Answer
$$\dfrac{5\pi}{2}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$\int^{3\pi/2}_0 \int^{\pi}_0 \int^1_0 5p^3 \sin \phi \space dp \space d\phi \space d\theta =\dfrac{5}{4} \times \int^{3\pi/2}_0 \int^{\pi}_0 \sin^3\phi \space d\phi d\theta \\=\dfrac{5}{4} \times \int^{3\pi/2}_0 ([-\dfrac{-\sin^2\phi \cos\phi}{3}]^\pi_0+\dfrac{2}{3} \times \int^\pi_0 \sin \phi \space d\phi \space d\theta \\=\dfrac{5}{6} \times \int^{3\pi/2}_0 [-cis\phi]^\pi_0 \space d\theta \\=\dfrac{5}{3} \times \int^{3\pi/2}_0 d\theta \\=\dfrac{5\pi}{2}$$
