Answer
$\dfrac{2\pi}{5}$
Work Step by Step
$I=\iiint_Ex^2dV=\int_0^{2\pi} \int_{0}^{1}\int_{0}^{2r} x^2 \times (r dz dr d\theta) $
$=\int_0^{2\pi} \int_{0}^{1}\int_{0}^{2r} (r\cos \theta)^2 \times (r dz dr d\theta) $
$=\int_0^{2\pi} \int_{0}^{1}[r^3 \times \cos^2 \theta(z)]_{0}^{2r} dr d\theta$
$=\int_0^{2\pi} \int_{0}^{1}2 \times r^4 \times \cos^2 \theta dr d\theta$
$=\int_0^{2\pi}(\dfrac{2}{5}) \times \cos^2 \theta d\theta$
$=0.2 \times \int_0^{2\pi} (1+\cos 2 \theta) d\theta$
$=0.2 \times \int_0^{2\pi} [\theta+\dfrac{\sin 2 \theta}{2})_0^{2\pi}$
$=\dfrac{2\pi}{5}$
