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