# Surface Area of $z = \sin(x)\sin(y)$

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!

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It seems highly unlikely that there's any closed form; substituting $u=2(x+y)$, $v=2(x-y)$ and then doing a double-Weierstrass substitution $s=\tan(u/2)$, $t=\tan(v/2)$ gives the integrand first as $\sqrt{6-\cos(u)-\cos(v)}$ (with some irrelevant constant factor) and then as ${1\over(1+s^2)(1+t^2)}\sqrt{6-{1-s^2\over1+s^2}-{1-t^2\over1+t^2}}$; normalizing the denominators under the square root by multiplying and dividing by $\sqrt{(1+s^2)(1+t^2)}$ will lead to an integrand that's some rational function of $s$ and $t$ times the square root of a term quartic in $s$ and $t$ (edit: actually, even worse than that; it's quartic in $s$ and $t$ individually, and of degree 8 as a polynomial in both since there are cross terms), so the general form is likely to be an elliptic integral.