Gauss Divergence theorem to find flux through sphere with cavity.

So your answer is correct.In summary, the Gauss Divergence Theorem can be used to find the outward flux of a vector field across the boundary of a region. In this particular case, the region is a sphere of radius 2 with a cavity of radius 1 at the center. The formula requires the surface to be oriented outwards, and the normal should be taken away from the interior of the region. The dot product is denoted as \cdot, not \bullet. The correct approach is to treat the whole region at once, rather than separately considering the outer and inner surfaces. Therefore, the answer for the outward flux is 48\pi.
  • #1
jameson2
53
0

Homework Statement


Use the divergence theorem to find the outward flux of a vector field [tex]
F=\sqrt{x^2+y^2+z^2}(x\hat{i}+y\hat{j}+z\hat{k})[/tex] across the boundary of the region [tex]1\leq x^2+y^2+z^2 \leq4
[/tex]


Homework Equations


The Gauss Divergence Theorem states [tex]\int_D dV \nabla \bullet F=\int_S F\bullet dA [/tex] where D is a 3d region and S is it's boundary.


The Attempt at a Solution


First, I sketched out the boundary, which I think is a sphere of radius 2 with a cavity f radius 1 at the centre. The formula requires that S is oriented outwards.
I basically need to know if the way to do this is first to treat it first as a sphere of radius 2 without a cavity, and work out the outward flux throught this. Then treat the cavity as a sphere of radius 1 and work out the flux going into this, and add the two.
If this is the right approach, I'm not sure how to treat an inward pointing area element. The formula seems to heavily stress that the outward normal is taken, and I don't know if taking an inward normal is allowed.
Alternatively, I think I might be able to take the region as a whole straightaway, and then when integrating over the volume just take the limits of the radius as 2 and 1.
This way, which is the only way can actually get an answer at the moment, gives me an answer for the flux as[tex]\int_S F\bullet dA=48\pi [/tex]
Any help would be much appreciated.
 
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  • #2
"Outward" means "pointing away from the interior of the region". The outward normal on the inner (radius-1) surface points toward the origin, not away from it.

Incidentally, dot product is \cdot, not \bullet: [tex]\int_S F \cdot dA[/tex]
 
  • #3
Ok, does that mean the way I got my answer isn't valid? Since by doing the whole thing in one go I only treated the normal pointing away from the outer surface and not the interior normal pointing towrads the origin?
 
  • #4
No; if you computed [tex]\int_D (\nabla\cdot F)\,dV[/tex], then the divergence theorem says that equals the flux with the normal taken in the correct (away from the region) direction.
 

What is the Gauss Divergence theorem?

The Gauss Divergence theorem, also known as the Gauss-Ostrogradsky theorem, is a fundamental theorem in vector calculus that relates the surface integral of a vector field over a closed surface to the volume integral of the divergence of the same vector field over the enclosed volume.

How is the Gauss Divergence theorem used to find flux through a sphere with a cavity?

The Gauss Divergence theorem can be used to find the flux through a sphere with a cavity by considering the closed surface as the outer surface of the sphere and the inner surface of the cavity. The volume integral of the divergence of the vector field over this enclosed volume is then equal to the surface integral of the same vector field over the outer surface of the sphere.

What is the significance of using a sphere with a cavity in the application of the Gauss Divergence theorem?

Using a sphere with a cavity allows for the calculation of the flux through a region that is not completely enclosed, as the cavity creates a "hole" in the surface. This is useful in many physical applications, such as calculating the electric flux through a charged spherical shell with a cavity.

Are there any limitations to using the Gauss Divergence theorem to find flux through a sphere with a cavity?

Yes, the Gauss Divergence theorem can only be applied to a closed surface with a continuous and differentiable vector field inside and outside the surface. Additionally, the surface must be orientable, meaning that there is a consistent way to assign an inward or outward normal vector to every point on the surface.

Can the Gauss Divergence theorem be extended to higher dimensions?

Yes, the Gauss Divergence theorem can be extended to higher dimensions through the use of differential forms and the generalized Stokes' theorem. In higher dimensions, the theorem relates the integral of the (n-1)-form representing the boundary of an n-dimensional region to the integral of the n-form representing the divergence of a vector field over the enclosed region.

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