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