Maximizing Flux: Finding the Optimal Surface for Vector Field F in R^3

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In summary: Thanks for trying though!In summary, the homework statement is trying to find the maximal flux through a surface in R^3. The attempt at a solution is to use the Gauss theorem to find the gradient of the vector field and set it equal to zero, which would give the maximal flux. However, the problem is integrating the gradient since that's the whole point of the question. The solution might be to find the limits of the surface which make the integral of the vector field equal to zero.
  • #1
AriAstronomer
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Homework Statement


Through what closed, oriented surface in R^3 does the vector field F = <(4x + 2zx^3), -y(x^2 + z^2), -(3x^2z^2 + 4y^2z)> have the greatest flux?


Homework Equations


Flux = double int F.ds
Gauss theorem perhaps (double int F.ds = triple int(DivF)dV)

The Attempt at a Solution


So I figured that since they want max flux, go with the gradient and set it equal to zero (if rate of change is 0, must be at a critical point). So Flux = double int F.ds, and taking the gradient we have grad(Flux) = 0. I thought Gauss might make it easier, so plugging in Gauss:
grad(Flux) = grad(triple int(DivF)dV)= 0.
Take the divergence of F and we get (after simplifying): divF = 4-4y^2 - x^2 -z^2. So now we have have:
grad(triple int(4-4y^2 - x^2 -z^2)dV)= 0

Now I'm stuck though, because I'm not sure how to integrate this since that's the whole point of the question. Is there a way to bring the gradient inside or something?
Thanks in advance
 
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  • #2
Hi AriAstronomer! :smile:

(have a grad: ∇ and an integral: ∫ and try using the X2 tag just above the Reply box :wink:)
AriAstronomer said:
divF = 4-4y^2 - x^2 -z^2. So now we have have:
grad(triple int(4-4y^2 - x^2 -z^2)dV)= 0

Now I'm stuck though, because I'm not sure how to integrate this since that's the whole point of the question …

Nooo … not integrating it is the whole point of the question! :biggrin:

You need to maximise ∫∫∫ (4-4y2 - x2 -z2) dx dy dz ¬

think … when will enlarging the boundary increase the integral? :smile:
 
  • #3
Hmm, well by the looks of things, since we have 4 -4y² -x² - z² (i.e. squared terms), it looks like any value of x,y,z is going make this integral decrease. So I guess the only solution is if x=y=z=0, i.e. if the surface was actually just a point at (0,0,0)? Can someone validate this if I'm right?
 
  • #4
Hi AriAstronomer! :smile:

(just got up :zzz: …)
AriAstronomer said:
Hmm, well by the looks of things, since we have 4 -4y² -x² - z² (i.e. squared terms), it looks like any value of x,y,z is going make this integral decrease. So I guess the only solution is if x=y=z=0 …

oh come on :rolleyes:

what about, for example, (1,0,-1)? :wink:
 
  • #5
I don't really follow. Do you mean if I set the limits (1,0,-1), or plug in values x=1, y=0, z=-1. If limits, you'll get 0 since y will be integrated from 0 to 0. If plugging in values, I get ∫∫∫ (2) dx dy dz, but I don't see how that will yield you a greater value than if you plugged in (0,0,0) and got ∫∫∫ (4) dx dy dz.

However, thinking about this a bit more: we are trying to maximize flux, and flux is rate of flow through a surface. Flux seems to decrease as a function of distance (e.g. Electric flux through a surface from a point charge). Flux will increase if your surface encloses more sources of flow though (e.g. more point charges). We took the divergence which finds sources/sinks, and get 4 -4y² -x² - z². Should I then try and find the limits which make 4 -4y² -x² - z² >= 0? All positive values represent positive flux, and I suppose it isn't until 4 -4y² -x² - z²< 0 that flow lines try to find their way back into the volume, and start decreasing the integral. Maybe I'm on a comlpete tangent here though...

Thus the limits would be y:0->1, x=z:0->2. I've tried a few other limits (x=y=z:0->1 and y:0->1, x:0->2, z:0->3) and get less values, so I think this might be right...what do you think?
 
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  • #6
Wait, since these are all squared terms, can't forget about negatives, so I guess the limits would be y:-1-->1, x=z:-2-->2
 
  • #7
AriAstronomer said:
Should I then try and find the limits which make 4 -4y² -x² - z² >= 0?

Yes! :smile:
We took the divergence which finds sources/sinks … …

All positive values represent positive flux, and I suppose it isn't until 4 -4y² -x² - z²< 0 that flow lines try to find their way back into the volume, and start decreasing the integral.

Well that's the correct result, but your reasons are weird …

the reason is simply that divF = 4 -4y² -x² - z², so the region for which ∫∫∫ divF is largest is the region for which 4 -4y² -x² - z² > 0
Thus the limits would be y:0->1, x=z:0->2. I've tried a few other limits (x=y=z:0->1 and y:0->1, x:0->2, z:0->3) and get less values, so I think this might be right...what do you think?

No, it won't be simple limits like that. It might be easier to change to spherical or cylindrical coordinates.
 
  • #8
OK. I converted to spherical co-ordinates and got:
4-4r²sin²θsin²φ - r²sin²θcos²φ - r²cos²θ >= 0. Assuming no algebra mistakes, I'm left with:
4>= r²(3sin²θsin²φ - 1)
Now the max values of sin²θsin²φ range from 0-1, so for r² the possible equations are:
sin²θsin²φ = 1
4 >= 2r²
r <= +/-root(2) so I guess r is between +/- root(2).

But if sin²θsin²φ = 0, get:
4>=-r², which is always true regardless of r...kind of confused there.
Also although I know that 0<=sin²θsin²φ<=1, can I just assume the usual limits (0<θ< Pi, 0<φ<2Pi)? I feel like not...Or, with cylindrical, i get:
4>= z² + s²(3sin²φ -1)

Doing the same thing (sin²φ between 0,1), sin²φ = 1:
4>= z² + 2s². If we let z be a minimum = 0, then s = +/- root(2)
If we let s be a minimum = 0 , then z = +/- 2

But if we let sin²φ = 0, we get that same problem:
4+s² = z² --> Any values can be used to make this true.

I feel like I'm getting close, but still need a bit of a hint..
Thanks for all the reply's tiny-tim.
 
  • #9
Hi AriAstronomer! :smile:

That's all a bit complicated. :redface:

It would be easier to use y as your "up" direction, not z …

then 4 -4y² -x² - z² = 4 -3y² -x² -y² - z² = 4 - 3y²- r² = 4 - 3r²cos²θ - r² …

carry on from there :smile:

(alternatively, 4 -4y² -x² - z² = 0 is obviously an ellipsoid, and you can easily see where it crosses the axes!)

(actually, why bother? :rolleyes: … the question only asked you for the surface, not the actual value :wink:)
 
  • #10
I get it now, thanks a lot for all the help!
 

1. What is a vector field?

A vector field is a mathematical concept used in physics and engineering to describe the direction and magnitude of a vector at every point in a given space.

2. Why is maximizing flux important?

Maximizing flux is important because it allows us to find the optimal surface for a vector field, which can help us understand the behavior and movement of objects in that space.

3. How is flux calculated?

Flux is calculated by taking the dot product of the vector field and the surface's normal vector at a given point. This tells us the amount of the vector field that passes through the surface at that point.

4. What is the optimal surface for a vector field?

The optimal surface for a vector field is one that maximizes the flux, meaning that it allows the most amount of the vector field to pass through it at every point.

5. How can we find the optimal surface for a vector field in R^3?

To find the optimal surface for a vector field in R^3, we can use mathematical techniques such as partial derivatives and optimization methods to determine the surface that maximizes the flux. This can also be visualized using computer software or physical models.

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