Find the volume of a body formed by a cylinder $x^2 + y^2 = 9$ and a hyperboloid $x^2 + y^2 + 9 = z^2$
My solution
Let's try to build the body data
When viewed from above, it will look like a circle with a radius of 3 and a center at the point $x = 0, y = 0$. The volume will be the same at the top and bottom due to symmetry, so we can only consider the part where $0 \leq z$. Let's move on to cylindrical coordinates using the replacement: $x = r\cos(\phi), y = r\sin(\phi), z = z$. Let's find the limits of integration. $$0 \leq \phi \leq 2\pi$$ $$0 \leq r \leq 3$$ A $z$ will be bounded from above by a hyperboloid $x^2 + y^2 + 9 = z^2$ $$r^2\cos^2(\phi) + r^2\sin^2(\phi) + 9 = z^2$$ $$r^2 + 9 = z^2$$ at the very beginning we chose the part where $0 \leq z$, therefore $$\sqrt{r^2 + 9} = z$$ As a result, we obtain the integral: $$\int_{0}^{2\pi}d\phi\int_{0}^{3}rdr\int_{0}^{\sqrt{r^2 + 9}}dz$$
And it needs to be counted. Please tell me if I did all this correctly? And at the end you will need to multiply by 2 to get the volume of the upper and lower parts
You need to solve for $z$, not for $z^2$, therefore I believe it should be $z=\sqrt{9+r^{2}}.$ Rest I believe is alright.