Answer
$(0,0, \dfrac{3(2+\sqrt 2)}{16}) $
Work Step by Step
The region $E$ using the point of intersection can be expressed 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,
$V=\iint_{E} dV=\int_0^{\pi/4} \int_0^{2 \pi} \int_0^1 \rho^2 \sin \phi d \rho d \theta d \phi =[\rho^3/3]_0^1 [-\cos\phi]_0^{\pi/4} [ \theta]{0}^{2\pi} = \dfrac{(2-\sqrt 2) \pi}{3}$
and the z-coordinate is $z=\dfrac{\pi}{8 V}$
Now,
$z=\dfrac{\pi}{8 \times \dfrac{(2-\sqrt 2) \pi}{3}}=\dfrac{3}{8(2-\sqrt 2)} \times \dfrac{2+\sqrt 2}{2+\sqrt 2} = \dfrac{3(2+\sqrt 2)}{16}$
Thus, the centroid is: $(0,0, \dfrac{3(2+\sqrt 2)}{16}) $
