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I have the answer. The hypobloid has parametrization = $x = acos^3(t)$ $y = asin^3(t)$ The explanation is you take a vector field $F(x,y) = (0, x) which has curl 1 than it says the area is equal to:

$$\int_{0}^{2\pi} (0,cos^3(t)) . (-3cos^2(t)sin(t),3asin^2(t)cos(t))$$

using Green's theorem you remake it.

What I don't understand is why create an artificial vector field to get the area? What's the logic begind that and how do I chose the right vector field?

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Pro tip: Known functions have a nice pretty-print version: Just put an \ before it :) – chubakueno May 14 '14 at 22:26

$\begin{align}x=a\cdot\cos^3t\\y=a\cdot\sin^3t\end{align}\quad=>\quad\bigg(\dfrac xa\bigg)^\tfrac23+\bigg(\dfrac ya\bigg)^\tfrac23=1$, which is a superellipse. A quarter of its area is

given by $\displaystyle\int_0^1\sqrt[n]{1-x^n}~dx$, where $n=\dfrac23$ . At the same time, we know that $\displaystyle\int_0^1\sqrt[m]{1-x^n}~dx=$

$=\dfrac{\Big(\frac1m\Big)!\cdot\Big(\frac1n\Big)!}{\Big(\frac1m+\frac1n\Big)!}$ , see beta function for more details. Replacing, and using the fact that $\Gamma\bigg(\dfrac12\bigg)$

is $~\sqrt\pi,~$ we have $A=4\cdot I=4\cdot\dfrac{3\pi}{32}=\dfrac{3\pi}8$

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