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

[AriAstronomer]

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

 

[tiny-tim]

Hi AriAstronomer!

(have a grad: ∇ and an integral: ∫ and try using the X² tag just above the Reply box )

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!

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

think … when will enlarging the boundary increase the integral? ​

 

[AriAstronomer]

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?

 

[tiny-tim]

Hi AriAstronomer!

(just got up :zzz: …)

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 …

what about, for example, (1,0,-1)? ​

 

[AriAstronomer]

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?

 

[AriAstronomer]

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

 

[tiny-tim]

Should I then try and find the limits which make 4 -4y² -x² - z² >= 0?

Yes!

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.

 

[AriAstronomer]

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.

 

[tiny-tim]

Hi AriAstronomer!

That's all a bit complicated.

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 ​

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

(actually, why bother? … the question only asked you for the surface, not the actual value )

 

[AriAstronomer]

I get it now, thanks a lot for all the help!