Calculating flux via divergence theorem.

[dustbin]

Homework Statement

Compute the flux of \vec{F} through z=e^{1-r^2} where \vec{F} = [x,y,2-2z]^T and r=\sqrt{x^2+y^2}.

EDIT: the curve must satisfy z\geq 0.

Homework Equations

Divergence theorem: \iint\limits_{\partial X} \Phi_{\vec{F}} = \iiint\limits_X \nabla\cdot\vec{F}\,dx\,dy\,dz

The Attempt at a Solution

For the given \vec{F}, we have \nabla\cdot\vec{F} = 0. So isn't the flux just zero by the divergence theorem? I am confused because there is a hint saying that I should change the given surface to a simpler one.

 

[tiny-tim]

hi dustbin!

For the given \vec{F}, we have \nabla\cdot\vec{F} = 0. So isn't the flux just zero by the divergence theorem?

correct

but that's only for a closed surface …

so find another (simpler) surface that you can join to this surface to make a closed surface

 

[dustbin]

Thanks for the tip tiny-tim! I should note that I forgot to put the restriction z\geq 1 on the given surface. I will think about your suggestion and post back!

 

[dustbin]

Since X must be a compact domain in \mathbb{R}^3, we must bound from below the region bounded above by the given surface. Since z\geq 1, setting 1=e^{1-r^2} gives the (simpler) surface x^2+y^2=1. The union of this disc and the given surface form the boundary, \partial X, of a compact region X of \mathbb{R}^3. Hence we may now apply the divergence theorem.

 

[tiny-tim]

hi dustbin!

yes, i think that's right

(except that I'm not sure which surface you mean … x²+y² = 1 is a cylinder )

 

[dustbin]

I meant x^2+y^2 = 1 to be confined to the plane z=1. Thank you!

To take it all the way:

Call S_1 the given surface and S_2 the new surface so that S_1\cup S_2 = \partial X. Observe that \Phi_{\vec{F}} = x\,dy\wedge dz - y\,dx\wedge dz + (2-2z)\,dx\wedge dy. Parametrize S_2 via x=r\cos\theta, \ y=r\sin\theta, \ z=1 such that \theta \in [0,2\pi) \ , \ 0\leq r \leq 1. Call this parametrization \gamma. Then \Phi_{\vec{F}}(\gamma) = (2-2)r = 0. Applying the theorem, we have

<br /> 0 = \iiint\limits_X \nabla\cdot\vec{F}\,dx\,dy\,dz = \iint\limits_{S_1} \Phi_{\vec{F}} + \iint\limits_{S_2} \Phi_{\vec{F}} = \iint\limits_{S_1} \Phi_{\vec{F}}.<br />

Thus the flux of \vec{F} across S_1 is 0.

 

[tiny-tim]

that's very complicated

isn't it simpler to keep to x,y,z. and say that F on the surface is (x,y,0)^T, and so … ?

 

[dustbin]

The way that I did it is the only way I know to calculate flux.

 

[tiny-tim]

The way that I did it is the only way I know to calculate flux.

flux is just F·ñ …

why do you need to convert to polar coordinates to calculate what (on this surface) is obviously 0 ??

 

[dustbin]

Is \hat{\textbf{n}} the orienting normal?

 

[tiny-tim]

yüp!

 

[dustbin]

I see. The text I use doesn't use standard notation, so when I look at other books or sites about vector calculus, I feel hopelessly lost with the notation!

Thank you for your help :-)