Question

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

• A. $\frac{\pi ab}{4}$
• B. $\frac{3\pi ab}{4}$
• C. $\frac{3\pi ab}{8}$
• D. None of these

Answer 1

The correct option is C $\frac{3\pi ab}{8}$

$x=a{\cos }^{3}t$
$\therefore$ $\frac{dx}{dt}=-3a{\cos }^{2}t\sin t$, $y=a{\sin }^{3}t$
$\therefore$ $\frac{dx}{dt}=+3a{\sin }^{2}t\cos t$
The closed curve is traced as t varies from 0 to $2\pi$
$\therefore$ $x\frac{dy}{dt}-y\frac{dx}{dt}=3ab{\sin }^{2}t{\cos }^{2}t$ (A)
$\therefore$ Required area $=\frac{1}{2}{\int }_0^{2\pi }(x\frac{dy}{dt}-y\frac{dx}{dt})dt$
$=\frac{3ab}{2}{\int }_0^{2\pi }{\sin }^{2}t{\cos }^{2}tdt$
$=\frac{3ab}{2}\times 4{\int }_0^{\pi /2}{\sin }^{2}t{\cos }^{2}tdt$
$=\frac{3ab}{2}\times 4\times \frac{1\times 1}{4\times 2}\times \frac{\pi }{2}$ ( Using Wallis formula).
$=\frac{3ab\pi }{8}$