Evaluate the integral
$$\int\int_{R}(x-y)^2 cos^2(x+y)dxdy$$ where $R$ is the rhombus with successive vertices as $(\pi,0), (2\pi,\pi), (\pi,2\pi), (0,\pi).$
My attempt- I tried doing this surface integral by performing integration by parts but the answer solution seems to be very lengthy, also am unable to get the final answer by this method. Just wanted to know if there is any other simpler method or can it be solved by using some theorem in vector calculus perhaps Green's theorem or something.
Thanks
1) Translate rhombus to origin: $$ x' = x - \pi \quad y' = y - \pi $$ $$ \int\int_{R}(x-y)^2 cos^2(x+y)dxdy \rightarrow \int\int_{R'}(x'-y')^2 cos^2(x'+y' + 2\pi)dx'dy' $$ 2) Change coordinate system: $$ x' - y' = u \quad x' + y' = v $$ Now: $$ -\pi \leq u \leq \pi \quad -\pi \leq v \leq \pi $$ $$ \int\int_{R'}(x'-y')^2 cos^2(x'+y')dx'dy' \rightarrow \int\int_{R''}u^2 cos^2(v)J(u,v)dudv $$ 3) Compute Jacobian: $J = 1/2$
4) Evaluate Integral: $$ \frac{1}{2}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}u^2 cos^2(v)dudv = \frac{1}{2}\int_{-\pi}^{\pi} u^2 du \int_{-\pi}^{\pi}cos^2(v)dv = \frac{1}{2}\frac{2\pi^3}{3} \cdot \pi = \frac{\pi^4}{3} $$