How do I integrate this double integral?

Question

$\iint_D \,(x+y)^2 e^{x-y} \,\,dxdy $

Where D is the region bounded by : $ x+y=1, \\ x+y=4, \\ x-y=-1 \\ x-y=1 $

I've tried to use $ x=r\cos(\theta) $ and $y=r\sin(\theta)$ but I haven't really got anywhere, how should I go about this?

Answer 1

Let $u=x+y,v=x-y$, then $$\frac{D(x,y)}{D(u,v)}=-1/2.$$ $$\iint_D \,(x+y)^2 e^{x-y} \,\,dxdy=\iint_{D_1} \,u^2 e^{v} \,\,\left|\frac{D(x,y)}{D(u,v)}\right|dudv$$ $$=1/2\iint_{D_1} \,u^2 e^{v} \,\,dudv=1/2\int_{1}^4u^2 du\int_{-1}^1 e^v dv =21/2(e-e^{-1}).$$ Here $D_1=[1,4]\times[-1,1]$.