Compute the flux of vector field through a sphere

[jerzey101]

Homework Statement

Compute the flux of the vector field F(x,y,z)=(z,y,x) across the unit sphere x²+y²+z²=1

Homework Equations

I believe the formula is ∫∫_D F(I(u,v))*n dudv
I do not know how to do the parameterization of the sphere and then I keep getting messed up with the normal vector.*
So you convert the sphere equation into spherical coordinates? I am so lost please help. Thank you!

The Attempt at a Solution

is I=(sin∅cosθ,sin∅sinθ,cos∅)?
Then to find n you compute the cross product of T_∅ X T_θ?
Then multiple F(I(u,v))*n?
Then integrate?

 

[DryRun]

Hi jerzey101. Welcome to PF!

Use the Gauss Divergence Theorem - it greatly simplifies the problem.
$$\iint_S \vec F \hat n \,.d\sigma= \iiint_D div \vec F\,.dV$$ where S is the smooth oriented surface and D is a simple closed region.
$$\vec F= z\hat i + y\hat j+x\hat k$$
The volume of the sphere is given by: $\iiint_S \,.dV$

In case you haven't yet done the Gauss Divergence Theorem, then you can also find it (although a little bit more work is involved) by surface integral using: $\iint_D \vec F \hat n \,.d\sigma$. You need to find $\hat n$ which is the outward unit normal from the surface of the sphere.

 

[Broccoli21]

There are a few ways to do this problem. Has your class covered the divergence theorem (Gauss' Theorem)? If so then that is a real fast way to do the problem.
If not then your procedure of parametrization looks good. But even then can you find the unit normal in an easier fashion? Geometrically, what vector is perpendicular to the surface of a sphere?

 

[jerzey101]

I believe I got it.
so div(F)=0+1+0=1
using spherical coordinates
∫^2∏₀∫^∏₀∫¹₀p²sin∅ dpd∅dθ
which I got to equal 4/3∏
That look correct?

 

[DryRun]

I believe I got it.
so div(F)=0+1+0=1
using spherical coordinates
∫^2∏₀∫^∏₀∫¹₀p²sin∅ dpd∅dθ
which I got to equal 4/3∏
That look correct?

It's the same volume as that of the sphere: $$\frac{4\pi r^3}{3}$$ where r=1.

Your answer is correct.

 

[jerzey101]

ooo ok. Thanks a lot!