Question

double integration of (x-y)^2sin^2(x+y)dxdy of a parelleogram with given vertix (0 π) (π 0) (2π π) (π 2π)​

Answer 1

Your calculation is correct, because if we look at the integrand f(x,y)=(x−y)2sin(x+y)f(x,y)=(x−y)2sin⁡(x+y), it obeys the property

f(2π−y,2π−x)=−f(x,y),f(2π−y,2π−x)=−f(x,y),

i.e., ff is antisymmetric upon reflection about the line x+y=2πx+y=2π. 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)2sin2(x+y),g(x,y)=(x−y)2sin2⁡(x+y),

then you would get the answer π4/3π4/3. So perhaps there is a typographical error.