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 $=\frac{1}{2}\underset{0}{\overset{\frac{\pi }{3}}{\int }}{r}^{2}d\theta$
$=\frac{1}{2}\underset{0}{\overset{\frac{\pi }{3}}{\int }}si{n}^2\varphi .\frac{1}{3}d\varphi$
$[Onputting,3\theta \varphi =\Rightarrow d\theta =\frac{1}{3}d\varphi ]$
$=\frac{a^2}{6}\underset{0}{\overset{\frac{\pi }{2}}{\int }}si{n}^2\varphi d\varphi$
$=\frac{a^2}{6}.\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{1-cos2\varphi }{2}d\varphi$
$[\because cos2\theta =1-2si{n}^2\theta ]$
$=\frac{a^2}{12}.[\varphi +\frac{sin2\varphi }{2}{]}_0^{\frac{\pi }{2}}$
$=\frac{a^2}{12}.[\frac{\pi }{2}+sin\pi ]=\frac{a^2\pi }{24}$