Answer
$\dfrac{\pi a^2 K}{8}$ and $(\overline{x},\overline{y},\overline{z})=(0,0,\dfrac{2a}{3})$
Work Step by Step
Since, $m=K v$
$m=K \int_0^{2\pi} \int_{0}^{\sqrt a/2}\int_{4r^2}^{a} r dz dr d\theta=K \int_0^{2\pi} \int_{0}^{(\sqrt a/2)} (ra-4r^3) dr d\theta=\dfrac{\pi a^2 K}{8}$
As per the symmetry, we get $M_{yz}=0$ and $M_{xz}=0$
This implies that
$M_{xy}=K \int_0^{2\pi} \int_{0}^{\sqrt a/2}\int_{4r^2}^{a} (zr) dz dr d\theta=K \int_0^{2\pi} (\dfrac{a^2r^2}{4}-\dfrac{4}{3}r^6)_{0}^{\sqrt a/2} dr d\theta$
and $K \int_0^{2\pi} (\dfrac{a^3}{16}-\dfrac{a^3}{48}) d\theta =\dfrac{\pi a^3 K}{12}$
The center of mass along the z-axis is given as:
$\overline{z}=\dfrac{M_{xy}}{m}=\dfrac{\pi a^3K/12}{\pi a^2 K/8}=\dfrac{2a}{3}$
Therefore, the co-ordinates for the center of mass are: $(\overline{x},\overline{y},\overline{z})=(0,0,\dfrac{2a}{3})$
