Volume between cone and sphere - First octant Find the volume between $z=\sqrt{x^2+y^2}$ and the sphere $x^2+y^2+z^2=1$ that lies in the first octant using cylindrical coordinates.
So I found the intersection and got $r=\frac{\sqrt2}{2}$.
I know theta has to be between $0$ and $\frac{\pi}{2}$ but not sure about z

We can do this as a single variable integration, along the $z$-axis. We integrate just the cone from $z=0$ to $z=\sqrt{2}/2$ and then just the sphere from $z=\sqrt{2}/2$ to $z=1$, because in those ranges the region is simply the part of the cone and the part of the sphere, respectively. We finally divide by $4$ because we are only interested in the first octant (which is $1$ of the $4$ octants with positive $z$ coordinate). 
$$V = \frac{1}{4}\left(\displaystyle\int_{0}^{\sqrt{2}/2} \pi z^2 \,\mathrm{d}z+\displaystyle\int_{\sqrt{2}/2}^1 \pi\left(1-z^2\right)\,\mathrm{d}z\right)$$
So the problem is now a straightforward single-variable integral. 
$$V=\frac{\pi}{4}\left(\displaystyle\int_{0}^{\sqrt{2}/2} z^2 \,\mathrm{d}z+\displaystyle\int_{\sqrt{2}/2}^1 \left(1-z^2\right)\,\mathrm{d}z\right)$$
$$V=\frac{\pi}{4}\left(1-\frac{\sqrt{2}}{2}+ \displaystyle\int_{0}^{\sqrt{2}/2} z^2 \,\mathrm{d}z - \displaystyle\int_{\sqrt{2}/2}^1 z^2\,\mathrm{d}z\right)$$
$$V=\frac{\pi}{4}\left(1-\frac{\sqrt{2}}{2}+ \frac{1}{3}\left(\frac{\sqrt{2}}{2}\right)^3  - \frac{1}{3}+\frac{1}{3}\left(\frac{\sqrt{2}}{2}\right)^3\right)$$
$$V=\frac{\pi}{4}\left(\frac{2}{3}-\frac{\sqrt{2}}{2}+ \frac{2}{3}\left(\frac{2\sqrt{2}}{8}\right)  \right)$$
$$V=\frac{\pi}{4}\left(\frac{2}{3}-\frac{\sqrt{2}}{3}  \right)$$
$$V= \boxed{\frac{2\pi-\pi\sqrt{2}}{12}}$$
