Answer
$\dfrac{\pi^2 ka^4}{4}$
Work Step by Step
The mass is given by: $m=\int_Bk\sqrt{x^2+y^2} dV=\int_0^{2\pi} \int_{0}^{a}\int_{-\sqrt{a^2-r^2}}^{\sqrt{a^2-r^2}} \times (kr) (r) dz dr d\theta$
$=\int_0^{2\pi} \int_{0}^{a}\int_{-\sqrt{a^2-r^2}}^{\sqrt{a^2-r^2}} kr^2 \times dz dr d\theta$
$=2\pi \int_{0}^{a} 2k \times r^2 \sqrt{a^2-r^2} dr$
Suppose $r=a\sin p \implies dr=a \cos p dp$
$=2\pi \times \int_{0}^{a} 2kr^2 \sqrt{a^2-r^2} dr$
$=2\pi \times \int_{0}^{\pi/2} 2ka^2 \times [\sin^2 p \sqrt{a^2-a^2 \sin^2 p} (a \cos p) ] dp$
$=\pi ka^4 \times \int_0^{\pi/2} \sin^2 (2p) dp$
$=(0.5) \pi \times ka^4 \times \int_0^{\pi/2} [1-\cos (4p)] dp$
$=(0.5) \times \pi ka^4 \times [p-\dfrac{\sin (4p) }{4}]_0^{\pi/2}$
$=\dfrac{\pi^2 ka^4}{4}$
