Answer
$I_x=I_y= 4 \sqrt{13} k \pi (1+6 \pi^2); I_z=8 \sqrt{13} k \pi$
Work Step by Step
Here, $ds=\sqrt{(dx/dt)^2+(dy/dt)^2+(dz/dt)^2}$
Thus, $ds=\sqrt{(2\cos t)^2+(-2\sin t)^2+3^2}dt=\sqrt{13} dt$
Now, $I_x=\int_{C} (y^2+z^2) \rho(x,y,z) ds=\sqrt {13} k \int_{0}^{2\pi} (4 \cos^2 t+9t^2) dt= 4 \sqrt{13} k \pi (1+6 \pi^2)$
$I_y=\int_{C} (x^2+z^2) \rho(x,y,z) ds=\sqrt {13} k \int_{0}^{2\pi} (4 \sin^2 t+9t^2) dt= 4 \sqrt{13} k \pi (1+6 \pi^2)$
$I_z=\int_{C} (x^2+y^2) \rho(x,y,z) ds=\int_{0}^{2\pi} (4 \sin^2 t+4 \cos^2 t)(k) \sqrt{13} dt=\sqrt {13} k \int_{0}^{2\pi} (4) dt=8 \sqrt{13} k \pi$
Hence, we get $I_x=I_y= 4 \sqrt{13} k \pi (1+6 \pi^2); I_z=8 \sqrt{13} k \pi$
