# Making Flux Negative in a Constant Vector Field Help!

1. Jun 16, 2012

### emzee1

1. The problem statement, all variables and given/known data I know I posted a question yesterday also, but this homework is getting on my nerves. My prof isnt the best out there. So basically the question is as follows:

http://img812.imageshack.us/img812/8130/problem6.png [Broken]

2. Relevant equations
Not sure, I sort of answered the questions with guessing/intuition

3. The attempt at a solution
Guessed on the attempt, I know 1 is correct but not sure which one. Any help on this is much appreciated!

Last edited by a moderator: May 6, 2017
2. Jun 16, 2012

### algebrat

There are several ways to do this.
1. Relate this surface flux to the flat surface flux in the x-y plane, using divergence theorem
2. If this vector field is a curl, globally, than you can use any surface with the same boundary curve
3. Intuitively, won't any y-component have to travel symmetrically in and out?
4. You could calculate it in spherical coordinates (or any coordinates maybe?).

Pick one and tell us what you think.

Last edited: Jun 16, 2012
3. Jun 16, 2012

### AlfredVioleta

Since a, b ,c are constants, shouldn't div F = (∂a/∂x)i + (∂b/∂y)i + (∂c/∂z)k = 0 and by Divergence theorem, the surface integral = 0 ?

4. Jun 16, 2012

### LCKurtz

You say "1 is correct". I don't see any "1" and your graphic looks like default choices for a,b, and c. Apparently you can choose drop down menus for the three constants and I suppose the choices are <, >, and =. It doesn't look like any choices have been made, unless you think the answer is all three constants are negative.

Here's what I would suggest. Close the surface by adding the circular disk in the xy plane along the bottom of the hemisphere. Call that surface B. Suppose for the moment that we consider the closed surface $S\cup B$ directed outward. You could apply the divergence theorem to that:$$\iint _S\vec F\cdot d\vec S +\iint_B \vec F\cdot d\vec S = \iiint_V\nabla\cdot \vec F dV =0$$since $\vec F\$is a constant vector. This tells you$$\iint _S\vec F \cdot d\vec S = -\iint_B \vec F \cdot d\vec S = -\iint_B \vec F \cdot (-\hat k) dydx=\iint_B \vec F \cdot (\hat k) dydx$$Now, remembering that this discussion was for the outward normal, which, for $S$ would be directed upwards. Taking that into account, what can you conclude about $\vec F\$ if the result is supposed to be negative?

Note: I was away for lunch while I was typing this and when I got back I see others have already responded so there may be some duplication of effort here.

Last edited: Jun 16, 2012
5. Jun 16, 2012

### emzee1

So, basically, at the end am I just looking at the dot product of F and the vector [0,0,1]?

If that is the case from what I've read in the posts above, wouldnt just the z-component (which is C) have to be less than zero, while the others are just less than or equal to zero?

6. Jun 16, 2012

### LCKurtz

How do you justify your claim that $a$ and $b$ are $\le 0$? And are you sure that $c$ isn't greater than 0?

7. Jun 16, 2012

### emzee1

My main reasoning behind that was since it's a dot product between F and <0,0,1>, technically a and b can be anything, but doesnt c have to be less than 0 to make the integrand negative?

8. Jun 16, 2012

### algebrat

the surface is oriented downward

9. Jun 16, 2012

### emzee1

So the exact opposite of what I just said?

10. Jun 16, 2012

### algebrat

the exact opposite is...?

11. Jun 16, 2012

### emzee1

a and b be anything and c be greater than zero?