# The volume bounded by a sphere and a plane

If I want to find the volume bounded below by the sphere $$x^2+y^2+z^2=9$$ and above by the plane z=1 .

Can we solve it as following?

$$V=\int\int_R\int_{z=-\sqrt{9-x^2-y^2}}^{z=1}dzdA$$ $$V=\int\int_R (1+\sqrt{9-x^2-y^2})dA$$ where R is the projection in xy plane of the intersection of the sphere and the plane z=1. $$V=\int_{\theta=0}^{2\pi}\int_{r=0}^{r=2\sqrt{2}}(1+\sqrt{9-r^2})rdrd\theta$$ If not , why ?

Note: I have a second solution :

Get the volume bounded below by the plane z=1 and above by the sphere , then subtract this volume from the volume of the sphere.

If you have any other solutions , I will be grateful if you provide them.

The problem is that your answer only includes the portion of the volume that lies within the cylinder $x^2+y^2=8$ so you need an extra piece if your volume is what I think it is from the description. I get \begin{align}V &= \int_{0}^{2\pi}\int_0^{2\sqrt2}\int_{-\sqrt{9-r^2}}^1dz\ rdrd\theta+\int_{0}^{2\pi}\int_{2\sqrt2}^3\int_{-\sqrt{9-r^2}}^{\sqrt{9-r^2}}dz\ rdrd\theta \\ & =2\pi\int_0^{2\sqrt2}\left(1+\sqrt{9-r^2}\right)rdr+2\pi\int_0^{2\sqrt2}2\sqrt{9-r^2}rdr \\ & =2\pi\left[\frac12r^2-\frac13\left(9-r^2\right)\right]_0^{2\sqrt2}+2\pi\left[-\frac23\left(9-r^2\right)\right]_{2\sqrt2}^3 \\ & =\frac{80\pi}3\end{align} As can be confirmed by fixing up the limits in @José Carlos Santos middle integral.
I don't understand. $R$ is the projection of what?
I would comput it in cylindrical coordinates:$$\int_0^{2\pi}\int_1^3\int_0^{\sqrt{9-z^2}}\rho\,\mathrm d\rho\,\mathrm dz\,\mathrm d\theta.$$
• Where did the $\rho$ come from? Why not $z,r,\theta$? – K Split X Dec 13 '17 at 22:59
• @KSplitX It's just the notation that I am used to. Use $r$ instead of $\rho$, if you wish. – José Carlos Santos Dec 13 '17 at 23:01