Solve Flux Across a Sphere: 4πa

[physics_197]

Homework Statement

T(x,y,z) = -ln(x^2+y^2+z^2), F = -grad(T); S is the sphere x^2+y^2+z^2=a^2

Homework Equations

The Attempt at a Solution

Given:
F = (2/a^2) <x,y,z>

and the sphere x^2+y^2+z^2 = a^2

F (dot) n = (2/a^2) <x,y,z> (dot) (x/z,y/z,1)

In the end I get:

the double integral of: 2(a^2 - x^2 - y^2)^-0.5

Changing to polar coords I get:
2r / (a^2-r^2 ) drd(theta)
with limits: 0 < r < a and 0 < theta < 2pi

Once I solve it, I get an answer of: 4(pi)a, but the answer in the back is 8(pi)a
Is the answer in the back wrong? or did I make a mistake somewhere.

 

[I like Serena]

F (dot) n = (2/a^2) <x,y,z> (dot) (x/z,y/z,1)

It seems to me that you assume that the normal vector n on the sphere is (x/z, y/z, 1) which is not true.
The normal vector has unit length.
Since any vector in the sphere has length a, the normal vector n is (x/a, y/a, z/a).
Does this answer your question?

 

[physics_197]

When I try to use that, (x/a, y/a, z/a), I get an answer without 'a' in it.

 

[I like Serena]

When I try to use that, (x/a, y/a, z/a), I get an answer without 'a' in it.

So we have:
$$F \cdot n = \frac 2 {a^2} \vec x \cdot \frac 1 a \vec x$$

And we have: $$\vec x \cdot \vec x = a^2$$

So how does 'a' disappear?

 

[physics_197]

So we have:
$$F \cdot n = \frac 2 {a^2} \vec x \cdot \frac 1 a \vec x$$

And we have: $$\vec x \cdot \vec x = a^2$$

So how does 'a' disappear?

Yeh, I retried it and now I get 2(pi)a

When I converted to polar coords, I forgot about the rdrd(theta) part, so when i integrated wrt to r, I had (a-0) and that a canceled with the one on the bottom. When it should have been (a^2-0)

 

[I like Serena]

Yeh, I retried it and now I get 2(pi)a

When I converted to polar coords, I forgot about the rdrd(theta) part, so when i integrated wrt to r, I had (a-0) and that a canceled with the one on the bottom. When it should have been (a^2-0)

Apparently you're integrating over r and theta, but to integrate over the sphere surface, you need to integrate over theta and phi (at constant r=a).

 

[physics_197]

Apparently you're integrating over r and theta, but to integrate over the sphere surface, you need to integrate over theta and phi (at constant r=a).

Thank you very much, I got 8(pi)a now.

Is there anyway of checking the correct answer? Because I have another question similar to it but this one doesn't have the answer at the back.

 

[I like Serena]

Thank you very much, I got 8(pi)a now.

Is there anyway of checking the correct answer? Because I have another question similar to it but this one doesn't have the answer at the back.

Yes, this seems to me a problem that is leading to Gauss' Theorem, which is:

With div F = 6/a^2 and the volume of the sphere being 4/3 pi a^3, you can see that

$$\int \nabla \cdot F dV = \frac 6 {a^2} \int 1 dV = \frac 6 {a^2} \times \frac 4 3 \pi a^3 = 8 \pi a$$
[EDIT]I've put in an extra step in the above formula[/EDIT]