Help with Integral (Blended Cylinder)

[minger]

Homework Statement

This problem goes into a paper I’m writing. The exact problem deals with acoustic radiation from a plunging piston. Because of the CFD code I’m using, I can’t have a discontinuous grid at the piston boundary, so I must blend it. The results I’m getting at slightly off and I think it’s due to the effective volume of the displaced cylinder. Unfortunately the blending region that I’m using is making it tricky for me to integrate.

Homework Equations

The volume of a cylinder can be give by:
$$V = \int_0^r \int_0^{2\pi} \int_0^H r\,dr$$
I know the volume of the actual piston and only want the volume of the blending region, from r0 to r’. The blending region is given by the function:
$$f(r) =1.0-3u^2 + 2u^3$$
Where:
$$u = \frac{r-r_0}{r_1-r_0}$$
My final integral is then:
$$V = \int_0^{2\pi} \int_{r_0}^{r_1} \left( 1-3\left(\frac{r-r_0}{r_1-r_0}\right)^2 + 2\left(\frac{r-r_0}{r_1-r_0}\right)^3 \right)r\,dr$$
Hope I typed that right.

The Attempt at a Solution

Without the additional r on the far right, this is a simple substitution where:
$$dr = (r_1-r_0)du$$
I eventually end up with:
$$V = \pi\left(r_1-r_0\right)$$

However, with the additional r term, I cannot seem to figure out how to either use substitution or approach this. I would like to have the derivation of this for an appendix without simply putting:
http://integrals.wolfram.com/index.jsp?expr=%28+1+-+3*%28%28x-b1%29%2F%28b2-b1%29%29^2+%2B+2*%28%28x-b1%29%2F%28b2-b1%29%29^3+%29+*x&random=false

If I could get some validation on my integral and then point me in the right direction, I would greatly appreciate it.

p.s. This isn’t homework per say, so feel free to move it to the general math section if needed (figured I’d get more hits here).

 

[minger]

OK, I think I got it figured out; I need to use both substitution and integration by parts. If the blending term is f(r ) and r is g(r ), then the volume is
$$V = 2\pi\left[ g(r )\int f(r ) – g'(r )\iint f(r )\right]$$
Letting
$$\begin{equation}<br /> \begin{split}<br /> r^* = r_2-r_1 \\<br /> u = \frac{r-r_1}{r_2-r_2}<br /> \end{split}<br /> \end{equation}$$
Then the solution is:
$$V = 2\pi\left[ rr^*\left(u-u^3+\frac{1}{2}u^4\right)-r^*^2\left(\frac{1}{2}u^2-\frac{1}{4}u^4+\frac{1}{10}u^5\right)\right]$$
In my case, my limits are from r1 to r2.
$$\begin{equation}<br /> \begin{split}<br /> r=r_2\rightarrow\,u=1 \\<br /> r=r_1\rightarrow\,u=0 <br /> \end{split}<br /> \end{equation}$$
This means that most of the damn thing cancels at r1. Since the only r term is “alone” and linear I can just substitute the limits in there and reduce to:
$$V = 2\pi\left[\frac{13}{20}\left(r_2-r_1\right)^2\right]$$
It should be right, thanks for anyone who considered the problem.