Find the area enclosed inside the curve parameterized by ($\cos^3 (t)$, $\sin^3 (t)$) for $0\leq t\leq 2\pi$

Question

When I do this I get $0$, which my online class returns as false. I used the formula for area, $\frac{1}{2}\oint -y dx + x dy$ . I used u sub and came up with $\frac{3}{2} \int_{0}^{2\pi} \sin^4(t) \cos(t)+ \cos^4(t)\sin(t) dt$ which evaluated to $0$.

Answer 1

The substitution of the curve equation into the formula for area ends up with: $$ \frac{3}{2} \int_{0}^{2\pi} \left[\sin^4(t) \cos^2(t)+ \cos^4(t)\sin^2(t)\right] dt =\frac{3}{2} \int_{0}^{2\pi}\frac{1-\cos(4t)}{8}dt =\frac{3\pi}{8}. $$ You just have lost a unit of power by differentiation of $\cos^3(t)$ and $\sin^3(t)$.

Answer 2

An alternative approach through Euler's Beta function.

Your curve is the astroid with equation $|x|^{2/3}+|y|^{2/3}=1$. By symmetry, the enclosed area is given by

$$ 4\int_{0}^{1}(1-x^{2/3})^{3/2}\,dx\stackrel{x\mapsto z^{3/2}}{=}6\int_{0}^{1}z^{1/2}(1-z)^{3/2}\,dz=6\,B\left(\tfrac{3}{2},\tfrac{5}{2}\right)=\frac{6\,\Gamma\left(\tfrac{3}{2}\right)\Gamma\left(\frac{5}{2}\right)}{\Gamma(4)}=\color{blue}{\frac{3\pi}{8}}. $$