Deriving the formula for a partially filled sphere using spherical polars.

[Azelketh]

Homework Statement

Deriving the formula for a partially filled sphere using spherical polars. Note this is not a homework problem , I have solved it using a cylindrical slice method, just been bugging me about how to obtain the same result using spherical polars.if the sphere has a radius a, and is intersected by a plane where its distance from the centre of the sphere is h.
I am interested in the volume of the smaller of the regions bounded by the plane at distance h from the origin.

Homework Equations

Now the volume element of a sphere in spherical polars is given by;
$$dv=r^2 \sin( \theta ) d\theta d\psi dr$$

giving volume as;

$$V= \iiint r^2 \sin( \theta ) d\theta d\psi dr$$

The Attempt at a Solution

now as far as I have got is setting the limits to
$$V= \int_{0}^{\cos(h/a)^-1} \int_{0}^{2\pi} \int_{0}^{a} r^2 \sin( \theta ) d\theta d\psi dr =2\pi (a^3 - h a^2 )$$
I obtained the $$\theta$$ limit from the hastily drawn diagram here:
http://img252.imageshack.us/img252/5101/hastilydrawnpaintdiagra.png

but this integral gives the volume of a conical slice.

so i think my question here is can anyone help with the setting of the limits of the integral to obtain the requiered volume?

 

[kurt.physics]

Thanks for any helpThe solution: The volume element of a sphere in spherical polars is given by; dv=r^2 \sin( \theta ) d\theta d\psi dr giving volume as; V= \iiint r^2 \sin( \theta ) d\theta d\psi dr V= \int_{0}^{\cos^{-1}(h/a) } \int_{0}^{2\pi} \int_{h}^{a} r^2 \sin( \theta ) d\theta d\psi dr =2\pi (a^3 - h a^2 )