How to calculate the area enclosed by an implicit function?

Question

I want to calculate the area bounded by the region $y^3 - x^4 - y^4 = 0$. What can I do with this? Thanks for any help.

Answer 1

You can express the half of the curve as $x(y)=(y^3-y^4)^{1/4}$ and calculate the area as integral:

$$ A = 2\int_0^1(y^3-y^4)^{1/4}dy=\frac{3\pi}{8\sqrt2} $$

Details of how to calculate the integral I leave to you.

Answer 2

Notice that $$x=\pm (y^3(1-y))^{1/4}, y \in [0,1]$$ so the required area is $$A= 2 \int_{0}^{1} (y^3(1-y))^{1/4} dy.$$ Use $y=\sin^2 t, dy =2\sin t \cos t dt. $ Then $$A=4 \int_{0}^{\pi/2} \sin ^{5/2} t \cos^{3/2} t dt = 2 \frac{\Gamma(7/4) \Gamma(5/4)}{\Gamma(4)}=2\frac{(3/4)(1/4) \Gamma(3/4) \Gamma(1/4)}{2!}=\frac{3\pi}{8\sqrt{2}}.$$ Here we have used $\Gamma(1+z)=z\Gamma(z), \Gamma(z) \Gamma(1-z)=\frac{\pi}{\sin \pi z}.$