### Example 24

Find the area of the circle of radius $R$ .

Figure 30

##### Solution

Let $A$ be the region bounded by a circle of radius $R$ centred at the origin. Then the area of this region is ${\int }_{A}dA$ . We will calculate this area by changing to polar coordinates, so consider the usual transformation $x=rcos\theta ,y=rsin\theta$ from cartesian to polar coordinates. First we require all the partial derivatives

$\phantom{\rule{2em}{0ex}}\frac{\partial x}{\partial r}=cos\theta \phantom{\rule{2em}{0ex}}\frac{\partial y}{\partial r}=sin\theta \phantom{\rule{2em}{0ex}}\frac{\partial x}{\partial \theta }=-rsin\theta \phantom{\rule{2em}{0ex}}\frac{\partial y}{\partial \theta }=rcos\theta$

Thus

$\begin{array}{rcll}J\left(r,\theta \right)=\left|\begin{array}{cc}\hfill \frac{\partial x}{\partial r}\hfill & \hfill \frac{\partial x}{\partial \theta }\hfill \\ \hfill \hfill & \hfill \hfill \\ \hfill \frac{\partial y}{\partial r}\hfill & \hfill \frac{\partial y}{\partial \theta }\hfill \end{array}\right|=\left|\begin{array}{cc}\hfill cos\theta \hfill & \hfill -rsin\theta \hfill \\ \hfill sin\theta \hfill & \hfill rcos\theta \hfill \end{array}\right|& =& cos\theta ×rcos\theta -\left(-rsin\theta \right)×sin\theta & \text{}\\ & =& r\left({cos}^{2}\theta +{sin}^{2}\theta \right)=r\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

This confirms the previous result for polar coordinates, $dxdy\to rdrd\theta$ . The limits on $r$ are $r=0$ (centre) to $r=R$ (edge). The limits on $\theta$ are $\theta =0$ to $\theta =2\pi$ , i.e. starting to the right and going once round anticlockwise. The required area is

$\phantom{\rule{2em}{0ex}}{\int }_{A}dA={\int }_{0}^{2\pi }{\int }_{0}^{R}\left|J\left(r,\theta \right)\right|drd\theta ={\int }_{0}^{2\pi }{\int }_{0}^{R}r\phantom{\rule{0.3em}{0ex}}drd\theta =2\pi \frac{{R}^{2}}{2}=\pi {R}^{2}$

Note that here $r>0$ so $\left|J\left(r,\theta \right)\right|=J\left(r,\theta \right)=r.$

##### Example 25

The diamond shaped region $A$ in Figure 31(a) is bounded by the lines $x+2y=2$ , $x-2y=2$ , $x+2y=-2$ and $x-2y=-2$ . We wish to evaluate the integral

$\phantom{\rule{2em}{0ex}}I=\int \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}{\int }_{A}{\left(3x+6y\right)}^{2}dA$

over this region. Since the region $A$ is neither vertically nor horizontally simple, evaluating $I$ without changing coordinates would require separating the region into two simple triangular regions. So we use a change of coordinates to transform $A$ to a square region in Figure 31(b) and evaluate $I$ .

Figures 31 (a) and (b)

##### Solution

By considering the equations of the boundary lines of region $A$ it is easy to see that the change of coordinates

$\phantom{\rule{2em}{0ex}}\phantom{\rule{0.3em}{0ex}}du=x+2y\phantom{\rule{2em}{0ex}}\left(1\right)\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}v=x-2y\phantom{\rule{2em}{0ex}}\left(2\right)$

will transform the boundary lines to $u=2$ , $u=-2$ , $v=2$ and $v=-2$ . These values of $u$ and $v$ are the new limits of integration. The region $A$ will be transformed to the square region ${A}^{\prime }$ shown above.

We require the inverse transformations so that we can substitute for $x$ and $y$ in terms of $u$ and $v$ . By adding (1) and (2) we obtain $u+v=2x$ and by subtracting (1) and (2) we obtain $u-v=4y$ , thus the required change of coordinates is

$\phantom{\rule{2em}{0ex}}x=\frac{1}{2}\left(u+v\right)\phantom{\rule{2em}{0ex}}y=\frac{1}{4}\left(u-v\right)$

Substituting for $x$ and $y$ in the integrand ${\left(3x+6y\right)}^{2}$ of $I$ gives

$\phantom{\rule{2em}{0ex}}{\left(\frac{3}{2}\left(u+v\right)+\frac{6}{4}\left(u-v\right)\right)}^{2}=9{u}^{2}$

We have the new limits of integration and the new form of the integrand, we now require the Jacobian. The required partial derivatives are

$\phantom{\rule{2em}{0ex}}\frac{\partial x}{\partial u}=\frac{1}{2}\phantom{\rule{2em}{0ex}}\frac{\partial x}{\partial v}=\frac{1}{2}\phantom{\rule{2em}{0ex}}\frac{\partial y}{\partial u}=\frac{1}{4}\phantom{\rule{2em}{0ex}}\frac{\partial y}{\partial v}=-\frac{1}{4}$

Then the Jacobian is

Then $d{A}^{\prime }=\left|J\left(u,v\right)\right|dA=\frac{1}{4}dA$ . Using the new limits, integrand and the Jacobian, the integral can be written

$\phantom{\rule{2em}{0ex}}I={\int }_{-2}^{2}{\int }_{-2}^{2}\frac{9}{4}{u}^{2}\phantom{\rule{0.3em}{0ex}}dudv.$

You should evaluate this integral and check that $I=48$ .

This Task concerns using a transformation to evaluate $\int \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\int \left({x}^{2}+{y}^{2}\right)\phantom{\rule{0.3em}{0ex}}dxdy$ .

1. Given the transformations $u=x+y$ , $v=x-y$ express $x$ and $y$ in terms of $u$ and $v$ to find the inverse transformations:

$\phantom{\rule{2em}{0ex}}u=x+y$ (1)

$\phantom{\rule{2em}{0ex}}v=x-y$ (2)

Add equations (1) and (2) $\phantom{\rule{1em}{0ex}}u+v=2x$

Subtract equation (2) from equation (1) $\phantom{\rule{1em}{0ex}}u-v=2y$

So $x=\frac{1}{2}\left(u+v\right)\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}y=\frac{1}{2}\left(u-v\right)$

2. Find the Jacobian $J\left(u,v\right)$ for the transformation in part (a):

Evaluating the partial derivatives, $\frac{\partial x}{\partial u}=\frac{1}{2}$ $\frac{\partial x}{\partial v}=\frac{1}{2}$ $\frac{\partial y}{\partial u}=\frac{1}{2}$ and $\frac{\partial y}{\partial v}=-\frac{1}{2}$ so the Jacobian $\left|\begin{array}{cc}\hfill \frac{\partial x}{\partial u}\hfill & \hfill \frac{\partial x}{\partial v}\hfill \\ \hfill \hfill \\ \hfill \frac{\partial y}{\partial u}\hfill & \hfill \frac{\partial y}{\partial v}\hfill \end{array}\right|=\left|\begin{array}{cc}\hfill \frac{1}{2}\hfill & \hfill \frac{1}{2}\hfill \\ \hfill \hfill \\ \hfill \frac{1}{2}\hfill & \hfill -\frac{1}{2}\hfill \end{array}\right|=-\frac{1}{4}-\frac{1}{4}=-\frac{1}{2}$

3. Express the integral $I=\int \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\int \left({x}^{2}+{y}^{2}\right)\phantom{\rule{0.3em}{0ex}}dxdy$ in terms of $u$ and $v$ , using the transformations introduced in 1. and the Jacobian found in 2.:

On letting $x=\frac{1}{2}\left(u+v\right)$ , $y=\frac{1}{2}\left(u-v\right)$ and $\phantom{\rule{0.3em}{0ex}}dxdy=\left|J\right|\phantom{\rule{0.3em}{0ex}}dudv=\frac{1}{2}\phantom{\rule{0.3em}{0ex}}dudv$ , the integral $\int \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\int \left({x}^{2}+{y}^{2}\right)\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}dxdy$ becomes

$\begin{array}{rcll}I& =& \int \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\int \left(\frac{1}{4}{\left(u+v\right)}^{2}+\frac{1}{4}{\left(u-v\right)}^{2}\right)×\frac{1}{2}\phantom{\rule{0.3em}{0ex}}dudv\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\\ & =& \int \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\int \frac{1}{2}\left({u}^{2}+{v}^{2}\right)×\frac{1}{2}\phantom{\rule{0.3em}{0ex}}dudv& \text{}\\ & =& \int \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\int \frac{1}{4}\left({u}^{2}+{v}^{2}\right)\phantom{\rule{0.3em}{0ex}}dudv& \text{}\end{array}$
4. Find the limits on $u$ and $v$ for the rectangle with vertices  $\left(x,y\right)=\left(0,0\right),\phantom{\rule{1em}{0ex}}\left(2,2\right),\phantom{\rule{1em}{0ex}}\left(-1,5\right),\phantom{\rule{1em}{0ex}}\left(-3,3\right)$ :

For ( $0,0$ ), $u=0$ and $v=0$

For ( $2,2$ ), $u=4$ and $v=0$

For ( $-1,5$ ), $u=4$ and $v=-6$

For ( $-3,3$ ), $u=0$ and $v=-6$

Thus, the limits on $u$ are $u=0$ to $u=4$ while the limits on $v$ are $v=-6$ to $v=0$ .

5. Finally evaluate $I$ :

The integral is

$\begin{array}{rcll}I& =& {\int }_{v=-6}^{0}{\int }_{u=0}^{4}\frac{1}{4}\left({u}^{2}+{v}^{2}\right)\phantom{\rule{0.3em}{0ex}}dudv& \text{}\\ & =& \frac{1}{4}{\int }_{v=-6}^{0}{\left[\frac{1}{3}{u}^{3}+u{v}^{2}\right]}_{u=0}^{4}\phantom{\rule{0.3em}{0ex}}dudv={\int }_{v=-6}^{0}\left[\frac{16}{3}+{v}^{2}\right]\phantom{\rule{0.3em}{0ex}}dv& \text{}\\ & =& {\left[\frac{16}{3}v+\frac{1}{3}{v}^{3}\right]}_{-6}^{0}=0-\left[\frac{16}{3}×\left(-6\right)+\frac{1}{3}×\left(-216\right)\right]=104\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$