Question

The area bounded by the curves $x=a{\cos }^{3}t,y=a{\sin }^{3}t$ is

• A. $\frac{3\pi a^2}{8}$
• B. $\frac{3\pi a^2}{16}$
• C. $\frac{3\pi a^2}{32}$
• D. None of the above

Answer 1

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

$x=a{\cos }^{3}t$ , $y=a{\sin }^{2}t$

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$

$A={\int }_0^{2\pi }xdy$

$A=a^2{\int }_0^{2\pi }{\cos }^{3}t\times 3{\sin }^{2}t\times \cos tdt$

$=3a^2{\int }_0^{2\pi }{\cos }^{2}t\times {\sin }^{2}tdt$ $\to \left(1\right)$

similarily $A={\int }_0^{2\pi }ydx$

$=3a^2{\int }_0^{2\pi }{\sin }^{2}t\times {\cos }^{2}tdt$ $\to \left(2\right)$

Adding (1) and (2)

$2A=3a^{2}\int_{0}^{2\pi}{\cos^{2}t\times\sin^{2}t}dt

$A=\frac{3a^2}{8}{\int }_0^{2\pi }{\sin }^{2}2dt$ $\to \left(3\right)$

Similarily $A=\frac{3a^2}{8}{\int }_0^{2\pi }{\cos }^{2}tdt$ $\to \left(4\right)$

By putting $t=t+\frac{\pi }{4}$ in (3)

Adding (3) and (4)

$2A=\frac{3a^2}{8}{\int }_0^{2\pi }dt$

$A=\frac{3a^2}{8}\pi$