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