Finding Flux of $\vec{F}$ Outward Across Sphere

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In summary, the flux of vector field F=(x, y, z) outward across the sphere x^2+y^2+z^2=a^2 is 4πa^2. This can be found by setting up the surface integral in polar coordinates and integrating from 0 to a for r and from 0 to 2π for theta. However, since the field is radial and constant in magnitude over the sphere, the flux can also be found simply by calculating the surface area of the sphere, which is 4πa^2. This is because the field cuts through the entire surface of the sphere.
  • #1
sandy.bridge
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Homework Statement


Find the flux of [itex]\vec{F}=(x, y, z)[/itex] outward across the sphere [itex]x^2+y^2+z^2=a^2[/itex].


I am able to get it to this point:
[tex]\int\int_Cadxdy[/tex] and I then convert it to polar coordinates, and integrate rdr from 0 to a, and theta from zero to 2pi. However, this does not give me the correct result, as the answer is 4a^3*pi, and I am getting a^3*pi.
 
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  • #2
hi sandy.bridge! :smile:
sandy.bridge said:
I am able to get it to this point:
[tex]\int\int_Cadxdy[/tex] and I then convert it to polar coordinates, and integrate rdr from 0 to a, and theta from zero to 2pi. However, this does not give me the correct result, as the answer is 4a^3*pi, and I am getting a^3*pi.

it's the surface of a sphere …

what does r have to do with it? :confused:

(and why are you integrating? surely you know the surface area of a sphere? :wink:)
 
  • #3
Okay, well I know that once I get it to this point, it's right:
[tex]\int\int_CadS=\int\int_Ca(1)dxdy[/tex]

The projection of the sphere on the xy-plane is a circle, no? So why can I not use
[tex]dxdy=rdrd\theta[/tex]?
 
  • #4
sandy.bridge said:
The projection of the sphere on the xy-plane is a circle, no?

projection? are you treating F as if it was a parallel field along one of the axes? :confused:

in that case, yes, the projection perpendicular to the field would be a circle

but the given F is radial ((x,y,z) = t), and constant in magnitude over the sphere,

so you just need the amount of surface it cuts through, which is 4πa2
 

What is flux?

Flux is a measure of the flow of a physical quantity through a surface. It is a vector quantity and is defined as the dot product of the vector field and the surface normal vector.

How do you find the flux of a vector field?

To find the flux of a vector field, you need to calculate the dot product of the vector field and the surface normal vector at each point on the surface. Then, integrate this product over the surface to get the total flux.

What is the outward flux?

The outward flux is the part of the total flux that is perpendicular to the surface and points outward from the surface. It represents the flow of the vector field through the surface in the outward direction.

What does the flux of a vector field outward across a sphere mean?

The flux of a vector field outward across a sphere is a measure of the flow of the vector field through the surface of a sphere in the outward direction. It takes into account the direction and magnitude of the vector field at each point on the sphere to determine the total flow through the surface.

What are some applications of finding the flux of a vector field outward across a sphere?

Finding the flux of a vector field outward across a sphere has many applications in physics and engineering, such as calculating the flow of electric or magnetic fields through a spherical surface, determining the rate of heat transfer through a spherical object, or analyzing fluid flow through a spherical container.

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