As before, first find the region [latex]R[/latex] and picture the transformation so it becomes easier to obtain the limits of integration after the transformations are made (Figure 5).

Figure 5. Transforming the region [latex]R[/latex] into the region [latex]S[/latex] to simplify the computation of an integral.

Given [latex]u=x-y[/latex] and [latex]v=x+y[/latex], we have [latex]x=\frac{u+v}2[/latex] and [latex]y=\frac{v-u}2[/latex] and hence the transformation to use is [latex]T(u,v)=\left(\frac{u+v}2,\frac{v-u}2\right)[/latex]. The lines [latex]x+y=1[/latex] and [latex]x+y=3[/latex] become [latex]v=1[/latex] and [latex]v=3[/latex], respectively. The curves [latex]x^{2}-y^{2}=1[/latex] and [latex]x^{2}-y^{2}=-1[/latex] become [latex]uv=1[/latex] and [latex]uv=-1[/latex], respectively.

Thus we can describe the region [latex]S[/latex] (see the second region Figure 5) as

[latex]\large{s=\left\{(u,v)|1\leq{v}\leq3,\frac{-1}v\leq{u}\leq\frac1v\right\}}[/latex].

The Jacobian for this transformation is

[latex]\large{J(u,v)=\frac{\partial(x,y)}{\partial(u,v)}=\begin{vmatrix}\frac{\partial{x}}{\partial{u}}&\frac{\partial{x}}{\partial{v}} \\ \frac{\partial{y}}{\partial{u}}&\frac{\partial{y}}{\partial{v}}\end{vmatrix}=\begin{vmatrix}1/2&-1/2 \\ 1/2&1/2\end{vmatrix}=\frac12}[/latex].

Therefore, by using the transformation [latex]T[/latex], the integral changes to

[latex]\large{\underset{R}{\displaystyle\iint}(x-y)e^{x^2-y^2}dA=\frac12\displaystyle\int_1^3\displaystyle\int_{-1/v}^{1/v}ue^{uv}du \ dv}[/latex].

Doing the evaluation, we have

[latex]\large{\frac12\displaystyle\int_1^3\displaystyle\int_{-1/v}^{1/v}ue^{uv}du \ dv=\frac4{3e}\approx0.490}[/latex].

1. For cylindrical coordinates, the transformation is [latex]T(r,\theta,z)=(x,y,z)[/latex] from the Cartesian [latex]r\theta{z}[/latex]–plane  to the Cartesian [latex]xyz[/latex]-plane (Figure 7). Here [latex]x=r\cos\theta[/latex], [latex]y=r\sin\theta[/latex], and [latex]z=z[/latex]. The Jacobian for the transformation is
   
   [latex]\hspace{1.5cm}\large{\begin{align}J(r,\theta,z)&=\frac{\partial(x,y,z)}{\partial(r,\theta,z)}=\begin{vmatrix}\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{vmatrix} \\&=\begin{vmatrix}\cos\theta&-r\sin\theta&0 \\ \sin\theta&r\cos\theta&0 \\ 0&0&1\end{vmatrix}=r\cos^2\theta+r\sin^2\theta=r(\cos^2\theta+\sin^s\theta)=r \end{align}}[/latex].
   
   We know that [latex]r\geq0[/latex], so [latex]|J(r,\theta,z)|=r[/latex]. Then the triple integral is
   
   [latex]\large{\underset{D}{\displaystyle\iiint}f(x,y,z)dV=\underset{G}{\displaystyle\iiint}f(r\cos\theta,r\sin\theta,z)r \ dr \ d\theta \ dz}[/latex].
   

   Figure 7. The transformation from rectangular coordinates to cylindrical coordinates can be treated as a change of variables from region [latex]G[/latex] in [latex]r\theta{z}[/latex]-space to region [latex]D[/latex] in [latex]xyz[/latex]-space.

2. For spherical coordinates, the transformation is [latex]T(\rho,\theta,\varphi)=(x,y,z)[/latex] from the Cartesian [latex]p\theta\varphi[/latex]–plane  to the Cartesian [latex]xyz[/latex]–plane  (Figure 8). Here [latex]x=\rho\sin\varphi\cos\theta[/latex], [latex]y=\rho\sin\varphi\sin\theta[/latex], and [latex]z=\rho\cos\varphi[/latex]. The Jacobian for the transformation is
   
   [latex]\large{J(\rho,\theta,\varphi)=\frac{\partial{(x,y,z)}}{\partial{(\rho,\theta\varphi)}}=\begin{vmatrix}\frac{\partial{x}}{\partial{\rho}}&\frac{\partial{x}}{\partial{\theta}}&\frac{\partial{x}}{\partial{\varphi}} \\ \frac{\partial{y}}{\partial{\rho}}&\frac{\partial{y}}{\partial{\theta}}&\frac{\partial{y}}{\partial{\varphi}} \\ \frac{\partial{z}}{\partial{\rho}}&\frac{\partial{z}}{\partial{\theta}}&\frac{\partial{z}}{\partial{\varphi}}\end{vmatrix}=\begin{vmatrix}\sin\varphi\cos\theta&-\rho\sin\varphi\sin\theta&\rho\cos\varphi\cos\theta \\ \sin\varphi\sin\theta&-\rho\sin\varphi\cos\theta&\rho\cos\varphi\sin\theta \\ \cos\theta&0&-\rho\sin\varphi\end{vmatrix}}[/latex]

   
   Expanding the determinant with respect to the third row:
   
   [latex]\hspace{2cm}\begin{align} &=\cos\varphi\begin{vmatrix}-\rho\sin\varphi\sin\theta&\rho\cos\varphi\cos\theta \\ \rho\sin\varphi\sin\theta&\rho\cos\varphi\sin\theta\end{vmatrix}-\rho\sin\varphi\begin{vmatrix}\sin\varphi\cos\theta&-\rho\sin\varphi\sin\theta \\ \sin\varphi\sin\theta&\rho\sin\varphi\cos\theta\end{vmatrix} \\ &=\cos\varphi(-\rho^2\sin\varphi\cos\varphi\sin^2\theta-\rho^2\sin\varphi\cos\varphi\cos^2\theta)-\rho\sin\varphi(\rho\sin^2\varphi\cos^2\theta+\rho\sin^2\varphi\sin^2\theta) \\ &=-\rho^2\sin\varphi\cos^2\varphi(\sin^2\theta+\cos^2\theta)-\rho^2\sin\varphi\sin^2\varphi(\sin^2\theta+\cos^2\theta) \\ &=-\rho^2\sin\varphi\cos^2\varphi-\rho^2\sin\varphi\sin^2\varphi \\ &=-\rho^2\sin\varphi(\cos^2\varphi+\sin^2\varphi)=-\rho^2\sin\varphi\end{align}[/latex]
   
   Since [latex]0\leq\varphi\leq\pi[/latex], we must have [latex]\sin\varphi\geq0[/latex]. Thus [latex]|J(\rho,\theta,\varphi)|=|-\rho^2\sin\varphi|=\rho^2\sin\varphi[/latex].

   

   Figure 8. The transformation from rectangular coordinates to cylindrical coordinates can be treated as a change of variables from region [latex]G[/latex] in [latex]\rho\theta\varphi[/latex]-space to region [latex]D[/latex] in [latex]xyz[/latex]-space.

   
   Then the triple integral becomes
   
   [latex]\underset{D}{\displaystyle\iiint}f(x,y,z)dV=\underset{G}{\displaystyle\iiint}f(\rho\sin\varphi\cos\theta,\rho\sin\varphi\sin\theta,\rho\cos\varphi)\rho^2\sin\varphi \ d\rho \ d\varphi \ d\theta[/latex].

As before, some kind of sketch of the region [latex]G[/latex] in [latex]xyz[/latex]-space over which we have to perform the integration can help identify the region [latex]D[/latex] in [latex]uvw[/latex]-space (Figure 9). Clearly [latex]G[/latex] in [latex]xyz[/latex]-space is bounded by the planes [latex]x=y/2[/latex], [latex]x=(y/2)+1[/latex], [latex]y=0[/latex],[latex]y=4[/latex], [latex]z=0[/latex], and [latex]z=4[/latex]. We also know that we have to use [latex]u=(2x-y)/2[/latex], [latex]v=y/2[/latex] and [latex]w=z/3[/latex] for the transformations. We need to solve for [latex]x[/latex], [latex]y[/latex], and [latex]z[/latex]. Here we find that [latex]x=u+v[/latex], [latex]y=2v[/latex], and [latex]z=3w[/latex].

Using elementary algebra, we can find the corresponding surfaces for the region [latex]G[/latex] and the limits of integration in [latex]uvw[/latex]-space. It is convenient to list these equations in a table.

Figure 9. The region [latex]G[/latex] in [latex]uvw[/latex]-space is transformed to region [latex]D[/latex] in [latex]xyz[/latex]-space

Now we can calculate the Jacobian for the transformation:

[latex]\large{J(u,v,w)=\begin{vmatrix}\frac{\partial{x}}{\partial{u}}&\frac{\partial{x}}{\partial{v}}&\frac{\partial{x}}{\partial{w}} \\ \frac{\partial{y}}{\partial{u}}&\frac{\partial{y}}{\partial{v}}&\frac{\partial{y}}{\partial{w}} \\ \frac{\partial{z}}{\partial{u}}&\frac{\partial{z}}{\partial{v}}&\frac{\partial{z}}{\partial{w}}\end{vmatrix}=\begin{vmatrix}1&1&0 \\ 0&2&0 \\ 0&0&3\end{vmatrix}=6}[/latex].

The function to be integrated becomes

[latex]\large{f(x,y,z)=x+\frac{z}3=u+v+\frac{3w}3=u+v+w}[/latex].

We are now ready to put everything together and complete the problem.

[latex]\hspace{3cm}\large{\begin{align} &\hspace{1cm}\displaystyle\int_0^3\displaystyle\int_0^4\displaystyle\int_{y/2}^{(y/2)+1}\left(x+\frac{z}3\right)dx \ dy \ dz \\ &=\displaystyle\int_0^1\displaystyle\int_0^2\displaystyle\int_0^1(u+v+w)|J(u,v,w)|du \ dv \ dw=\displaystyle\int_0^1\displaystyle\int_0^2\displaystyle\int_0^1(u+v+w)|6|du \ dv \ dw \\ &=6\displaystyle\int_0^1\displaystyle\int_0^2\displaystyle\int_0^1(u+v+w)du \ dv \ dw=6\displaystyle\int_0^1\displaystyle\int_0^2\left[\frac{u^2}2+vu+wu\right]_0^1dv \ dw \\ &=6\displaystyle\int_0^1\displaystyle\int_0^2\left(\frac12+v+w\right)dv \ dw=6\displaystyle\int_0^1\left[\frac12v+\frac{v^2}2+wv\right]_0^2dw \\ &=6\displaystyle\int_0^1(3+2w)dw=6\left[3w+w^2\right]_0^1=24 \end{align}}[/latex].