Find volume of sphere $ x^2+y^2+z^2=9$ bounded by planes $z=0$ and $z=2$ using double integral Find volume of sphere $x^2+y^2+z^2=9 $ bounded by planes $z=0$ and $z=2$ using double integral
I tried to take the total volume of the bigger hemisphere but i get zero, i managed to take the volume of the smaller hemisphere only.
 A: We need to use the fact that $x^2+y^2=r^2$, so we can convert to polar coordinates. That is, $x^2+y^2+z^2=9$ implies $z = \pm \sqrt{9-(x^2+y^2)}=\pm\sqrt{9-r^2}$.  
However, since the volume of the solid we want is above $z=0$, we only need to consider the top half of the sphere, that is, $z=+\sqrt{9-r^2}$.
Now here's the trick:
Find the volume of the half-sphere of radius $3$ (and you can figure out why the radius is $3$, and subtract the double integral below (which you should evaluate):
\begin{align}
\int_0^{2\pi} \int_2^{\sqrt{9-r^2}} r \, dr \, d\theta
\end{align}
So we can find the volume:
$$V=\underbrace{\frac{1}{2}\left(\frac{4}{3} \pi r^3 \right)}_{\text{half sphere}} - \underbrace{\int_0^{2\pi} \int_2^{\sqrt{9-r^2}} r \, dr \, d\theta}_{\text{double integral to subtract}}$$
Basically, we should try to employ as much geometry as possible, to simplify things whenever we can. 
A: $V = \displaystyle \int_{0}^{2\pi} \int_{r=0}^{\sqrt{5}} \int_{z=0}^{\sqrt{9-r^2}} rdzdrd\theta$
I mistyped it.
The downvote was a joke !
