How to evaluate $\iint_S(x-y)^2\sin^2(x+y)dxdy$ , where $S$ is the parallelogram with vertices $(\pi,0);(2\pi,\pi);(\pi,2\pi);(0,\pi)$ ?

Question

How to evaluate $\iint_S(x-y)^2\sin^2(x+y)dxdy$ , where $S$ is the parallelogram with vertices $(\pi,0);(2\pi,\pi);(\pi,2\pi);(0,\pi)$ ? Do I have to use a linear transformation of co-ordinates ? Please help . Thanks in advance

Answer 1

Sure. You'd really like to do the integral first with respect to one variable, then another. But your domain parallelogram is "tilted." So it's a good idea to do a substitution like

$$ u = x + y \\ v = x - y $$ because these $uv$ coordinates align with the edges of your domain.

Once you do that, what does the domain (in $uv$ coordinates) become? What does the integrand become? And finally, what does $dx ~dy$ become, in terms of $du~ dv$?

[If you try to work this much out, and show the work as edits in your question, I'll follow up with more detail in my answer.]

Addition: I'll get you started on the limits of integration. First, let's write the vertices of the domain in $uv$ coordinates.

The first vertex is at $(x,y) = (\pi, 0)$. Since $u = x + y$, that makes $u = \pi + 0 = \pi$. Similarly, $v = x - y = \pi - 0$. So the $uv$ coordinates of the first point are $(\pi, \pi)$. What are the coordinates of the other three points?

Also: consider the edge between the first two points. That's a line segment in $xy$-coordinates (i.e., it's a connected subset of a line, i.e., something determined by a single linear equation, in this case $x-y = \pi$). Will it still be a line segment in $uv$-coordinates? Why or why not?