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