Answer
$$5\pi $$
Work Step by Step
Our aim is to integrate the integral as follows:
$$\int^{2\pi}_0 \int^{\pi/3}_0 \int^2_{\sec (\phi)} (3) p^2 \sin\phi \space dp \space d\phi \space d\theta =\int^{2\pi}_0 \int^{\pi/3}_0 (8-sec^3\phi) \sin\phi \space d\phi \space d\theta \\=\int^{2\pi}_0 [-8 \cos\phi-\dfrac{1}{2} \sec^2\phi]^{\pi/3}_0 \space d\theta \\=\int^{2\pi}_0[(-4-2)-(-8-\dfrac{1}{2})] \space d\theta \\ =\dfrac{5}{2} \times \int^{2\pi}_0 d\theta \\=5\pi $$
