\(\iint_R r^2drd\theta\)
Region R bounded by curves r = a sin θ & r = 2a sin θ
\(\displaystyle \int\limits^\pi_0\left(\int\limits^{2a\sin \theta}_{a\sin \theta}r^2\,dr\right)d\theta\)
\(=\frac{1}{3}\displaystyle\int\limits^\pi_0 (r^3)^{2a\sin\theta}_{a\sin\theta}d\theta\)
\(=\frac{1}{3}\displaystyle\int\limits^\pi_0(8a^3\sin^3\theta-a^3\sin^3\theta)d\theta\)
\(=\frac{7}{3}\displaystyle\int\limits^\pi_0\sin^3\theta \,d\theta\)
\(=\frac{14}{3}\displaystyle\int\limits^{\frac{\pi}{2}}_0\sin^3\theta \,d\theta\)
\(=\frac{14}{3}\displaystyle\int^{\frac{\pi}{2}}_0\left(\frac{3\sin\theta-\sin 3\theta}{4}\right)d\theta\) \(\begin{pmatrix}\because \sin 3\theta =3\sin\theta-4\sin^3\theta\\\Rightarrow \sin^3\theta=\frac{3\sin\theta-\sin 3\theta}{4}\end{pmatrix}\)
\(=\frac{7}{6}\left[-3\cos\theta+\frac{\cos3\theta}{3}\right]^{\frac{\pi}{2}}_0\)
\(=\frac{7}{6}\Big[-3\cos\frac{\pi}{2}+\frac{\cos\frac{3\pi}{2}}{3}+3\cos 0-\frac{\cos 0}{3}\Big]\)
\(=\frac{7}{6}\left(3-\frac{1}{3}\right)\) \((\because \cos\frac{\pi}{2}=\cos\frac{3\pi}{2}=0 \) & cos 0 = 1)
\(=\frac{7}{6}\times \frac{8}{3}=\frac{28}{9}\)
