EDIT :

V = ???dV =

[(pSin

^{2}(Phi)] dpd(Phi)d?

For an entire sphere centered at (0,0,0) around the z-axis :

0 ? p ? R

0 ? ? ? 2(Pi)

0 ? Phi ? Pi

I assume you know this already.

For a portion of a sphere, however, Phi & p alone change. You can graph the sphere and see this yourself. Notice : Phi changes from 0 to an angle , call it 'a'.

z = pCos(Phi)

Suppose the sphere is cut at z = h , then h = pCos(Phi). Where h = height of the portion of a sphere with respect to the origin along the z-axis.

p = h / Cos(Phi) , this will make the integration a bit more difficult but will give a precise volume of the sphere.

The Variables become :

0 ? p ? h / Cos(Phi)

0 ? ? ? 2(Pi)

0 ? Phi ? a

V =

[(p

^{2}Sin(Phi)] dpd(Phi)d? = 1/3 ?? [R

^{3} - ( h

^{3} / Cos

^{3}(Phi) ](Sin(Phi)) d(Phi)d?

0 ? Phi ? a

V = - ? 1/3 [ R

^{3}Cos(a) + h

^{3} tan

^{2}(a)/2 - R

^{3} ] d?

Cos(a) = h/R and tan

^{2}a = ( 1 - Cos

^{2}a / Cos

^{2}a ) = ( R

^{2} - h

^{2} ) / h

^{2}V = - ? 1/3 [ R

^{2}h + h(R

^{2} - h

^{2}) - R

^{3} ] d?

Integrate w.r.t ? : 0 ? ? ? 2(Pi)

V = 2(Pi)/3 [ R

^{2}(R - h) + h( h

^{2} - R

^{2} ) / 2 ]

This is the gen eq for the volume of a portion of a sphere cut by a plane perpendicular to z. You can manipulate it the way u want. All u have to do is to plug in the value of "h" and radius "R".