Answer
$$\int_{0}^{3} \int_{0}^{\pi / 2} t^{2} \sin ^{3} \phi d \phi d t=6 $$
Work Step by Step
Given $$\int_{0}^{3} \int_{0}^{\pi / 2} t^{2} \sin ^{3} \phi d \phi d t$$So we have
\begin{aligned}I&= \int_{0}^{3} \int_{0}^{\pi / 2} t^{2} \sin ^{3} \phi d \phi d t\\
&= \left[\int_{0}^{\pi / 2} \sin ^{3} \phi d \phi\right]\left[\int_{0}^{3} t^{2} d t\right] \\ &=\left[\int_{0}^{\pi / 2} \sin ^{2} \phi \cdot \sin \phi d \phi\right]\left[\frac{t^{3}}{3}\right]_{0}^{3}
\\ &=\left[\int_{0}^{\pi / 2} \sin ^{2} \phi \cdot \sin \phi \ d \phi\right]\left[\frac{3^{3}}{3}-0\right]
\\ &=9 \int_{0}^{\pi / 2}\left(1-\cos ^{2} \phi\right) \cdot \sin \phi \ d \phi \\
& =9 \int_{0}^{\pi / 2}\left(\sin \phi-\sin \phi \cos ^{2} \phi\right) d \phi
\\
& =9 \left(-\cos \phi+\frac{1}{3}\cos ^{3} \phi\right)_{0}^{\pi / 2}
\\
& =9 \left(-\cos \frac{\pi}{2}+\frac{1}{3}\cos ^{3} \frac{\pi}{2}\right) - 9 \left(-\cos 0+\frac{1}{3}\cos ^{3} 0\right) \\
&=-9 \left(-1+\frac{1}{3}\right) \\
&=-9 \left(-\frac{2}{3}\right)\\
&=6
\end{aligned}
