# Finding the volume of a solid inside $x^2+y^2+z^2=49$ and upper nappe of the cone $z^2=x^2+y^2$

I'm having a problem in setting the integral bounds in the questions which is

Find the volume of a solid inside $x^2 + y^2 + z^2 = 49$ and upper nappe of a cone $z^2 = x^2 + y^2$.

I see a lot of similar types of problems using spherical coordinates, but I have now idea how to approach this using the cylindrical coordinates - which I have to.

Can I get some help here? Thank you!

From your equations it's clear the cone is below the sphere, so you will vary $z$ from the cone's $z = \sqrt{x^2+y^2}$ to the sphere's $z = \sqrt{49-x^2-y^2}$.
The question to answer is what is the intersection of the curves, projected onto the $xy$-plane so we can integrate over the region it bounds? To find the intersection note that we have the same point $(x,y,z)$ on both curves, which yields $$x^2+y^2 = z^2 = 49 - x^2 - y^2$$ and equating the left and the right sides, we get $$x^2+y^2 = 49 - x^2 - y^2\\ 2x^2 + 2y^2 = 49\\ x^2 + y^2 = \left(\frac{7\sqrt2}{2}\right)^2,$$ wich is a familiar circle of radius $7\sqrt{2}/2 = 7/\sqrt{2}$. To set this up in cylindrical coordinates, $z$ will be left as is, and we will need to integrate over that disc in the $xy$-plane, so the radius will vary from $0 \le r \le 7/\sqrt{2}$ and the angle will be $0 \le \theta \le 2\pi$.
Last thing to do is convert the above bounds on $z$ using $r^2=x^2+y^2$, and noticing that $\sqrt{r^2} = |r| = r$ (last equality since $r\ge 0$), so we end up with $$\begin{split} V &= \int_0^{2\pi} \int_0^{7/\sqrt{2}}\int_{|r|}^\sqrt{49-r^2} rdzdrd\theta \\ &= 2\pi \int_0^{7/\sqrt2} \left[ \sqrt{49-r^2} - r\right] r dr \end{split}$$ Can you take it from here?