Answer
$\dfrac{\pi a^2 K}{8}$ ; $(0,0,\dfrac{2a}{3})$
Work Step by Step
Conversion of rectangular to cylindrical coordinate system gives: $r^2=x^2+y^2 \\ \tan \theta=\dfrac{y}{x} \\z=z$
Here, $x=r \cos \theta; y=r \sin \theta, z=z$
Since, $M=K v$
Then, we have $M=K \int_0^{2\pi} \int_{0}^{\sqrt a/2}\int_{4r^2}^{a} r dz dr d\theta$
or, $=K \int_0^{2\pi} \int_{0}^{\sqrt a/2} (ra-4r^3) dr d\theta=K \int_0^{2\pi} \dfrac{a^2}{8}-\dfrac{a^2}{16} d\theta$
or, $=\dfrac{\pi a^2 K}{8}$
Due to symmetry, we have $M_{yz}=0$ and $M_{xz}=0$
Then, we have $M_{xy}=K \int_0^{2\pi} \int_{0}^{\sqrt a/2}\int_{4r^2}^{a} (zr) dz dr d\theta$
or, $=K \int_0^{2\pi} (\dfrac{a^2r^2}{4}-\dfrac{4}{3}r^6)_{0}^{\sqrt a/2} dr d\theta=K \int_0^{2\pi} \dfrac{a^3}{16}-\dfrac{a^3}{48} d\theta$
or, $=\dfrac{\pi a^3 K}{12}$
Now, the center of mass along the z-axis is: $\overline{z}=\dfrac{M_{xy}}{M}=\dfrac{\dfrac{\pi a^3Kk}{12}}{\dfrac{\pi a^2 K}{8}}=\dfrac{2a}{3}$
The co-ordinates for the center of the mass are: $(\overline{x},\overline{y},\overline{z})=(0,0,\dfrac{2a}{3})$
