The circle $r = 2 \sin \theta $ i.e $ r^2 = 2r\sin\theta$ becomes in Cartesian system

$X^2 +y^2 = 2y $ i.e $ x^2 + (y - 2)^2 = 4$

In the given region a varies from $2 \sin \theta$ to $4\sin\theta$ and $\theta$ varies from o to $\pi$

$I=\int\limits_0^{\pi}\int\limits_{2\sin \theta}^{4\sin\theta}r^3drdo$

$=\int\limits_0^{\pi}\Bigg[\dfrac {r^4}4\Bigg]_{2\sin \theta}^{4\sin\theta} d\theta$

$=\dfrac 14\int\limits_0^{\pi} 4^4\sin^4\theta -2^4\sin^4\theta d\theta$

$=\dfrac {40}4\Bigg[\int\limits_0^{\pi}\sin^4\theta d\theta\Bigg]$

$=60\times2 \int\limits_0^{\dfrac 12}\sin^4\theta d\theta ----\text{Direct Formula}$

$=120\times\dfrac 34\times \dfrac 12\times \dfrac \pi2=\dfrac {45\pi}2$