Question

What is the area of a loop of the curve r = a sin3 $\theta$?

• A. $\frac{\pi a^2}{6}$
• B. $\frac{\pi a^2}{8}$
• C. $\frac{\pi a^2}{12}$
• D. $\frac{\pi a^2}{24}$

Answer 1

Answer: (D)

If curve r = a sin 3$\theta$
To trace the curve, we consider the following table :

$3\theta=$	$0$	$\frac{\pi}{2}$	$\pi$	$\frac{3\pi}{2}$	$2\pi$	$\frac{5\pi}{2}$	$3\pi$
$\theta=$	$0$	$\frac{\pi}{6}$	$\frac{\pi}{3}$	$\frac{\pi}{2}$	$\frac{2\pi}{3}$	$\frac{5\pi}{6}$	$\pi$
$r=$	$0$	$a$	$0$	$-a$	$0$	$a$	$0$

Thus there is a loop between
$\theta$ = 0 & $\theta = \frac{\pi}{3}$ as r varies from r = 0 to r = 0

Hence, the area of the loop lying in the
positive quadrant $=\frac{1}{2} \, \int\limits_{0}^{\frac{\pi}{3}} \, r^2 \, d\theta$
$=\frac{1}{2} \, \int\limits_{0}^{\frac{\pi}{3}} \, \, sin^2 \phi. \frac{1}{3} d\phi$
$[On \, putting, \, 3\theta \,\phi = \, \, \Rightarrow \, d\theta \, =\frac{1}{3} d\phi]$
$= \frac{a^2}{6} \, \int \limits_{0}^{\frac{\pi}{2}} \, sin^2 \phi \, d\phi$
$=\frac{a^2}{6}.\int\limits_{0}^{\frac{\pi}{2}} \frac{1-cos 2\phi }{2 } \, d\phi$
$ [\because \, cos 2 \theta = 1 - 2sin^2\theta]$
$=\frac{a^2}{12} . [ \phi+ \frac{sin 2\phi }{2}]^{\frac{\pi}{2}}_{0}$
$=\frac{a^2}{12}.\bigg[{\frac{\pi}{2} + sin \pi} \bigg] \, = \frac{a^2 \pi}{24}$