March 22nd, 2016, 07:43 AM  #1 
Newbie Joined: Mar 2016 From: GĂ¤vle 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 $ $\displaystyle 2<= z <=2$ $\displaystyle F(x,y,z) = (x, 2y, 13z)$ 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  
Senior Member Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 Math Focus: Electrical Engineering Applications 
Hi 9robban6, and welcome to the forum. Quote:
Quote:
 First, we know that the total flow through the cylinder is 0 as mentioned above.  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: $\displaystyle 20 \pi$  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: $\displaystyle 28\pi$  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 $\displaystyle \text{Flow} \ = \ +48 \pi$  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?  

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calculus, divergence, flow, integral, vectors 
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