Answer
$$11\pi \sqrt{3}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$\int^{\pi/2}_{\pi/6} \int^{\pi/2}_{-\pi/2} \int^2_{\csc\phi}5p^4 \sin\phi \space dp \space d\theta \space d\phi =\int^{\pi/2}_{\pi/6} \int^{\pi/2}_{-\pi/2} (32-\csc^5\phi) \sin^3\phi \space d\theta \space d\phi \\=\int^{\pi/2}_{\pi/6} \int^{\pi/2}_{-\pi/2} (32 \sin^3\phi-\csc^2\phi) \space d\theta \space d\phi\\=\pi \int^{\pi/2}_{[\pi/6}(32 \sin^3\phi-\csc^2\space \phi) \space d\phi \\=\pi(\dfrac{32\sqrt{3}}{24})-\dfrac{64\pi}{3}[\cos\phi]^{\pi/2}_{\pi/6}-\pi(\sqrt{3}) \\=\dfrac{\sqrt{3}}{3}\pi+(\dfrac{64\pi}{3}(\dfrac{\sqrt{3}}{2})) \\=\dfrac{33\pi\sqrt{3}}{3} \\ =11\pi \sqrt{3}$$
