Flux integral on a radial vector field

In summary: If not, could you please explain what this theorem states?In summary, the flux integral over S1 has a value of 4. The Flux integral can be evaluated by multiplying the magnitude of F times the surface area of the sphere: 4pi r^2. For part A, I just used this magnitude because it matches the fact that it has to be inversely proportional to the square of the distance from the radius. And so I got the right answer which is 4. For part B however just increasing the proportionality to be (1/pi r^3) doesn't work, so I don't know what I'm missing here.
  • #1
Raziel2701
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Homework Statement



(The S1 after the double integral is supposed to be underneath them btw, I just can't seem to do it right using LaTeX right now so bear with me please.)

Suppose F is a radial force field, S1 is a sphere of radius 9 centered at the origin, and the flux integral ∫∫S1F⋅dS=4.
Let S2 be a sphere of radius 81 centered at the origin, and consider the flux integral ∫∫S2F⋅dS.

(A) If the magnitude of F is inversely proportional to the square of the distance from the origin,what is the value of ∫∫S2F⋅dS?

(B) If the magnitude of F is inversely proportional to the cube of the distance from the origin, what is the value of ∫∫S2F⋅dS?


2. The attempt at a solution

So the flux integral over S1 has a value of 4. The Flux integral can be evaluated by multiplying the magnitude of F times the surface area of the sphere: 4pi r^2. Solving for the magnitude of F I get that it's (1/pi r^2).

For part A, I just used this magnitude because it matches the fact that it has to be inversely proportional to the square of the distance from the radius. And so I got the right answer which is 4.

For part B however just increasing the proportionality to be (1/pi r^3) doesn't work, so I don't know what I'm missing here.

I'd like to know first, how we know that the flux integral is equal to the magnitude of F times the surface area of whatever given surface there is. I was just told this, and I can't find it on the textbook, so I'd like to first understand what is the methodology used behind solving this problem.

Thank you. Also, does anyone know of a resource similar to Paul's Online Math Notes that would cover Surface Integral more thoroughly?
 
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  • #2
Raziel2701 said:
The Flux integral can be evaluated by multiplying the magnitude of F times the surface area of the sphere: 4pi r^2.

This is only true if [itex]\textbf{F}[/itex] is both radial (so that it is normal to the surface) and spherically symmetric. Otherwise, [itex]\int_{\mathcal{S}_1}\textbf{F}\cdot d\textbf{S}\neq 4\pi r^2F[/itex].
 
  • #3
the 1/r potential leads to a 1/r^2 radial vector field which is the form of many physical fields (gravity, electric etc.). This is a special field as you've shown as ttal flux is conserved, under certian condition, in this case for any bounded surface containing the origin.

Then as you've found the 1/r^3 magnitude radial field does not have that property...

Do you know about divergence, and the divergence theorem?
 

FAQ: Flux integral on a radial vector field

What is a flux integral on a radial vector field?

A flux integral on a radial vector field is a mathematical calculation used in physics and engineering to measure the flow of a vector field through a surface. It is often used to determine the rate of energy transfer or the amount of fluid passing through a specific area.

How is a flux integral on a radial vector field calculated?

The flux integral on a radial vector field is calculated by integrating the dot product of the vector field and the unit normal vector of the surface over the surface. This can be represented mathematically as ∫∫S F⋅n dS, where F is the vector field and n is the unit normal vector.

What does a positive flux integral on a radial vector field mean?

A positive flux integral on a radial vector field indicates that the vector field is flowing outwards through the surface. This means that there is a net outward flow of energy or fluid through the surface.

Can a flux integral on a radial vector field be negative?

Yes, a flux integral on a radial vector field can be negative. This would indicate that the vector field is flowing inwards through the surface, resulting in a net inward flow of energy or fluid.

What are some real-world applications of flux integrals on radial vector fields?

Flux integrals on radial vector fields have many practical applications, such as calculating the rate of heat transfer in thermodynamics, determining the amount of fluid passing through a pipe in fluid dynamics, and measuring the electric or magnetic field through a surface in electromagnetism.

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