I would like to calculate the area of the eye-shaped curve created by the following equation:
$$
\sin^4(x) + (\cos(y) - 3)^2 - 16 = 0
$$
If we plot this equation we get:
So the idea is to calculate the area of one of those "eyes" in the image.
The first step for me is to isolate the $y$ variable and so I did:
$\sin^4(x) + (\cos(y) - 3)^2 - 16 = 0$
$(\cos(y) - 3)^2 = 16 - \sin^4(x)$
$\cos(y) = 3 \pm \sqrt{16 - \sin^4(x)}$
$y = \arccos\left(3 \pm \sqrt{16 - \sin^4(x)}\right)$
If we plot $y = \arccos\left(3 - \sqrt{16 - \sin^4(x)}\right)$ we get half of the shape as depicted in this image:
So my idea was to move that curve close to the $x$ axis and then integrate from $-\pi/2$ to $\pi/2$. Using some approximation, the shifted equation is:
$$ y = \arccos\left(3 - \sqrt{16 - \sin^4(x)}\right) - 2.632 $$
And thus we have:
$$ Area_{eye} = 2 \left(\int_{-\pi/2}^{\pi/2} \arccos\left(3 - \sqrt{16 - \sin^4(x)}\right) - 2.632 \; dx\right) \approx 1.61256 $$
(multiplying by 2 in order to get the area of the entire shape)
I am sure there are more efficient ways to solve this, so how would you have solved this problem?
There is no reason to translate the curve, because the given relation is clearly doubly periodic with periods $(\pi, 2\pi)$. Moreover, if $(x,y)$ is a solution, so is $(x,-y)$, so the area of a single "eye" can be expressed as $$2 \left(\pi^2 - \int_{x=0}^\pi \cos^{-1} \left( 3 - \sqrt{16 - \sin^4 x} \right) \, dx \right) \approx 1.589300092423124478717962637727100126858 \ldots.$$