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)

$\frac{\pi a^2}{24}$

Answer 2

If curve $r = a \sin \, 3\theta$ To trace the curve, we consider the following table : 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 $= \frac{1}{2} \int^{\frac{\pi}{3}}_{0} r^{2} d \theta$ $= \frac{1}{2} \int ^{\frac{\pi }{3}}_{0}\sin^{2} \phi . \frac{1}{3} d\phi$ [On putting, $3\theta = \phi \Rightarrow d\theta = \frac{1}{3} d\phi$] $= \frac{a^{2}}{6} \int ^{\frac{\pi }{2}}_{0} \sin^{2} \phi d\phi$ $= \frac{a^{2}}{6}. \int^{\frac{\pi}{2}}_{0} \frac{1 - \cos 2\phi}{2} d\phi \left[ \because \, \, \cos 2\theta = 1 - 2 \sin^{2} \theta\right]$ $= \frac{a^{2}}{12} . \left[\phi + \frac{\sin 2 \phi}{2}\right]^{\frac{\pi}{2}}_{0}$ $= \frac{a^{2}}{12}. \left[ \frac{\pi}{2} + \sin \pi\right] = \frac{a^{2} \pi}{24}$