This seems like an easy problem, but I can't seem to get a closed form solution: What is the surface area of the surface defined by the equation $z = \sin(x)\sin(y)$, over some rectangular region bounded by the following equations:
$y = -x+a$
$y = x+a$
$y = x-a$
$y = 2\pi-x-a$.
Other useful information: $x\in[0,\pi]$, $y\in[0,\pi]$, $a\in[0,\pi]$. You can see the function on wolfram alpha here.
I've tried $A = \int\int|\vec{r_x}\times\vec{r_y}|dxdy$, but taking $\vec{r} = \langle x,y,\sin(x)\sin(y) \rangle$, I can't get a closed form anti-derivative of the integrand:
$\begin{align}|\vec{r_x}\times\vec{r_y}| &= \sqrt{-2s\sin^2(x)\sin^2(y)+\sin^2(x)+\sin^2(y)+1} \\ &= \frac{1}{2}\sqrt{6-\cos(2(x+y))-\cos(2(x-y))}\end{align}$
Thank you in advance for your time, it is much appreciated!