Question

What is the area of a loop of the curve $r=a \sin 3 \theta$ ?

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