Volume inside sphere bounded by plane in double integrals

Question

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.

Answer 1

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!

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.

Answer 2

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.
$$\int_0^c\pi(9 - z^2)dz = \pi(9c - \frac{c^3}{3})$$