find area of the region $x=a\cos^3\theta$ $y=a\sin^3\theta$

Question

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?

Answer 1

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 $(3/8)a^2\pi$.

Answer 2

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 $$ \begin{align} 16A &=3a^2\int_0^{2\pi}1\,\mathrm{d}\theta\\[4pt] &=6\pi a^2\\[8pt] A&=\frac{3\pi a^2}{8}\tag{6} \end{align} $$