Find the Volume of a solid block bounded by $z^2=x^2+y^2$ and $x^2+y^2+z^2=1$

Question

I'm wondering whether I did it correctly. I see it as a region bounded from above by cone and by a sphere. I used polar form to make it easier to calculate: $$2 \int_{0}^{2\pi}d\theta \int_{0}^{\frac{\sqrt{2}}{2}}dr \int_{r}^{\sqrt{1-r^2}}r\ dz.$$

Answer 1

The two surfaces, the double cone $z^2=x^2+y^2$ and the sphere $x^2+y^2+z^2=1$, divide space into two bounded regions inside the sphere: the one inside the double cone $$V_i=2\int_{r=0}^{1}\int_{\theta=0}^{2\pi}\int_{\phi=0}^{\frac{\pi}{4}}r^{2}\sin(\phi ) d\phi d\theta dr=\frac{4\pi-2\sqrt{2}\pi}{3}$$ and the one outside the double cone $$V_o=\int_{r=0}^{1}\int_{\theta=0}^{2\pi}\int_{\phi=\frac{\pi}{4}}^{\frac{3\pi}{4}}r^{2}\sin(\phi ) d\phi d\theta dr=\frac{2\sqrt{2}\pi}{3}.$$