Answer
$\dfrac{9 \pi}{8}$
Work Step by Step
$\iiint_E \dfrac{z}{x^2+z^2} dV=\int_{1}^{4} \int_{y}^{4} \int_{0}^{z}\dfrac{z}{x^2+z^2} dx dzdy$
and $\int_{1}^{4} \int_{y}^{4} [(z)\dfrac{1}{z}\arctan(\dfrac{x}{z})]_{0}^{z} dzdy=(\dfrac{\pi}{4})\int_{1}^{4} \int_{y}^{4} dzdy$
$= \int_{1}^{4} [\dfrac{1}{4}(\pi z) ]_y^4dy$
$= \int_{1}^{4} [\pi-\dfrac{\pi y}{(2)(4)} ] dy$
$=[\pi y-\dfrac{\pi}{8}y^2]_1^4$
$=[\pi (4-1)-\dfrac{\pi}{8}(4-1)^2]_1^4$
$=\dfrac{9 \pi}{8}$
