# area enclosed by level curve

Fix $$a\in [0,1]$$ and consider the closed curve

$$C_a:= \{ (x,y) : \sin(x) \sin(y) = a,\ (x,y)\in [0,\pi]\times[0,\pi]\}.$$

See contour plot of function $$\sin(x) \sin(y)$$. I am interested in computing the area of the enclosed by $$C_a$$ as a function of $$a$$. Is there a simple analytical form?

First, change your variables $$x \to \pi/2 - x$$ and $$y \to \pi/2 - y$$ so your integral becomes:

$$4 \int\limits_{x = 0}^{\cos^{-1} a} \cos^{-1}\left( \frac{a}{\cos x} \right)\ dx$$

which should be clear from the plot:

Unfortunately, there seems to be no closed form for arbitrary $$a$$. (Of course you can perform the integration numerically for an assigned value of $$a$$.)

You seem to be Mathematica literate, so use this instead of your code and insert your value for $$a$$ in the Contour level:

ContourPlot[Cos[x] Cos[y],
{x, -π/2, π/2},
{y, -π/2, π/2},
AxesOrigin -> {0, 0},
Axes -> True,
Contours -> {0.1}]