Answer
$\dfrac{\pi a^2 K}{8}$ and The co-ordinates for the center of the mass is: $(\overline{x},\overline{y},\overline{z})=(0,0,\dfrac{2a}{3})$
Work Step by Step
We know that $M=K v$
$M=K\times \int_0^{2\pi} \int_{0}^{\sqrt a/2}\int_{4r^2}^{a} r \times dz dr d\theta$
$=K \times \int_0^{2\pi} \int_{0}^{\sqrt a/2} (ra-4r^3) dr d\theta$
$=K \times \int_0^{2\pi} \dfrac{a^2}{8}-\dfrac{a^2}{16} d\theta$
Thus, $M=\dfrac{\pi a^2 K}{8}$
$M_{yz}=0$ and $M_{xz}=0$ ( Symmetry)
That is, $M_{xy}=K \times \int_0^{2\pi} \int_{0}^{\sqrt a/2}\int_{4r^2}^{a} (zr) \times dz dr d\theta=K \times \int_0^{2\pi} \times (\dfrac{a^2r^2}{4}-\dfrac{4}{3}r^6)_{0}^{\sqrt a/2} dr d\theta=K \times \int_0^{2\pi} [\dfrac{a^3}{16}-\dfrac{a^3}{48}] d\theta=\dfrac{\pi a^3}{12}k$
Now, the center of mass along the z-axis is given by: $\overline{z}=\dfrac{M_{xy}}{M}=\frac{\dfrac{\pi a^3Kk}{12}}{\dfrac{\pi a^2 K}{8}}=\dfrac{2a}{3}$
Therefore, the co-ordinates for the center of the mass becomes: $(\overline{x},\overline{y},\overline{z})=(0,0,\dfrac{2a}{3})$
