2 The Jacobian

Given an integral of the form $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_{A}f\left(x,y\right)dxdy$

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

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

Notation

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

$J\left(u,v\right)=J\left(x,y;u,v\right)=J\left(\frac{x,y}{u,v}\right)=\left|\frac{\partial \left(x,y\right)}{\partial \left(u,v\right)}\right|$

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

$dxdy\to \left|J\left(u,v\right)\right|dudv$

that is, $\left|J\left(u,v\right)\right|dudv$ plays the role of $dxdy$ .

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 ${\int \; <!--nolimits\; -->}_{A}dA$ . 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

$\frac{\partial x}{\partial r}=cos\theta \frac{\partial y}{\partial r}=sin\theta \frac{\partial x}{\partial \theta }=-rsin\theta \frac{\partial y}{\partial \theta }=rcos\theta$

Thus

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

${\int \; <!--nolimits\; -->}_{A}dA={\int \; <!--nolimits\; -->}_0^{2\pi }{\int \; <!--nolimits\; -->}_0^R\left|J\left(r,\theta \right)\right|drd\theta ={\int \; <!--nolimits\; -->}_0^{2\pi }{\int \; <!--nolimits\; -->}_0^{R}rdrd\theta =2\pi \frac{R^2}{2}=\pi R^2$

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

$I=\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_A{\left(3x+6y\right)}^{2}dA$

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

$du=x+2y\left(1\right)v=x-2y\left(2\right)$

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

$x=\frac{1}{2}\left(u+v\right)y=\frac{1}{4}\left(u-v\right)$

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

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

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

$\frac{\partial x}{\partial u}=\frac{1}{2}\frac{\partial x}{\partial v}=\frac{1}{2}\frac{\partial y}{\partial u}=\frac{1}{4}\frac{\partial y}{\partial v}=-\frac{1}{4}$

Then the Jacobian is

$J\left(u,v\right)=\left|\begin{array}{cc}\hfill \frac{1}{2}\hfill & \hfill \frac{1}{2}\hfill \\ \hfill \hfill & \hfill \hfill \\ \hfill \frac{1}{4}\hfill & \hfill -\frac{1}{4}\hfill \end{array}\right|=-\frac{1}{4}$

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

$I={\int \; <!--nolimits\; -->}_{-2}^2{\int \; <!--nolimits\; -->}_{-2}^2\frac{9}{4}{u}^{2}dudv.$

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

Task!

This Task concerns using a transformation to evaluate $\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->\left(x^2+y^2\right)dxdy$ .

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:

   Answer

   $u=x+y$ (1)

   $v=x-y$ (2)

   Add equations (1) and (2) $u+v=2x$

   Subtract equation (2) from equation (1) $u-v=2y$

   So $x=\frac{1}{2}\left(u+v\right)y=\frac{1}{2}\left(u-v\right)$

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

   Answer

   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 $I=\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->\left(x^2+y^2\right)dxdy$ in terms of $u$ and $v$ , using the transformations introduced in 1. and the Jacobian found in 2.:

   Answer

   On letting $x=\frac{1}{2}\left(u+v\right)$ , $y=\frac{1}{2}\left(u-v\right)$ and $dxdy=\left|J\right|dudv=\frac{1}{2}dudv$ , the integral $\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->\left(x^2+y^2\right)dxdy$ becomes

   $$\begin{array}{rcll}I& =& \int \; <!--nolimits\; -->\int \; <!--nolimits\; -->\left(\frac{1}{4}{\left(u+v\right)}^2+\frac{1}{4}{\left(u-v\right)}^2\right)\times \frac{1}{2}dudv& \\ & =& \int \; <!--nolimits\; -->\int \; <!--nolimits\; -->\frac{1}{2}\left(u^2+v^2\right)\times \frac{1}{2}dudv& \\ & =& \int \; <!--nolimits\; -->\int \; <!--nolimits\; -->\frac{1}{4}\left(u^2+v^2\right)dudv& \end{array}$$
4. Find the limits on $u$ and $v$ for the rectangle with vertices  $\left(x,y\right)=\left(0,0\right),\left(2,2\right),\left(-1,5\right),\left(-3,3\right)$ :

   Answer

   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 $I$ :

   Answer

   The integral is

   $$\begin{array}{rcll}I& =& {\int \; <!--nolimits\; -->}_{v=-6}^0{\int \; <!--nolimits\; -->}_{u=0}^4\frac{1}{4}\left(u^2+v^2\right)dudv& \\ & =& \frac{1}{4}{\int \; <!--nolimits\; -->}_{v=-6}^0{\left[\frac{1}{3}{u}^3+u{v}^2\right]}_{u=0}^{4}dudv={\int \; <!--nolimits\; -->}_{v=-6}^0\left[\frac{16}{3}+v^2\right]dv& \\ & =& {\left[\frac{16}{3}v+\frac{1}{3}{v}^3\right]}_{-6}^0=0-\left[\frac{16}{3}\times \left(-6\right)+\frac{1}{3}\times \left(-216\right)\right]=104& \end{array}$$