Answer
$$\dfrac{11\pi}{6}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$ Volume=\int^{2\pi}_0 \int^{\pi/2}_0 \int^2_{\cos\phi} p^2 \sin \phi d p \space d\phi \space d\theta \\=\dfrac{1}{3} \int^{2\pi}_{0}\int^{\pi/2}_0 (3 \cos\phi+3 \cos^2 (\phi) +\cos^3 (\phi)] \sin \phi \space d\phi \space d\theta \\=\dfrac{1}{3}\int^{2\pi}_0 [-\dfrac{3}{2} \cos^2\phi- \cos^3 (\phi)-\dfrac{1}{4} \cos^4\phi]^{\pi/2}_0 \space d\theta \\=\dfrac{1}{3} \int^{2\pi}_0(\dfrac{3}{2}+1+\dfrac{1}{4} \space d\theta) \\=\dfrac{11}{12}\int^{2\pi}_0 (1) \space d\theta \\=\dfrac{11\pi}{6}$$
