Compute $$\iint_D(x^2-y^2)^{10}dxdy,$$ where $D$ is bounded by $|x|+|y|\leq 1.$
I'm supposed to solve this by introducing new variables by $u=x-y$ and $v=x+y.$ I get that $$\iint_D(x^2-y^2)^{10}dxdy=\iint_E(uv)^{10}\cdot \frac{d(x,y)}{d(u,v)} dudv.$$
I have that $$\frac{d(x,y)}{d(u,v)}=\begin{vmatrix} x'_{u} & x'_{v} \\ y'_{u} & y'_{v} \end{vmatrix} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}=1=\text{no scaling.}$$
So it boils down to $$\iint_E(uv)^{10}dudv.$$
I'm having trouble understanding the bounds. What are the bounds in $D$ and in $E$? I did the following:
$$|x|+|y|\leq 1 \Leftrightarrow -1+|x| \leq y \leq 1-|x|,$$ So my area $D$ is a triangle bounded by $x=0, \ y = -1+x$ and $y=1-x.$ How do I translate these bounds w.r.t $(u,v)?$
If to move forward, you want, then reflect upon your past work. Needs attention first, your Jacobian does. Find $x$ and $y$ in terms of $u$ and $v$, you must. Solve linear system $$u=x-y,\qquad v=x+y$$ to find $$x=\frac{u+v}{2},\qquad y=\frac{v-u}{2}$$ Find a different Jacobian, you will $$\frac{d(x,y)}{d(u,v)}=\begin{vmatrix}\frac{1}{2}&\frac{1}{2}\\-\frac{1}{2}&\frac{1}{2}\end{vmatrix}=\frac{1}{2}$$ Not triangular, but diamond shape, your original domain $D$ is. To image shape of new domain, you must imagine reshaping your original diamond through a linear transformation, hmm? $$|x|+|y|\leq1\implies|u+v|+|u-v|\leq2$$ $\qquad\qquad\qquad$
$\quad\implies$ 
A rotation, the transformation is. Finish the problem now, can you?