Question

The area bounded by $x=a{\cos }^3\theta ,y=a{\sin }^3\theta$ is:

• A. $\frac{3\pi a^2}{16}$
• B. $\frac{3\pi a^2}{8}$
• C. $\frac{3\pi a^2}{32}$
• D. $3\pi a^2$

Answer 1

The correct option is B $\frac{3\pi a^2}{8}$

Eliminating $\theta$, we have
$x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}x=0\Rightarrow y=\pm ay=0\Rightarrow x=\pm a$
Symmetric about both axis
Required area $=4{\int }_0^{a}ydx=4{\int }_{\frac{\pi }{2}}^{0}y\frac{dx}{dt}dt$
$y=a{\sin }^{3}t,x=a{\cot }^{3}t\Rightarrow \frac{dx}{dt}=-3a{\cos }^{3}t\sin t$
Area $=4{\int }_{\frac{\pi }{2}}^{0}a{\sin }^{3}t.(-3a{\cos }^{2}t\sin t)dt$
$=12a^2{\int }_0^{\frac{\pi }{2}}{\sin }^{4}t{\cos }^{2}tdt$
$=\frac{{12a}^2\frac{5}{2}!\frac{3}{2}!}{2\frac{8}{2}!}=\frac{{6a}^2\times \frac{3}{2}\times \frac{1}{2}\times \sqrt{\pi }.\frac{1}{2}\sqrt{\pi }}{3\times 2}$
$=\frac{3}{8}{\pi a}^2$