Answer
$\dfrac{9 \pi}{8}$
Work Step by Step
Let us consider that $I=\iiint_E \dfrac{z}{x^2+z^2} dV$
$ I=\int_{1}^{4} \int_{y}^{4} \int_{0}^{z}\dfrac{z}{x^2+z^2} dx dzdy$
Further, $\int_{1}^{4} \int_{y}^{4} [(z)\dfrac{1}{z}\tan^{-1}(\dfrac{x}{z}]_{0}^{z} dzdy=\int_{1}^{4} \int_{y}^{4} (\dfrac{\pi}{4}) dzdy$
and $ \int_{1}^{4} (\dfrac{\pi}{4})[z]_y^4dy= \int_{1}^{4} [\pi-\dfrac{\pi y}{4} ] dy$
Hence$\iiint_E \dfrac{z}{x^2+z^2} dV=[\pi y-(\dfrac{\pi}{8})y^2]_1^4=\dfrac{9 }{8}\pi$
