Finding the volume between sphere and hyperboloid

Question

I am trying to find the volume between the sphere: $x^2+y^2+z^2=9$ and the hyperboloid $x^2+y^2-z^2=1$. I set the integral as: $$\int_{0}^{2\pi}\int_{-3}^{3}\int_{\sqrt{1+z^2}}^{\sqrt{9-z^2}}rdrdzd\theta$$ but it does not give me the correct answer. Can somebody tell me what is wrong with my calculations?

Answer 1

Your parametrization of the solid is wrong. Notice that $\sqrt{z^2+1}<\sqrt{9-z^2}$ when $-2<z< 2$, but $\sqrt{z^2+1}>\sqrt{9-z^2}$ when $|z|>2$.

When $-2\leq z\leq 2$, the part of the solid looks like a pipe and it is parametrized as $-2\leq z\leq 2$, $0\leq r\leq\sqrt{z^2+1}$, $0\leq\theta\leq 2\pi$. We can find the volume of this part by cross-section method: $V_1=\int_{-2}^2A(z)dz=2\int_0^2\pi r^2dz=2\pi\int_0^2(z^2+1)dz=\frac{28}3\pi$.

When $2\leq z\leq 3$, the part of the solid is a spherical cap parametrized by $2\leq z\leq\sqrt{9-r^2}$, $0\leq r\leq\sqrt 5$, $0\leq\theta\leq2\pi$. Hence, its volume is $V_2=\int_0^{2\pi}\int_0^{\sqrt5}(\sqrt{9-r^2}-2)rdrd\theta=\frac83\pi$. Bottom cap is also $V_3=\frac83\pi$. The total volume $V=V_1+V_2+V_3=\frac{28+8+8}{3}\pi=\frac{44}3\pi.$