How Do You Calculate the Flux of a Vector Field Through a Hemisphere?

[PsychonautQQ]

Homework Statement

F has a vector potential A = <x,y,x^2+y^2>. Find the flux of F through the upper hemisphere x^2+y^2+z^2=1 z≥0 oriented with upward pointing normal vector

The Attempt at a Solution

So if F has a vector potential A, would you take the gradient of A to get F?
in which case F would be <1,1,0>.

Then to find the flux would I use stokes theorem? In which case I would have ∫∫∫(∇ F)dV

then what would dV equal..? Looking for the volume you would use (p^2)sin∅dpd∅dθ.. for surface area would you just drop the dp?

 

[vela]

Homework Statement

F has a vector potential A = <x,y,x^2+y^2>. Find the flux of F through the upper hemisphere x^2+y^2+z^2=1 z≥0 oriented with upward pointing normal vector

The Attempt at a Solution

So if F has a vector potential A, would you take the gradient of A to get F?
in which case F would be <1,1,0>.

Nope, so the first thing you should do is look up how to get $\vec{F}$ from a vector potential $\vec{A}$.

Then to find the flux would I use stokes theorem? In which case I would have ∫∫∫(∇ F)dV

then what would dV equal..? Looking for the volume you would use (p^2)sin∅dpd∅dθ.. for surface area would you just drop the dp?

 

[PsychonautQQ]

Ahh so I take the cross product of the del operator and A to get F? And then after I do that is the rest of what I wrote correct?

 

[vela]

To calculate the flux, you can do it directly by calculating the appropriate surface integrals, or you can use the divergence theorem to turn it into a volume integral.
$$\oint \vec{F}\cdot \hat{n}\,dS = \int \nabla\cdot\vec{F}\,dV.$$ Keeping in mind that $\vec{F}$ is the curl of a $\vec{A}$, what can you say about $\nabla\cdot\vec{F}$?

 

[PsychonautQQ]

That would mean that del dot F is equal to 0? So the answer I'm looking for is zero? That being said the problem wants me to solve it using divergence theorem.

 

[PsychonautQQ]

oh wait... so the answer is going to just be zero?

 

[vela]

Yup, it's zero, which you can easily see by using the divergence theorem.

 

[PsychonautQQ]

Even if the divergence itself is zero, I still have to integrate that against 3 integrals to find the volume. Couldn't the volume be non zero because of the constants that pop up?

 

[vela]

Well, you can try it and find out.

 

[vela]

Oh, wait! The problem asks for only the flux through the top hemisphere, so the answer may not be 0. If you calculate the flux through the entire closed surface, it would be 0, but you're only asked to find the flux through the top and not the bottom.

 

[PsychonautQQ]

But wouldn't that only change the parameters of the integration? Del dot F is still going to be zero and integrating against 0 will just give zero right?

 

[vela]

Well, the volume integral isn't relevant unless you have a closed surface. It only equals the flux through a closed surface. If you don't have a closed surface, as in this case, the divergence theorem doesn't apply. You can't say how the volume integral is related to the surface integral.

If you make a closed surface S enclosing volume V by adding a bottom, you can say that
$$\oint_S \vec{F}\cdot\hat{n}\,dS = \int_\text{top} \vec{F}\cdot\hat{n}\,dS + \int_\text{bottom} \vec{F}\cdot\hat{n}\,dS = \int_V \nabla\cdot\vec{F}\,dV$$

 

[PsychonautQQ]

Right on, and sorry for not using LaTex, learning it is on my to do list.

so I have to

∫(F)(n)ds over the top half of the surface of the sphere to get the answer?

If so, should it be a double integral because it's a surface area?
turning dS into something useful when dealing with a sphere is something I can't figure out how to do anywhere, know any good resources or a good understanding to explain it to me? Also the normal vector to the top half of a sphere? it says there is an upward pointing normal, does that mean completely up? If it doesn't specify the exact location doesn't a sphere have infinite normal lines?

 

[vela]

If you were told to use the divergence theorem, you're probably supposed to calculate the surface integral over the bottom, which is likely easier to do than over the hemisphere, and the volume integral. From those, you can infer what the integral over the top should be.

 

[PsychonautQQ]

actually it says I have to turn it from a vector line integral to a surface integral, which would be stokes theorem. So I already did del x A to get F, and then to use stokes theorem I would need to again take a cross product? this time of del x F? How do I parameterize a spheres surface? In terms of theta and phi?

 

[vela]

Ah, okay. Consider
$$\iint_S \vec{F}\cdot\hat{n}\,dS = \iint_S (\nabla \times \vec{A})\cdot\hat{n}\,dS.$$ Use Stoke's theorem to convert that last integral into a line integral.

 

[PsychonautQQ]

Oh okay, so I turn
∫∫∇xA dS
=
∫F dS

where F = ∇xA
∫ goes from 0 and 2∏
and dS = c'(t)dt
where c'(t) = (-sinθ,cosθ,0)

and that shall give me the correct answer?

 

[PsychonautQQ]

and since del x A = 2yi - 2xj
I make y = sin(theta) and x = cos(theta)

 

[PsychonautQQ]

I did all this and got -8pi

 

[vela]

No, you're saying that
$$\iint \vec{F}\cdot d\vec{S},$$ which is a surface integral of vector field $\vec{F}$, is equal to
$$\int \vec{F}\cdot d\vec{r},$$ which is a line integral of the same vector field. Stoke's theorem doesn't say that.

What does Stoke's theorem say?

 

[PsychonautQQ]

∫∫(∇xA) ds = ∫A ds

where A = <cos(t),(sin(t), (1)>
and
ds = c'(t)drdt
ds = <-sin(t),cos(t),0>drdt
so A dot dS will just give zero? is my answer zero?

 

[vela]

The second integral isn't a surface integral, so you shouldn't be integrating with respect to dS.

 

[PsychonautQQ]

ahh this is what my notes say from class. we always write ds even though my textbook goes to dr.

still though, is what I translated the dS into accurate? I mean I could call it dr, it doesn't really matter since either way i'd turn it into c'(t)drdt, and either way it will give me zero?

 

[vela]

No, it's not. $d\vec{r}$ is a line element while $d\vec{S}$ is an area element.

 

[PsychonautQQ]

∫∫(∇xA) ds = ∫A dr

where A = <cos(t),(sin(t), (1)>
and
dr = c'(t)dt
c(t) = <cos(t),sin(t),0>
dr = <-sin(t),cos(t),0>dt
so A dot dr will just give zero? is my answer zero?

 

[vela]

Looks good.

 

[PsychonautQQ]

Is there a way we can turn this into a scalar surface integral? My group isn't sure we are doing this correctly and that the answer is 0. Can somebody double check this for me?

 

[PsychonautQQ]

Okay not into a scalar surface integral, but are we sure this is the correct way to do this and the answer is zero? my group is skeptical but they aren't as great as mathematicians are all you guys :D.

THE ANSWER IS ZERO RIGHT ??

 

[vela]

Yes, the answer is 0.

 

[PsychonautQQ]

thank you, sorry I'm stressed lol