Answer
$\dfrac{16 \pi}{3}$
Work Step by Step
$\iiint_E x dV=\iint_D[\int_{4y^2+4z^2}^4 x dx]$
$= \iint_{D}(x^2/2)_{4y^2+4z^2}^4 dA$
$=\iint_{D}8-8(y^2+z^2)^2 dA$
$= \int_{0}^{2 \pi}\int_0^1[8-8(r^2)^2] r dr d\theta$
$=\dfrac{8}{3} \int_{0}^{2 \pi} 1 d \theta$
$=\dfrac{16 \pi}{3}$
