Answer
The centroid is $(0,0, \dfrac{3(2+\sqrt 2)}{16}) $ and $V= \dfrac{(2-\sqrt 2) \pi}{3}$
Work Step by Step
We can define the region $E$ using the point of intersection as follows:
$E=\left\{ (\rho, \theta, \phi) | 0 \leq \rho \leq 1, 0 \leq \theta \leq 2 \pi, 0 \leq \phi \leq \dfrac{\pi}{4} \right\}$
Now,
$Volume=V=\iint_{E} dV \\ =\int_0^{\pi/4} \int_0^{2 \pi} \int_0^1 \rho^2 \sin \phi \ d \rho \ d \theta \ d \phi \\=[\rho^3/3]_0^1 [-\cos\phi]_0^{\pi/4} [ \theta]{0}^{2\pi} \\= \dfrac{(2-\sqrt 2) \pi}{3}$
Also, the z-coordinate is equal to: $z=\dfrac{\pi}{8 V}$
$\implies z=\dfrac{\pi}{8 \times \dfrac{(2-\sqrt 2) \pi}{3}}\\=\dfrac{3}{8(2-\sqrt 2)} \times \dfrac{2+\sqrt 2}{2+\sqrt 2} \\= \dfrac{3(2+\sqrt 2)}{16}$
Therefore, the centroid is $(0,0, \dfrac{3(2+\sqrt 2)}{16}) $ and $V= \dfrac{(2-\sqrt 2) \pi}{3}$
