3 The Jacobian in 3 dimensions

When changing the coordinate system of a triple integral

$I=\int \; <!--nolimits\; -->\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_{V}f\left(x,y,z\right)dV$

we need to extend the above definition of the Jacobian to 3 dimensions.

The volume element $dV=dxdydz$ becomes $dV=\left|J\left(u,v,w\right)\right|dudvdw$ . As before the limits and integrand must also be transformed.

Example 26

Use spherical coordinates to find the volume of a sphere of radius $R$ .

Figure 32

Solution

The change of coordinates from Cartesian to spherical polar coordinates is given by the transformation equations

$x=rcos\theta sin\varphi y=rsin\theta sin\varphi z=rcos\varphi$

We now need the nine partial derivatives

$\begin{array}{ccc}\frac{\partial x}{\partial r}=cos\theta sin\varphi \hfill & \frac{\partial x}{\partial \theta }=-rsin\theta sin\varphi \hfill & \frac{\partial x}{\partial \varphi }=rcos\theta cos\varphi \hfill \\ \frac{\partial y}{\partial r}=sin\theta sin\varphi \hfill & \frac{\partial y}{\partial \theta }=rcos\theta sin\varphi \hfill & \frac{\partial y}{\partial \varphi }=rsin\theta cos\varphi \hfill \\ \frac{\partial z}{\partial r}=cos\varphi \hfill & \frac{\partial z}{\partial \theta }=0\hfill & \frac{\partial z}{\partial \varphi }=rsin\varphi \hfill \end{array}$

Hence we have

$J\left(r,\theta ,\varphi \right)=\left|\begin{array}{ccc}\hfill cos\theta sin\varphi \hfill & \hfill -rsin\theta sin\varphi \hfill & \hfill rcos\theta cos\varphi \hfill \\ \hfill sin\theta sin\varphi \hfill & \hfill rcos\theta sin\varphi \hfill & \hfill rsin\theta cos\varphi \hfill \\ \hfill cos\varphi \hfill & \hfill 0\hfill & \hfill -rsin\varphi \hfill \end{array}\right|$

$J\left(r,\theta ,\varphi \right)=cos\varphi \left|\begin{array}{cc}\hfill -rsin\theta sin\varphi \hfill & \hfill rcos\theta cos\varphi \hfill \\ \hfill rcos\theta sin\varphi \hfill & \hfill rsin\theta cos\varphi \hfill \end{array}\right|+0-rsin\varphi \left|\begin{array}{cc}\hfill cos\theta sin\varphi \hfill & \hfill -rsin\theta sin\varphi \hfill \\ \hfill sin\theta sin\varphi \hfill & \hfill rcos\theta sin\varphi \hfill \end{array}\right|$

Check that this gives $J\left(r,\theta ,\varphi \right)=-r^{2}sin\varphi$ . Notice that $J\left(r,\theta ,\varphi \right)\le 0$ for $0\le \varphi \le \pi$ , so $\left|J\left(r,\theta ,\varphi \right)\right|=r^{2}sin\varphi$ . The limits are found as follows. The variable $\varphi$ is related to ‘latitude’ with $\varphi =0$ representing the ‘North Pole’ with $\varphi =\pi ∕2$ representing the equator and $\varphi =\pi$ representing the ‘South Pole’.

The variable $\theta$ is related to ‘longitude’ with values of $0$ to $2\pi$ covering every point for each value of $\varphi$ . Thus limits on $\varphi$ are $0$ to $\pi$ and limits on $\theta$ are $0$ to $2\pi$ . The limits on $r$ are $r=0$ (centre) to $r=R$ (surface).

To find the volume of the sphere we then integrate the volume element $dV=r^{2}sin\varphi drd\theta d\varphi$ between these limits.

Example 27

Find the volume integral of the function $f\left(x,y,z\right)=x-y$ over the parallelepiped with the vertices of the base at

$\left(x,y,z\right)=\left(0,0,0\right),\left(2,0,0\right),\left(3,1,0\right)$ and $\left(1,1,0\right)$

and the vertices of the upper face at

$\left(x,y,z\right)=\left(0,1,2\right)$ , $\left(2,1,2\right)$ , $\left(3,2,2\right)$ and $\left(1,2,2\right)$ .

Figure 33

Solution

This will be a difficult integral to derive limits for in terms of $x$ , $y$ and $z$ . However, it can be noted that the base is described by $z=0$ while the upper face is described by $z=2$ . Similarly, the front face is described by $2y-z=0$ with the back face being described by $2y-z=2$ . Finally the left face satisfies $2x-2y+z=0$ while the right face satisfies $2x-2y+z=4$ .

The above suggests a change of variable with the new variables satisfying $u=2x-2y+z$ , $v=2y-z$ and $w=z$ and the limits on $u$ being $0$ to $4$ , the limits on $v$ being $0$ to $2$ and the limits on $w$ being $0$ to $2$ .

Inverting the relationship between $u$ , $v$ , $w$ and $x$ , $y$ and $z$ , gives

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

The Jacobian is given by

$J\left(u,v,w\right)=\left|\begin{array}{ccc}\hfill \frac{\partial x}{\partial u}\hfill & \hfill \frac{\partial x}{\partial v}\hfill & \hfill \frac{\partial x}{\partial w}\hfill \\ \hfill \hfill \\ \hfill \frac{\partial y}{\partial u}\hfill & \hfill \frac{\partial y}{\partial v}\hfill & \hfill \frac{\partial y}{\partial w}\hfill \\ \hfill \hfill \\ \hfill \frac{\partial z}{\partial u}\hfill & \hfill \frac{\partial z}{\partial v}\hfill & \hfill \frac{\partial z}{\partial w}\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill \frac{1}{2}\hfill & \hfill \frac{1}{2}\hfill & \hfill 0\hfill \\ \hfill \hfill \\ \hfill 0\hfill & \hfill \frac{1}{2}\hfill & \hfill \frac{1}{2}\hfill \\ \hfill \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right|=\frac{1}{4}$

Note that the function $f\left(x,y,z\right)=x-y$ equals $\frac{1}{2}\left(u+v\right)-\frac{1}{2}\left(v+w\right)=\frac{1}{2}\left(u-w\right)$ . Thus the integral is

Task!

Find the Jacobian for the following transformation:

$x=2u+3v-w$ , $y=v-5w$ , $z=u+4w$

Answer