First we need to find the region of integration. This region is bounded below by [latex]y=0[/latex] and above by [latex]y=\sqrt{2x-x^2}[/latex] (see the following figure).

Squaring and collecting terms, we find that the region is the upper half of the circle [latex]x^{2}+y^{2}-2x=0[/latex] that is, [latex]y^{2}+(x-1)^{2}[/latex]. In polar coordinates, the circle is [latex]r=2\cos\theta[/latex] so the region of integration in polar coordinates is bounded by [latex]0\leq{r}\leq\cos\theta[/latex] and [latex]0\leq\theta\leq\frac{\pi}2[/latex].

The Jacobian is [latex]J(r,\theta)=r[/latex], as shown in Example “Finding the Jacobian”. Since [latex]r\geq0[/latex], we have [latex]|J(r,\theta)|=r[/latex].

The integrand [latex]\sqrt{x^2+y^2}[/latex] changes to [latex]r[/latex] in polar coordinates, so the double iterated integral is

[latex]\displaystyle\int_0^2\displaystyle\int_0^{\sqrt{2x-x^2}}\sqrt{x^2+y^2}dy \ dx=\displaystyle\int_0^{\pi/2}\displaystyle\int_0^{2\cos\theta}r|J(r,\theta)|dr \ d\theta=\displaystyle\int_0^{\pi/2}\displaystyle\int_0^{2\cos\theta}r^2 \ dr \ d\theta[/latex].

First, we need to understand the region over which we are to integrate. The sides of the parallelogram are [latex]x-y+1=0[/latex], [latex]x-y-1=0[/latex], [latex]x-3y+5=0[/latex], and [latex]x-3y+9=0[/latex] (Figure 4). Another way to look at them is [latex]x-y=-1[/latex], [latex]x-y=1[/latex], [latex]x-3y=-5[/latex], and [latex]x-3y=-9[/latex].

Clearly the parallelogram is bounded by the lines [latex]y=x+1[/latex], [latex]y=x-1[/latex], [latex]y=\frac13(x+5)[/latex], and [latex]y=\frac13(x+9)[/latex].

Notice that if we were to make [latex]u=x-y[/latex] and [latex]v=x-3y[/latex], then the limits on the integral would be [latex]-1\leq{u}\leq1[/latex] and [latex]-9\leq{v}\leq-5[/latex].

To solve for [latex]x[/latex] and [latex]y[/latex], we multiply the first equation by [latex]3[/latex] and subtract the second equation, [latex]3u-v=(3x-3y)-(x-3y)=2x[/latex]. Then we have [latex]x=\frac{3u-v}2[/latex]. Moreover, if we simply subtract the second equation from the first, we get [latex]u-v=(x-y)-(x-3y)=2y[/latex] and [latex]y=\frac{u-v}2[/latex].

Thus, we can choose the transformation

[latex]\large{T(u,v)=\left(\frac{3u-v}2,\frac{u-v}2\right)}[/latex]

and compute the Jacobian [latex]J(u, v)[/latex]. We have

[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}3/2&-1/2 \\ 1/2&-1/2\end{vmatrix}=-\frac34+\frac14=-\frac12}[/latex].

Therefore, [latex]|J(u,v)|=\frac12[/latex]. Also, the original integrand becomes

[latex]\large{x-y=\frac12[3u-v-u+v]=\frac12[3u-u]=\frac12[2u]=u}[/latex].

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

[latex]\large{\underset{R}{\displaystyle\iint}(x-y)dy \ dx=\displaystyle\int_{-9}^{-5}\displaystyle\int_{-1}^{1}J(u,v)u \ du \ dv=\displaystyle\int_{-9}^{-5}\displaystyle\int_{-1}^{1}\left(\frac12\right)u \ du \ dv}[/latex],

which is much simpler to compute. In fact, it is easily found to be zero. And this is just one example of why we transform integrals like this.

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).

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].