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 March 22nd, 2016, 07:43 AM #1 Newbie   Joined: Mar 2016 From: Gävle Sweden Posts: 1 Thanks: 0 Divergence & flow integral Hi! I got this task: S = the mantle area of the cylinder. F = the vector field. $\displaystyle x^2+y^2=4$ $\displaystyle -2<= z <=2$ $\displaystyle F(x,y,z) = (x, 2y, 1-3z)$ As div(F) = 0, shouldn't the flow through the mantle area S also 0 ? Because in one solution I got, which I don't really believe in, the take: flow(total cylinder) - flow(top part cylinder) - flow(bottom part cylinder) = flow(mantle cylinder) I get why they subtract like that, but shouldnt it be zero as div(F) = 0 ? Last edited by 9robban6; March 22nd, 2016 at 07:47 AM. March 26th, 2016, 07:53 PM   #2
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Hi 9robban6, and welcome to the forum.

Quote:
 As div(F) = 0, shouldn't the flow through the mantle area S also 0 ?
No, I don't think so. I think that div(F)=0 means that the flow through the whole cylinder is zero (this follows from the Divergence Theorem).

Quote:
 Because in one solution I got, which I don't really believe in, the take: flow(total cylinder) - flow(top part cylinder) - flow(bottom part cylinder) = flow(mantle cylinder)
Let's calculate each term.

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First, we know that the total flow through the cylinder is 0 as mentioned above.

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Next, for the flow through the top of the cylinder, the x and the y components of F do not play a role since they are in a direction perpendicular to the normal of the surface. The z component is obviously parallel to the normal so in the dot product, cos(angle) = 1 (you probably know all of this but I thought I would state it to be sure).

So on the top of the cylinder z=2 (constant) and F(z)=-5. The area is $\pi \left(2^2 \right) \ = \ 4\pi$. The normal is in the +z direction so the flow is:

$\displaystyle -20 \pi$

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For the bottom part of the cylinder, again the x and y components do not play a role. Here, z=-2 (constant), so F(z)=7. The area is $4\pi$ and the normal is in the -z direction. So the flow is:

$\displaystyle -28\pi$

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For the mantle of the cylinder, the z component of F does not play a role. Also, I found it easier to convert to cylindrical components using $\rho$ and $\phi$ (at least I think the following is correct; someone may correct me if my procedure is wrong).

$\text{x} \ = \ \rho \cos(\phi)$ and $\text{y} \ = \ \rho \sin(\phi)$, and the normal is in the $+\phi$ direction, so the angle between the x component and the normal is $\cos(\phi)$ and the angle between the y component and the normal is $\sin(\phi)$. With $A$ as the area this gives the flow as:

$\displaystyle \text{Flow(x)}=\rho \cos^2(\phi) \ dA \qquad \text{and} \qquad \text{Flow(y)}=2\rho \sin^2(\phi) \ dA$

$\rho \ = \ 2$ (constant), so the flow is given by:

$\displaystyle \text{Flow} \ = \int_0^{2\pi}2\cos^2(\phi) \ dA + \int_0^{2\pi}4\sin^2(\phi) \ dA$ where

$\displaystyle dA \ = \Delta z \ \rho \ d \phi = 8 d \phi$ giving

$\displaystyle \text{Flow} \ = \int_0^{2\pi}16\cos^2(\phi) \ d \phi + \int_0^{2\pi}32\sin^2(\phi) \ d \phi$

Since

$\displaystyle \int_0^{2\pi} \cos^2(\phi) \ d \phi \ = \ \int_0^{2\pi} \sin^2(\phi) \ d \phi \ = \ \pi$ the flow is

$\displaystyle \text{Flow} \ = \ +48 \pi$

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So the flow equation becomes:

$\displaystyle 0-(-20 \pi)-(-28 \pi) \ = \ 48 \pi$ which holds per the Divergence Theorem.

In the solution that you referenced, was the flow $48 \pi$ for the mantle area? Tags calculus, divergence, flow, integral, vectors Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post KJ17 Calculus 1 October 5th, 2015 05:04 PM james777 Calculus 2 December 2nd, 2014 08:47 AM alex369 Math Software 1 July 17th, 2013 07:44 AM leroy Calculus 3 November 25th, 2011 11:26 AM OSearcy4 Calculus 7 November 6th, 2009 02:53 AM

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