# Calculate the Flux through the Pipe

1. Dec 29, 2008

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

Water flows with a speed v down a rectangular pipe of dimensions s and l as shown. What is the rate $\Phi=volume\, per\, unit\, time$
at which water accumulates in the bucket?

2. Relevant equations
$\Phi=\int_S\mathbf{v}\cdot d\mathbf{a}$

3. The attempt at a solution

I am confused as to how to compute this integral. I do not see how could use the divergence theorem to simplify it since I only have the dimensions of the 2-dim surface of the rectangular pipe.

I am not sure how to evaluate the surface integral based solely on its definition. Can someone help to get me started?

Do I just use the definition $\Phi=\int_S\mathbf{v}\cdot d\mathbf{a}$ for this?

2. Dec 29, 2008

Any suggestions would be appreciated.

3. Dec 29, 2008

### Dick

Sure, flux is integral v.dA. If the velocity is uniform that's just v*A. Isn't it? Why do you want to use the divergence theorem? This problem sure looks simple to me. Is this some kind of a trick?

4. Dec 29, 2008

I just don't know anymore :sad: I think I am getting stupider and stupider.

I don't know what he is looking for (Griffith's Intro to Electrodynamics problem 1.32 2nd ed.).

I don't want to just memorize formulas. I want to compute the integrals. But I get all effed up when I get a surface integral no matter how simple. If I have to use the definition, I get all flustered.

How do I compute the integral?

I get,

$\phi=\int_S v \cdot da$
$=\int_S v*da\cos\theta$
$=\int_S v*da$
$=\int_s\int_l v*(dsdl)=v*s*l$

is that right?

The reason I want to do this right is that the next three parts of the problem have angles and \$hit that I'll and non-uniform velocities that I will have to deal with.

5. Dec 29, 2008

### Dick

Yes, that's right. It's v*A=v*s*l. cos(theta)=1.