Answer
$\dfrac{\pi^2 ka^4}{4}$
Work Step by Step
Since, $m=kv=\\\int_Bk\sqrt{x^2+y^2} dV$
This implies that $m=\int_0^{2\pi} \int_{0}^{a}\int_{-\sqrt{a^2-r^2}}^{\sqrt{a^2-r^2}} (kr) \times (r) dz dr d\theta=\int_0^{2\pi} \int_{0}^{a}\int_{-\sqrt{a^2-r^2}}^{\sqrt{a^2-r^2}} kr^2 dz dr d\theta$
and $\int_0^{2\pi} \int_{0}^{a}\int_{-\sqrt{a^2-r^2}}^{\sqrt{a^2-r^2}} kr^2 dz dr d\theta =2\pi \int_{0}^{a} 2kr^2 \sqrt{a^2-r^2} dr$
Consider $r=a\sin b$ and $dr=a \cos b db$
Thus, $m=2\pi \int_{0}^{\pi/2} 2ka^2 \sin^2 b \sqrt{a^2-a^2 \sin^2 b} \times a \cos b db=\pi ka^4\int_0^{\pi/2} \times (\sin^2 2b db) $
and, $\dfrac{1}{2} \pi ka^4\int_0^{(\pi/2)} 1-\cos 4b db=\dfrac{ \pi ka^4}{2}[b-\dfrac{\sin 4b}{4}]_0^{\pi/2}$
Hence, $m=\dfrac{1}{4}\pi^2 a^4K$
