Answer
$$2\pi $$
Work Step by Step
Our aim is to integrate the integral as follows:
$$\int^2_0 \int^0_{-\pi} \int^{\pi/2}_{\pi/4} p^3 \sin (2\phi) \space d\phi \space d\theta \space dp =\int^2_0 \int^0_{-\pi} \space
p^3[\dfrac{-[\cos (2\phi)]}{2}]^{\pi/2}_{\pi/4} \space d\theta \space dp \\=\int^2_0 \int^0_{-\pi} \dfrac{p^3}{2} \space d\theta \space dp \\=\pi \times \int^2_0 \dfrac{p^2}{2} \space dp \\=\pi [\dfrac{p^4}{8}]^2_0 \\=2\pi $$
