The following problem appeared in a Multivariable Calculus final and I can't solve it:
Let $g$ and $h$ be two continuous single variable functions. Let $f(x,y) = g(x)h(y)$ and D the paralellogram of vertices $(1,0), (0,1), (-1/2,1/2), (1/2,-1/2)$. Calculate:
$$\iint_{D}f(x+y,x-y)\,dx\,dy$$
I have tried the obvious change of variables $u=x+y$ and $v=x-y$, which maps the parallelogram bounded by the lines $y=-x, y = -x+1, y = x+1, y = x-1$ to the rectangle $[0,1]\times[-1,1]$. The Jacobian of the transformation is $-1/2$. This is what I have so far:
\begin{align*}\iint_{D}f(x+y,x-y)\,dx\,dy &= \int_{-1}^1\int_0^1f(u,v)\,|-\frac{1}{2}|\,du\,dv\\ &= \frac{1}{2}\int_{-1}^1\int_0^1g(u)h(v)\,du\,dv \\ &= \frac{1}{2}\int_{-1}^1 h(v) \, dv \int_0^1g(u)\,du. \end{align*}
And then I don't know what else to do. I think I'm supposed to find a value, but I don't know anything else about the functions. I thought that maybe one of the parts should be zero; for example, if $h$ was an even function integrated between $(-1,1)$, but the problem doesn't specify this. Anyway, I've run out of ideas. I would really appreciate any help.
You've done it perfectly (except that the Jacobian is actually $-1/2$, but the absolute value of the Jacobian is $1/2$). You cannot go any further without some knowledge of $g$ and $h$. (Actually, you wish that $h$ were odd, and then $\int_{-1}^1 h(v)dv$ would vanish.)