Answer
$\dfrac{2\pi}{5}$
Work Step by Step
Consider $I=\iiint_Ex^2dV$
Here, we have $x=r \cos \theta; y=r \sin \theta, z=z$
Then, we get
$\int_0^{2\pi} \int_{0}^{1}\int_{0}^{2r} x^2 r dz dr d\theta=\int_0^{2\pi} \int_{0}^{1}\int_{0}^{2r} (r\cos \theta)^2 r dz dr d\theta$
and
$=\int_0^{2\pi} \int_{0}^{1}[r^3 \cos^2 \theta \times (z)]_{0}^{2r} dr d\theta$
$=\int_0^{2\pi} \int_{0}^{1}2r^4 \times (\cos^2 \theta ) dr d\theta$
$=\int_0^{2\pi} \times (\dfrac{2}{5}) \times (\cos^2 \theta) d\theta$
$=(\dfrac{2}{10}) \int_0^{2\pi} (1+\cos 2 \theta) d\theta$
$=(\dfrac{2}{10}) [\theta+\dfrac{\sin 2 \theta}{2}]_0^{2\pi}$
$=\dfrac{2\pi}{5}$
