Flux through a paraboloid?

  • Thread starter Raen
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  • #1
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The Problem: I have a paraboloid open along the positive z-axis, starting at the origin and ending at z = 100. At z=100, the horizontal surface is a circle with a radius of 20. Water is flowing through the paraboloid with the velocity F = 2xzi - (1100 + xe^-x^2)j + z(1100 - z)k. I'm asked to find the flux through the paraboloid using the divergence theorem.


Equations: divF = dF1/dx + dF2/dy + dF3/dz
Flux = [tex]\int divF dV[/tex]



My attempt: I started by finding the divergence.

2z + 0 + (1100 - 2z) = 1100

Next, I found the equation for the paraboloid.

z = r^2/x
100 = 20^2/x
100 = 400/4
z = r^2/4

Then I iterated the integral.

Flux = [tex]\int^{2\pi}_{0} \int^{20}_{0} \int^{100}_{r^2/4} 1100r dz dr d\theta[/tex]

When I solve the integral, I end up with -11000000pi, but the answer is supposed to be 10000pi. Where am I going wrong?

If I have it correct thus far and my problem is in my integration, please let me know and I'll type out the integration step by step, as well. Thank you!
 
Last edited:

Answers and Replies

  • #2
tiny-tim
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Hi Raen! :smile:

(try using the X2 icon just above the Reply box :wink:)
The Problem: I have a paraboloid open along the positive z-axis, …

(i haven't checked your figures, but …)

The paraboloid is open, so you'll need to subtract the flux through the open end. :wink:
 
  • #3
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The flux through the top is positive, so subtracting the flux through the open circle from what I already had gives me an even larger negative number, -51000000pi.
 
  • #4
tiny-tim
Science Advisor
Homework Helper
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oooh, I didn't notice that minus sign! :redface:

but how did you manage to get a negative result from integrating something that's everywhere positive? :confused:
 

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