I am new to calculus and was hoping for some feedback re the following question and my proposed answer. Many thanks in advance.
Parallelogram $D$ in the first quadrant has corners at $(0,0)$, $(2,2)$, $(3,4)$ and $(1,2)$. Calculate $$\iint_D xy \,\mathrm dx\,\mathrm dy.$$
My answer:
The parallelogram is bordered by:
$$y = 2x,\ y = 2x-2,\ y = x,\ y = x+1.$$
Therefore,
$$2x-2\leq y \leq 2x,$$ $$-2\leq y -2x \leq 0$$
and
$$x\leq y \leq x+1,$$ $$0\leq y - x \leq 1.$$
Let's define $u=y-2x,\ v=y-x$, then
$$\frac{\partial(u,v)}{\partial(x,y)}=\begin{vmatrix}-2 & 1 \\ -1 &1\end{vmatrix}=-1,$$ so the Jacobean is also $-1$, with absolute value $1$.
$$xy=(-u+2v)(v-u)=u^2-3uv+2v^2$$
$$\int^1_0dv\int^0_{-2}u^2-3uv+2v^2du=\int^1_0dv[\frac{1}{3}u^3-\frac{3}{2}u^2v+2v^2u]^0_{-2}$$
Therefore $$\int^1_0\frac{8}{3}+6v+4v^2dv=[\frac{8}{3}v-3v^2-\frac{4}{3}v^3]^1_0=7$$
Your result is correct, although you should fix some minor problems already pointed out by Christoph. In this specific case you could also do very well without the change of variables, just by splitting the integral: $$ \iint_D xy dx dy = \int_0^2\int_{y/2}^y xy \,dx dy + \int_2^4 \int_{y-1}^{y/2+1} xy \,dx dy. $$