Answer
$$\dfrac{-4+3\pi}{18}$$
Work Step by Step
Our aim is to integrate the integral as follows:
$$ Volume =\int^{\pi/2}_{0} \int^{sin\theta}_0 \int^{\sqrt{1-r^2}}_0$ \space dz r \space dr \space d\theta \\= \int^{\pi/2}_{0} \int^{sin\theta}_0 r\sqrt{1-r^2} \space dr \space d\theta \\=-\dfrac{1}{3}\int^{\pi/2}_{0} [(1-sin^2 \theta)^{3/2}-1] \space d\theta \\=-\dfrac{1}{3} \times \int^{\pi/2}_{0} (\cos^3 \theta-1)\space d\theta \\=-\dfrac{1}{3} \times ([\dfrac{\cos^2\theta \sin \theta}{3}]^{\pi/2}_0+\dfrac{2}{3} \times \int^{\pi/2}_{0} \space \cos\theta \space d\theta)+[\dfrac{\theta}{3}]^{\pi/2}_0 \\=-\dfrac{2}{9}[\sin \space \theta]^{\pi/2}_0+\dfrac{\pi}{6} \\=\dfrac{-4+3\pi}{18}$$