Question

Volume of a conical frustum using a triple integral:

Answer 1

I'm trying to find the volume of a frustum using strictly a triple integration in cylindrical coordinates. I've been able to find the volume through several other methods, but whenever I try to do it using a triple integration, it fails to produce the correct result.

This is what my frustum looks like:
I've tried setting up several integrations, but they have all failed. For me, the difficult part comes in setting the limits of each integral. From what I understand, the general form of the integral will look like this:\[V= {\int\limits\limits}{\int\limits\limits}{\int\limits\limits}r\ dr\ d\theta\ dz\]And my attempt at putting in limits is...\[V=\int\limits\limits_{z=0}^{\frac{h}{2}}\int\limits_{r=0}^{\frac{R}{h}z+R}\int\limits_{\theta=0}^{2\pi}r\ d\theta\ dr\ dz\]The limits of r are derived from this:Where the function of that line can be described as:\[y=\frac{R}{h}z+R\]And the radius (r) is directly related to y in that \[y=r\]I've tried doing this integration, and I don't know what my final answer is as I've gone through so many iterations of different integrations that things have been blurred. All I know is that the answer is wrong :P.

Any help on the limits, or the integrand in general? Thanks!