Surface Integral of a Vector Field

Hi guys, need some assurance here in evaluating this surface integral.

$\displaystyle \iint_{S} \mathbf{F}\cdot d\mathbf{S}$

where

$\displaystyle \mathbf{F} = \left \langle x^2+y^2+z^2, -2xy+\sin(z), 2z \right \rangle, (x,y,z)\in \mathbb{R}^3$

and

$\displaystyle S:x^2+y^2+(z-5)^2=1$ with positive orientation (outward pointing normal).

Here's what I did:

Using Green's theorem (Divergence theorem)

$\displaystyle \textup{div }\mathbf{F}=2x-2x+2=2$

$\displaystyle \iint_{S} \mathbf{F}\cdot d\mathbf{S}= \iiint_{E} \textup{div }\mathbf{F } dV$

$\displaystyle =2\iiint_{E} dV=2(\textup{volume of sphere})=2( \frac{4}{3} \pi r^3 )=\frac{8\pi}{3}$

Anything wrong with what I did?

Re: Surface Integral of a Vector Field

Bump... Not even a 'OK' or 'Good' or 'Wrong'?

Lol I've done the question already, only need someone to assure me this is the right way to go.

Re: Surface Integral of a Vector Field

I thought I had responded to this, yesterday. Did you post on more than one forum?

Yes, what you have done is completely correct. I considered suggesting you check by doing the surface integral directly but that "sin(z)" makes things horrible!

Re: Surface Integral of a Vector Field

Quote:

Originally Posted by

**HallsofIvy** I thought I had responded to this, yesterday. Did you post on more than one forum?

Yes, what you have done is completely correct. I considered suggesting you check by doing the surface integral directly but that "sin(z)" makes things horrible!

Thanks! :)

Yes I posted on another forum, on which you replied as well. Thanks for both instances! :)