Answer
$\dfrac{\pi^2 ka^4}{4}$
Work Step by Step
Here, we have $m=\\\int_Bk\sqrt{x^2+y^2} dV=\int_0^{2\pi} \int_{0}^{a}\int_{-\sqrt{a^2-r^2}}^{\sqrt{a^2-r^2}} (kr) \times (r) dz dr d\theta$
or, $=\int_0^{2\pi} \int_{0}^{a}\int_{-\sqrt{a^2-r^2}}^{\sqrt{a^2-r^2}} kr^2 dz dr d\theta$
or, $=2\pi \int_{0}^{a} 2kr^2 \sqrt{a^2-r^2} dr$
Plug in $r=a\sin p$ and $dr=a \cos p dp$
Thus, $m=2\pi \int_{0}^{a} 2kr^2 \sqrt{a^2-r^2} dr=2\pi \int_{0}^{\pi/2} 2ka^2 \sin^2 p \sqrt{a^2-a^2 \sin^2 p} \times a \cos p dp$
or, $=\pi ka^4\int_0^{\pi/2} \sin^2 2p dp$
or, $=(1/2) \pi ka^4\int_0^{\pi/2} 1-\cos 4p dp$
or, $=(1/2) \pi ka^4[p-\dfrac{\sin 4p}{4}]_0^{\pi/2}$
or, $m=\dfrac{\pi^2 ka^4}{4}$
