## Calculus: Early Transcendentals 8th Edition

$\dfrac{16 \pi}{3}$
Consider $I=\iiint_E x dV$ $I=\iint_D[\int_{4y^2+4z^2}^4 x dx]= \iint_{D}(x^2/2)_{4y^2+4z^2}^4 dA$ or, $=\iint_{D}8-8(y^2+z^2)^2 dA$ We need to use polar co-ordinates. or, $= \int_{0}^{2 \pi}\int_0^1[8-8(r^2)^2] r dr d\theta$ or, $=int_{0}^{2 \pi}\int_0^1[8r-8r^5] dr d\theta$ or, $=int_{0}^{2 \pi}[4-\dfrac{4}{3}] d\theta$ or, $=(8/3) \int_{0}^{2 \pi} d \theta$ or, $=\dfrac{16 \pi}{3}$