Answer
$\dfrac{\ln 2}{3}$
Work Step by Step
Here, we have $ \int_{0}^1 \int_{0}^{1} \int_{0}^{\sqrt{1-z^2}} \dfrac{z}{y+1} dx dz dy= \int_{0}^1 \int_{0}^{1} [\dfrac{xz}{y+1}]_{0}^{(\sqrt{1-z^2})} dz dy$
or, $=\int_{0}^1 (y+1)^{-1} dy \times \int_{0}^{1} z\sqrt{1-z^2} dz$
Plug $1-z^2=u \implies z dz=-\dfrac{1}{2}du$
Thus, we get
$\implies \int_{0}^1 (y+1)^{-1} dy \times \int_{0}^{1} z\sqrt{1-z^2} dz=[\ln |y+1|]_{0}^1 \int_{0}^{1} \sqrt u [-\dfrac{du}{2}]$
and
$ [\dfrac{-(\ln 2)}{2}] \int_0^1 (u)^{1/2} du= \dfrac{\ln 2}{3}$
