Answer
$(2-\sqrt 2)\pi$
Work Step by Step
Region of integration:
$R=${$ (r,\theta, z) | r \lt z \leq 1, 0 \leq r \leq 1, 0 \leq \theta \leq 2 \pi$}
Consider $I= \iint_{R} \dfrac{z}{(x^2+y^2+z^2)^{3/2}} \ dV$
or, $=\int^{0}_{2 \pi} \int_{0}^{1} \int_{r}^1 \dfrac{zr}{(r^2+z^2)^{3/2}} ( dz \ dr \ d \theta) $
We will use a calculator to find the triple integral.
$\int^{0}_{2 \pi} \int_{0}^{1} \int_{r}^1 \dfrac{zr}{(r^2+z^2)^{3/2}} ( dz \ dr \ d \theta) =(2-\sqrt 2)\pi$
