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
 A: 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. 
$$V=\frac{\pi}{4}\left(\displaystyle\int_{0}^{\sqrt{2}/2} z^2 \,\mathrm{d}z+\displaystyle\int_{\sqrt{2}/2}^1 \left(1-z^2\right)\,\mathrm{d}z\right)$$
$$V=\frac{\pi}{4}\left(1-\frac{\sqrt{2}}{2}+ \displaystyle\int_{0}^{\sqrt{2}/2} z^2 \,\mathrm{d}z - \displaystyle\int_{\sqrt{2}/2}^1 z^2\,\mathrm{d}z\right)$$
$$V=\frac{\pi}{4}\left(1-\frac{\sqrt{2}}{2}+ \frac{1}{3}\left(\frac{\sqrt{2}}{2}\right)^3  - \frac{1}{3}+\frac{1}{3}\left(\frac{\sqrt{2}}{2}\right)^3\right)$$
$$V=\frac{\pi}{4}\left(\frac{2}{3}-\frac{\sqrt{2}}{2}+ \frac{2}{3}\left(\frac{2\sqrt{2}}{8}\right)  \right)$$
$$V=\frac{\pi}{4}\left(\frac{2}{3}-\frac{\sqrt{2}}{3}  \right)$$
$$V= \boxed{\frac{2\pi-\pi\sqrt{2}}{12}}$$
A: Dividing into a cone and a spherical cap,
The height of the cone is $\frac{\sqrt2}{2}$, because the side angle of the cone is $\frac{\pi}{4}$ (as $y=0$ gives $z=x$ in the 1st equation).
The radius of the bottom of the cone is also $\frac{\sqrt2}{2}$.
$$\text{Volume of the cone}=\frac{\pi\left(\frac{\sqrt2}{2}\right)^3}{3}=\frac{\sqrt2 \pi}{12}$$
For the volume of the spherical cap, refer to https://en.wikipedia.org/wiki/Spherical_cap
And use $a=\frac{\sqrt2}{2},h=1-a$
$$\text{Volume of the spherical cap}=\frac{\pi h}{6}(3a^2+h^2)=\frac{\pi}{6}\left(1-\frac{\sqrt2}{2}\right)\left(\frac{3}{2}+\left(1-\frac{\sqrt2}{2}\right)^2\right)=...$$
A: Using Cylindrical coordinates:
$\displaystyle\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{1}{\sqrt{2}}}\int_{r}^{\sqrt{1-r^2}}rdzdrd\theta=\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{1}{\sqrt{2}}}{r(\sqrt{1-r^2}-r)}drd\theta=\int_{0}^{\frac{\pi}{2}}[-\frac{1}{3}(1-r^2)^{\frac{3}{2}}-\frac{r^3}{3}]_{0}^{\frac{1}{\sqrt{2}}}=\frac{\pi}{2}[(-\frac{1}{3(2\sqrt{2})}-\frac{1}{6\sqrt{2}})-(\frac{1}{3})]=\frac{\pi}{2}[\frac{-2\sqrt{2}}{12}+\frac{1}{3}]=\frac{\pi(2-\sqrt{2})}{12}$
