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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: eye shaped curve

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:

half-eye shaped curve

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:

$$ y = \arccos\left(3 - \sqrt{16 - \sin^4(x)}\right) - 2.632 $$

shifted shape

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?

HugoTeixeira
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  • There seems to be an error in your calculations. Centering the eye leads to the equation $\cos^4 x+(3+\cos y)^2=16$, which is solved by $y=\arccos \bigl(\sqrt{16-\cos^4 x}-3\bigr)$. This leads to an area of $\approx1.5893$. – Christian Blatter Sep 11 '18 at 14:35
  • @ChristianBlatter The curve $y = \arccos(3 - \sqrt{16 - \sin^4(x)})$ is just the lower section of the eye shape, which means half of the shape. You can see both equations together here: https://www.desmos.com/calculator/jpfjxgdulb – HugoTeixeira Sep 11 '18 at 14:52

1 Answers1

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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 $$2 \left(\pi^2 - \int_{x=0}^\pi \cos^{-1} \left( 3 - \sqrt{16 - \sin^4 x} \right) \, dx \right) \approx 1.589300092423124478717962637727100126858 \ldots.$$

heropup
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  • Thanks, your solution seems to be the correct one. I was wondering what went wrong with my calculation since I got a different result. After calculating those areas over and over again using different approaches, I realize that my mistake was that I tried to calculate the area using the section under the curve, which is actually outside the "eye" shape. The area under the curve is different, which makes me believe that those shapes cannot be grouped side by side to create a pattern. – HugoTeixeira Sep 24 '18 at 02:06