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

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

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 :-)