Answer
$I_x= 4 \sqrt{13} k \pi (1+6 \pi^2)$; $I_y= 4 \sqrt{13} k \pi (1+6 \pi^2)$ and $ I_z=8 \sqrt{13} k \pi$
Work Step by Step
$I_x=\int_{0}^{2\pi} (4 \cos^2 t+9t^2)(k) \sqrt{13} dt=\sqrt {13} k \int_{0}^{2\pi} (4 \cos^2 t+9t^2) dt= 4 \sqrt{13} k \pi (1+6 \pi^2)$
and $I_y=\int_{0}^{2\pi} (4 \sin^2 t+9t^2)(k) \sqrt{13} dt=\sqrt{13} k \int_{0}^{2\pi} 2 (1-\cos 2t)+9t^2= 4 \sqrt{13} k \pi (1+6 \pi^2)$
Also, we have
$I_z=\int_{0}^{2\pi} (4 \sin^2 t+4 \cos^2 t)(k) \sqrt{13} dt=(\sqrt {13}) k \int_{0}^{2\pi} (4) dt=(8 \sqrt{13}) k \pi$
Hence, we have $I_x= 4 \sqrt{13} k \pi (1+6 \pi^2)$; $I_y= 4 \sqrt{13} k \pi (1+6 \pi^2)$ and $ I_z=8 \sqrt{13} k \pi$
