Krdytkyu

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

$$int_0^{2pi}int_0^{sqrt{9-c^2}} (c-2sqrt{9-x^2-y^2})r drdtheta$$
? 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 $36pi - int_0^{2pi}int_0^{sqrt{9-c^2}}sqrt{9-x^2-y^2} - cr$ $drdtheta$ and got $pi(18 + 9a - frac{a^3}{3})$ which according to the text book is the right answer.

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
$$4int_0^frac{pi}{2}int_0^3int_0^{operatorname{arcsin}frac{c}{3}}r^2cos{phi}d{phi}drd{theta} = 6{pi}c$$
Then the total volume is
$$pi(9c - frac{c^3}{3})$$
In fact it is easier to use xyz co-ordinates.
$$int_0^cpi(9 - z^2)dz = pi(9c - frac{c^3}{3})$$