find area of the region $x=a\cos^3\theta$ $y=a\sin^3\theta$ Find the area of the region enclosed by $x=a\cos^3\theta$ and $y=a\sin^3\theta$
What steps should I take in order to find the area?
 A: To get the area:
$$
\begin{align}
A
&=\int_{}^{}x\,\mathrm{d}y\\
&=\int_0^{2\pi}a\cos^3(\theta)\,3a\sin^2(\theta)\cos(\theta)\,\mathrm{d}\theta\\
&=3a^2\int_0^{2\pi}\sin^2(\theta)\cos^4(\theta)\,\mathrm{d}\theta\tag{1}
\end{align}
$$
We can compute $A$ without using any trigonometric integrals.
Substituting $\theta\mapsto\theta+\pi/2$ in $(1)$, we get
$$
A=3a^2\int_0^{2\pi}\cos^2(\theta)\sin^4(\theta)\,\mathrm{d}\theta\tag{2}
$$
Adding $(1)$ and $(2)$, remembering that $\sin^2(\theta)+\cos^2(\theta)=1$, yields
$$
2A=3a^2\int_0^{2\pi}\cos^2(\theta)\sin^2(\theta)\,\mathrm{d}\theta\tag{3}
$$
Multiplying $(3)$ by $4$ gives
$$
8A=3a^2\int_0^{2\pi}\sin^2(2\theta)\,\mathrm{d}\theta\tag{4}
$$
Substituting $\theta\mapsto\theta+\pi/4$ in $(4)$, we get
$$
8A=3a^2\int_0^{2\pi}\cos^2(2\theta)\,\mathrm{d}\theta\tag{5}
$$
Adding $(4)$ and $(5)$ yields
$$
\begin{align}
16A
&=3a^2\int_0^{2\pi}1\,\mathrm{d}\theta\\[4pt]
&=6\pi a^2\\[8pt]
A&=\frac{3\pi a^2}{8}\tag{6}
\end{align}
$$
A: Hint:
Use this fact that if $x=x(t),~~~y=y(t),~~\alpha\leq t\le\beta$ then:
$$S=\frac{1}2\int_{\alpha}^{\beta}(xy'-yx')dt$$ Here we have $0\le t\le 2\pi$ and the definite integral is not so hard. Think of $(3/8)a^2\pi$. 
