Evaluate $\int_{A}(x-y)^{2} \cos (x+y) d(x, y)$ where $A$ is the square region in the plane with vertices $(0,1),(1,2),(2,1)$ and $(1,0).$

Question

Evaluate (Use a linear change of variables)

$$\int_{A}(x-y)^{2} \cos (x+y) d(x, y)$$

where $A$ is the square region in the plane with vertices $(0,1),(1,2),(2,1)$ and $(1,0).$

I can compute followwing integral, that is I know what i'd do like this: $\iint_{R}\left(4 x^{2}-y^{2}\right)^{4} d x d y$, but I don't know what i'd do in uppermost question, may you add answer? Thanks...

Answer 1

Let $w=x-y$ and $z = x+y$ then solving for $x$ and $y$ we get $x= \dfrac{z+w}{2}$ and $y = \dfrac{z-w}{2}$. Then just compute the Jacobian and the integral becomes $$\int_{A'} w^2 \cos(z) \left\|\begin{bmatrix} \frac{\partial{x}}{\partial{z}} \ \frac{\partial{x}}{\partial{w}} \\ \frac{\partial{y}}{\partial{z}} \ \frac{\partial{y}}{\partial{w}}\end{bmatrix} \right\|\,\mathrm dw\,\mathrm dz.$$