Flux through a paraboloid? The Divergence Theorem and Integration Error

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The discussion centers on calculating the flux through a paraboloid using the divergence theorem, with a specific velocity field provided. The user initially calculates the divergence of the velocity field and sets up the integral for flux, but encounters a discrepancy between their result of -11000000pi and the expected 10000pi. A key point raised is the need to account for the flux through the open end of the paraboloid, which may contribute to the negative result. Additionally, there is confusion about obtaining a negative value from an integral that should yield a positive result. Clarification on the integration steps and the treatment of the open surface is requested to resolve the issue.
Raen
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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 = \int divF dV



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

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!
 
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Hi Raen! :smile:

(try using the X2 icon just above the Reply box :wink:)
Raen said:
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:
 
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.
 
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:
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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