Question

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

• A. $\frac{3}{8}\pi ab$
• B. $\frac{5}{8}\pi ab$
• C. $\frac{1}{8}\pi ab$
• D. None of these

Answer 1

Answer: (A)

$\frac{3}{8}\pi ab$

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

Whole area$=A=\left|{\int }_0^{2\pi }ydx\right|$

$=\left|{\int }_0^{2\pi }b{\sin }^{3}td\left(a{\cos }^{3}t\right)\right|$

$=|-3ab{\int }_0^{2\pi }{\cos }^{2}t{\sin }^{4}tdt|$

$=6ab{\int }_0^{\pi }{\cos }^{2}t{\sin }^{4}tdt[\because {\int }_0^{2m}f\left(\theta \right)d\theta =2{\int }_0^{m}f\left(\theta \right)d\theta ;iff\left(\theta \right)=f(2m-\theta )]$

$=12ab{\int }_0^{\pi /2}{\cos }^{2}t{\sin }^{4}tdt$

$A=12ab{\int }_0^{\pi /2}{\cos }^{2}t{\sin }^{4}tdt=12ab{\int }_0^{\pi /2}{\sin }^{2}t{\cos }^{4}tdt$

$\therefore 2A=12ab{\int }_0^{\pi /}({\cos }^{2}t{\sin }^{4}t+{\sin }^{2}t{\cos }^{4}t)dt$

$2A=12ab{\int }_0^{\pi /2}{\cos }^{2}t{\sin }^{2}tdt(\because {\cos }^{2}t+{\sin }^{2}t=1)$

$2A=3ab{\int }_0^{\pi /2}{\sin }^{2}2tdt$

$2A=3ab{\int }_0^{\pi /2}\left(\frac{1-\cos 4t}{2}\right)dt$

$2A=\frac{3ab}{2}{[t-\frac{\sin 4t}{4}]}_0^{\pi /2}$

$2A=\frac{3ab\pi }{4}$

$\therefore A=\frac{3}{8}\pi ab$