Find the area of the region bounded by $\sin(x)\sin(y)=k$ where $0 \leq x \leq \pi$, $0 \leq y \leq \pi$, and $0 \leq k \leq 1$

Question

On the coordinate plane, the equation $\sin(x)\sin(y)=k$ where $0 \leq x \leq \pi$, $0 \leq y \leq \pi$, and $0 \leq k \leq 1$ forms a closed region. Find the area of this region in terms of $k$.

I've had some progress, but I've hit a dead end with an integral that appears to not be solvable.

Solving in terms of $x$, you get that $y=\arcsin(\dfrac{k}{\sin(x)})$. When you integrate from the bounds given by the minimum and maximum $x$ of this region, you get $\int_{\arcsin(k)}^{\pi-\arcsin(k)} \arcsin(\dfrac{k}{\sin(x)}) dx$. Thus, considering the fact that this integral gives the region below the closed region, the expression for the closed region is $$\pi(\pi-2\arcsin(k))-2\int_{\arcsin(k)}^{\pi-\arcsin(k)} \arcsin\left(\dfrac{k}{\sin(x)}\right).$$ However, I need help with the integral as WolframAlpha cannot evaluate it.

Help would be appreciated (or perhaps even a better solution than this!)

Thanks

Answer 1

Let's shift the problem over so that it's centered around the origin. We can examine the relation $$\sin\left(x-\frac{\pi}{2}\right)\sin\left(y-\frac{\pi}{2}\right)=C$$ In the region $(x,y)\in [-\pi/2,\pi/2]\times[-\pi/2,\pi/2].$ Now the characteristic curve is rotationally symmetric about the origin. We can use a property of trig functions that $\sin(t-\pi/2)=-\cos(t)$ and see that this is equivalent to $$\cos\left(x\right)\cos\left(y\right)=C$$ Thus $$y=\pm\arccos\left(\frac{C}{\cos(x)}\right)$$ Therefore $$A=2\int_{-\arccos(C)}^{\arccos(C)}\arccos\left(\frac{C}{\cos(x)}\right)\mathrm{d}x$$ This is quite a difficult integral and I doubt it has any reasonable closed form expressions. Numerical methods are the only option here. Below is a graph:

Answer 2

Starting from @K.defaoite's answer, $$A=2\int_{-\arccos(C)}^{\arccos(C)}\arccos\left(\frac{C}{\cos(x)}\right)\,dx$$ it seems that we could have a "reasonable" approximation expanding the integrand as a Taylor series around $x=0$ $$\arccos\left(\frac{C}{\cos(x)}\right)=\arccos(C)+\sum_{n=1}^\infty a_n\,x^{2n}$$ where $$a_n= \frac {C\, P_{n}(C)}{(2n)! \,\,\left(1-C^2\right)^{n-\frac{1}{2}}}$$ The first polynomials are listed below $$\left( \begin{array}{cc} n & P_n(C) \\ 1 & -1 \\ 2 & 2 C^2-5 \\ 3 & -16 C^4+32 C^2-61 \\ 4 & 272 C^6-984 C^4+522 C^2-1385 \\ 5 & -7936 C^8+35584 C^6-74256 C^4-2096 C^2-50521 \\ 6 & 353792 C^{10}-1946368 C^8+4184192 C^6-7738688 C^4-1973438 C^2-2702765 \end{array} \right)$$ Using these terms, the results are $$\left( \begin{array}{ccc} C & \text{approximation} & \text{exact} \\ 0.00 & 9.86960 & 9.86960 \\ 0.05 & 8.94015 & 8.79304 \\ 0.10 & 8.13310 & 7.99304 \\ 0.15 & 7.41526 & 7.29659 \\ 0.20 & 6.76475 & 6.66673 \\ 0.25 & 6.16666 & 6.08584 \\ 0.30 & 5.61055 & 5.54351 \\ 0.35 & 5.08889 & 5.03283 \\ 0.40 & 4.59611 & 4.54891 \\ 0.45 & 4.12799 & 4.08806 \\ 0.50 & 3.68129 & 3.64744 \\ 0.55 & 3.25342 & 3.22477 \\ 0.60 & 2.84231 & 2.81820 \\ 0.65 & 2.44627 & 2.42618 \\ 0.70 & 2.06389 & 2.04741 \\ 0.75 & 1.69397 & 1.68078 \\ 0.80 & 1.33550 & 1.32533 \\ 0.85 & 0.98759 & 0.98022 \\ 0.90 & 0.64947 & 0.64472 \\ 0.95 & 0.32048 & 0.31817 \\ 1.00 & 0.00000 & 0.00000 \end{array} \right)$$ which seems to be quite decent.

Edit

When $c$ is small,expanding the integrand as a Taylor series $$\arccos\left(\frac{C}{\cos(x)}\right)=\frac{\pi }{2}-C \sec (x)-\frac{1}{6} C^3 \sec ^3(x)-\frac{3}{40} C^5 \sec^5(x)+O\left(C^7\right)$$ we have for $A$ the approximation $$A=2\pi \cos ^{-1}(C)+$$ $$\frac{1}{240} \left(C \left(27 C^4+80 C^2+960\right) \log \left(\frac{\cos (t)-\sin (t)}{\sin (t)+\cos (t)}\right)-C \sqrt{1-C^2} \left(27 C^2+98\right)\right)$$ where $t=\frac{1}{2} \cos ^{-1}(C)$.

For $C=0.1$, this gives $8.00120$.