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 $=\frac{1}{2}\underset{0}{\overset{\frac{\pi }{3}}{\int }}{r}^{2}d\theta$
$=\frac{a^2}{2}\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^2\varphi \cdot \frac{1}{3}d\varphi$
[On putting, $3\theta =\varphi \Rightarrow d\theta =\frac{1}{3}d\varphi$ ]
$=\frac{a^2}{6}\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^2\varphi d\varphi$
$=\frac{a^2}{6}\cdot \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1-\cos 2\varphi }{2}d\varphi$
$\left[\because \cos 2\theta =1-2{\sin }^2\theta \right]$
$=\frac{a^2}{12}\cdot {\left[\varphi +\frac{\sin 2\varphi }{2}\right]}_0^{\frac{\pi }{2}}$
$=\frac{a^2}{12}\cdot \left[\frac{\pi }{2}+\frac{\sin \pi }{2}\right]=\frac{a^2\pi }{24}.$