Flux through a spherical surface

[1MileCrash]

Homework Statement

Calculate the flux of vector field F = -3r through sphere radius 5 at the origin.

Homework Equations

The Attempt at a Solution

Since the orientation is always exactly opposite of the orientation of the surface, I expect a negative answer.

Also, since they are along the same direction, and n is a unit vector the dot product must reduce to a multiplication of magnitudes.

(I can't find the vector arrow in itex..! F is a vector and has an arrow above!)

F \bullet \widehat{n} \Delta A = || F ||\Delta A

Which is 300pi over the entire surface area, and since it's in the opposite direction, -300pi.

What am I doing wrong?

 

[1MileCrash]

Pardon the bump, but, any ideas? Should I go back to square 1?

 

[HallsofIvy]

Why do you think you are doing anything wrong?

 

[1MileCrash]

Well, because my answer was marked incorrect.

 

[DryRun]

F = -3r? I'm not familiar with this format for the vector field. Did you type it correctly? For a vector field, you normally need to specify if -3r is in which direction?
For example: \vec F = -3r \vec i

Normally, i would use the Gauss Divergence theorem to solve this easily.
\iiint\ div\vec F \,.dV

 

[1MileCrash]

F = -3r? I'm not familiar with this format for the vector field. Did you type it correctly? For a vector field, you normally need to specify if -3r is in which direction?
For example: \vec F = -3r \vec i

Normally, i would use the Gauss Divergence theorem to solve this easily.

No, r is in the direction of radial increase. These aren't Cartesian vectors.

Gauss Divergence is in the next chapter, the professor said it would cut the work down considerably but I think I should learn the basics first.

 

[DryRun]

OK, so i think what you need to learn is how to use surface integral to find flux through a surface, S.

\iint_S \vec F. \hat n \,.d \sigma
where d\sigma is the differential area and \hat n is the unit normal vector outwards from the surface.

I would suggest finding only the upper half of the sphere, where \hat n is positive, and then you multiply your final result (flux) by 2 to get the total flux. But maybe there's a simpler way.

However, according to what i know, you normally first need to have (or in this case, derive) the vector field \vec F in terms of its unit direction vectors.

In your case, the equation of the sphere would be: x^2+y^2+z^2=25

 

[1MileCrash]

OK, so i think what you need to learn is how to use surface integral to find flux through a surface.

\iint \vec F. \hat n \,.d \sigma
where d\sigma is the differential area and \hat n is the unit normal vector outwards from the surface.

Well, yes I know that. That dot product is just a multiplication of magnitudes. F and the sphere's orientation are the same, but opposite, at all points.

And since the vector field is constant magnitude, I can pull it out of integration.

Which leaves me with the integral of differential area multiplied by the constant vector field magnitude, which is just the total surface area multiplied by the constant vector field magnitude. Which is what I've done.

So what's wrong with that thinking?

 

[1MileCrash]

However, according to what i know, you normally first need to have (or in this case, derive) the vector field \vec F in terms of its unit direction vectors.

In your case, the equation of the sphere would be: x^2+y^2+z^2=25

Why convert this to cartesian coordinates? It's already in unit vector notation, and I'm dealing with a spherical surface. Wouldn't this only complicate the problem?

 

[DryRun]

Then, since F=-15, you can find out \hat n=\frac{∇ \phi }{\left| ∇ \phi \right|} and d\sigma, from its surface area formula.
Let \phi (x,y,z)=x^2+y^2+z^2-25

I think \vec F=\left[ 0\,, 0\,, -15 \right]^T

 

[1MileCrash]

I don't really understand why I'd need any equation for this sphere at all here.

 

[tiny-tim]

Calculate the flux of vector field F = -3r through sphere radius 5 at the origin.
…
Which is 300pi over the entire surface area, and since it's in the opposite direction, -300pi.

area = 4π(5)², flux = field x area (since it's normal to the area), = -300π …

looks ok to me ​

 

[1MileCrash]

area = 4π(5)², flux = field x area (since it's normal to the area), = -300π …

looks ok to me ​

Wiley Plus strikes again, it seems.

Thank you very much.

 

[1MileCrash]

area = 4π(5)², flux = field x area (since it's normal to the area), = -300π …

looks ok to me ​

*Sigh*

I think I see where we erred.

The field is 3r, which actually has a magnitude of 15 at all points of the surface, not 3, because of the sphere's radius of 5.

So the answer should be -1500π

 

[tiny-tim]

oh!

-3\mathbf{r}, not -3\mathbf{\hat{r}} ​