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Flux across surface

  • Thread starter JaysFan31
  • Start date
JaysFan31
1. Homework Statement
Consider the vector field
F= r + grad(1/magnitude(r)).
Compute the flux (double integral F ndS) of F across the surface of the sphere x^2+y^2+z^2=a^2 where a>0. ndS is the vector element of surface with n the unit normal which here is assumed to point away from the enclosed volume.

2. Homework Equations
Flux equals the double integral of F*n dS where * is the dot product between the vector field F and the unit normal n.

3. The Attempt at a Solution
If I break it into two parts, I get F=r and F= grad(1/magnitude(r))

F=r just has flux 4*pi*a^3 since F=xi+yj+zk and F*N=a where * is the dot product. Note that N=(x/a)i+(y/a)j+(z/a)k. I know this just by looking at it, but how would I set up this double integral to get 4*pi*a^3?

F=grad(1/magnitude(r))=(-x)/(x^2+y^2+z^2)^{3/2}i-y/(x^2+y^2+z^2)^{3/2}j-z/(x^2+y^2+z^2)^{3/2}k

I can simplify it by using a^2=x^2+y^2+z^2 and use polar coordinates, but how do I do this?

I get that grad(1/magnitude(r))=
[(-x)/a^(3/2)]i + [(-y)/a^(3/2)]j + [(-z)/a^(3/2)]k

I know that the unit normal N=(x/a)i + (y/a)j + (z/a)k.

Thus the dot product of grad(1/magnitude(r)) and the unit normal=
(-x^2)/a^(5/2) + (-y^2)/a^(5/2) + (-z^2)/a^(5/2) =
(-a^2)/(a^5/2)= -1/a^(3/2).

Now how do I integrate this to get the flux of the surface?

Is it,
integral from 0 to 4pi, integral from 0 to a^2 of (-a^(-3/2))dr dtheta?
 

Answers and Replies

JaysFan31
Is spherical coordinates easier? How do I this?

OK, does an answer of (4*pi*a^3)-(4*pi*a^(-1/2))? make sense for the final answer?
 
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