# Volume inside sphere bounded by plane in double integrals

I am trying to solve for volume below a plane bounded by a sphere given by

$$x^2+y^2+z^2 = 9$$ below a plane z $$\in$$ [-3,3] using a double integral with polar coordinates. If the plane is given by z=c, should the integral be

$$\int_0^{2\pi}\int_0^{\sqrt{9-c^2}} (c-2\sqrt{9-x^2-y^2})r drd\theta$$ ? I am wondering how to set up the double integral so that it handles the fact that z could be both positive and negative. Thanks in advance!

EDIT: I ended up taking the volume $$36\pi - \int_0^{2\pi}\int_0^{\sqrt{9-c^2}}\sqrt{9-x^2-y^2} - cr$$ $$drd\theta$$ and got $$\pi(18 + 9a - \frac{a^3}{3})$$ which according to the text book is the right answer.

• Would you mind explaining your thought process on this issue? After solving the above integral, I would get an expression in terms of $c$, $x$, and $y$. Commented Nov 20, 2018 at 16:27
• I actually ended up taking the volume of the entire sphere and subtracted the volume of the spherical cap bounded by the part above the plane and below the sphere! Commented Nov 22, 2018 at 8:36

The following is to calculate the volume of a sphere $$x^2+y^2+z^2=9$$ bounded by the planes z=0 and z=c. For polar co-ordinates we have the sphere r=3 and the plane $$rsin{\phi} = c$$. Then the intersection of the plane and the sphere is $$sin{\phi} = \frac{c}{r}$$ It is better to split the volume into 2 parts. The first is the up side down cone with the base at z = c which has a radius of $$\sqrt{9-c^2}$$. The volume is $$\frac{1}{3}\pi(9-c^2)c$$ The second is the volume between the cone and z = 0 which can be obtained by $$4\int_0^\frac{\pi}{2}\int_0^3\int_0^{\operatorname{arcsin}\frac{c}{3}}r^2cos{\phi}d{\phi}drd{\theta} = 6{\pi}c$$ Then the total volume is $$\pi(9c - \frac{c^3}{3})$$ In fact it is easier to use xyz co-ordinates.
$$\int_0^c\pi(9 - z^2)dz = \pi(9c - \frac{c^3}{3})$$
I ended up taking the volume and got $$\pi(18+9a-\frac{a^{3}}{3})$$ which according to the text book is the right answer.