Evaluate the following integral $$\iint_{R_{xy}} (x-y)^2\sin(x+y)\,dx\,dy$$ where $R_{xy}$ is the parallelogram with the successive vertices $(\pi,0),(2\pi,\pi),(\pi,2\pi),(0,\pi)$, using $u = x-y, v=x+y$.
The Jacobian is $\frac 12$, therefore: $$\iint_{R_{xy}} (x-y)^2\sin(x+y)\,dx\,dy = \frac 12 \int_{-\pi}^{\pi} \int_\pi^{3\pi} u^2\sin v \,dv\,du = 0.$$
The correct answer is $\frac{\pi^4}3$. Any thoughts on the problem? Am I missing something?
Your calculation is correct, because if we look at the integrand $f(x,y) = (x-y)^2 \sin (x+y)$, it obeys the property $$f(2\pi - y, 2\pi - x) = -f(x,y),$$ i.e., $f$ is antisymmetric upon reflection about the line $x + y = 2\pi$. Since the domain of integration is also symmetric about this line, it follows that the integral is zero.
However, if we modify the integrand to be $$g(x,y) = (x-y)^2 \sin^2 (x+y),$$ then you would get the answer $\pi^4/3$. So perhaps there is a typographical error.