# Calculation of particles by integral

1. Mar 12, 2015

### nazmulislam

Hi,

I would like to calculate total number of particles in one periodic wave-section whose radius varies by h(z)=1+A*sin(2*pi*z/L), where A is the profile amplitude and L is the wavelength of one wave-section.
I thought, I will do by volume integration, say integral over V (volume of one wave-section) and the integrand is the concentration C(z,r,t), where z & r are the axial and radial direction respectively, and I want to integrate w.r. to V, i.e., dV.

2. Mar 12, 2015

### Overt

I'm having trouble figuring out what you're trying to calculate. I'm assuming you have a separate function for concentration you haven't listed? C(z,r,t)

In this case, you are correct. C(z,r,t) will be the integrand and you will integrate over the entire wave-section volume. The next step in this is figuring out exactly what is in that infinitesimal element dV. What's a convenient way to express volume given the coordinate system?

3. Mar 12, 2015

### nazmulislam

Thanks for your response, Overt. I want to express dV=dA*dz, where dA is the area of a circle and dz is the step size in z-direction. Now, the question is, how can I express dA? Will it be dA=2*pi*r*dr or dA=pi*h(z)^2, where h(z) is the tube radius at z?

4. Mar 12, 2015

### Staff: Mentor

The correct formula would be dv = A * dz. In other words, it wouldn't be dA * dz. Based on what you have said, the cross-sectional area A would be $\pi (h(z))^2$.

5. Mar 12, 2015

### nazmulislam

Thanks, Mark.

6. Mar 12, 2015

### nazmulislam

So, finally the integral will come into the form \int^{}_{V} C(z,r,t) dV=\pi \int^{}_{z} C(z,r,t) h^2(z) dz.