Answer
$\dfrac{16 \pi}{3}$
Work Step by Step
Let us consider that $I=\iiint_E x dV$
$I=\iint_D[\int_{4y^2+4z^2}^4 x dx]=\iint_{D}[8-8(y^2+z^2)^2]dA$
In polar co-ordinates, we have $r^2=x^2+y^2 \implies r=\sqrt{x^2+y^2}$ and $x=r \cos \theta \\ y=r \sin \theta$
$\int_{0}^{2 \pi}\int_0^1[8-8(r^2)^2] r dr d\theta=\int_{0}^{2 \pi}\int_0^1[8-8r^4] r dr d\theta=\int_{0}^{2 \pi}\int_0^1[8r-8r^5] dr d\theta$
and $\int_{0}^{2 \pi}[4-\dfrac{4}{3}] d\theta=(\dfrac{8}{3}) \int_{0}^{2 \pi} d \theta$
Hence, $\iiint_E x dV=\dfrac{16 \pi}{3}$
