Volume of Frustum Using Triple Integration

1. Nov 3, 2012

Xishem

Volume of Frustum Using Triple Integral [Solved]

1. The problem statement, all variables and given/known data

Edit: I've solved the issue! My limits of r were wrong. Instead of this:
$$V=\int_{z=0}^{\frac{h}{2}}\int_{r=0}^{\frac{R}{h}z+R}\int_{\theta=0}^{2\pi}r\ d\theta\ dr\ dz$$It should have been this:
$$V=\int_{z=0}^{\frac{h}{2}}\int_{r=0}^{\frac{-R}{h}z+R}\int_{\theta=0}^{2\pi}r\ d\theta\ dr\ dz$$

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.

The semi-specific frustum I'm looking at has the following properties:

$height = \frac{h}{2}$
$radius_{bottom} = R$
$radius_{top}=\frac{R}{2}$

2. Relevant equations

$$dV=r\ dr\ d\theta\ dz$$

3. The attempt at a solution

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:

$V=\int\int\int\ r\ dr\ d\theta\ dz$

And my attempt at putting in limits is...

$$V=\int_{z=0}^{\frac{h}{2}}\int_{r=0}^{\frac{R}{h}z+R}\int_{\theta=0}^{2\pi}r\ d\theta\ dr\ dz$$

The limits of r are derived from the slope of one of the sides of the frustum projected onto a 2D plane.

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. The answer I get is correct in terms of variables, but incorrect in terms of a constant factor, if I remember correctly.

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

Last edited: Nov 3, 2012