Using Stoke's Theorem..

**calciihelp** · August 3rd 2009, 10:34 AM

I'm having difficulty to find the surface integral using Stoke's Theorem, can anyone please help? Help will be much appreciated!

**arbolis** · August 3rd 2009, 11:49 AM

I'm learning exactly the same subject so I might be wrong.
I believe the following : the surface of the sphere is $x^2+y^2+z^2=4$ with the restriction of the first octant.
So $z=\sqrt {4-x^2-y^2}$ . You're integration over the circle $x^2+y^2=4$ so the boundaries of the surface integral would be $\int _0^{\frac{\pi}{4}} \int _0^{2} F dA$ . Notice that I used polar coordinates.
I'm not sure though, so I'll wait any further help.

**calciihelp** · August 3rd 2009, 12:52 PM

I'm thinking you'd somehow have to use the integral of region S of Curl F multiply by dS..not sure if i'm on the right track..if anyone else can provide more help, please do!

**Calculus26** · August 3rd 2009, 12:58 PM

The integration limits are correct, however in using Stokes theorem

the integrand is( curlF*N dA) where N = -dz/dx i -dz/dy j +k where

d/dx,d/dy are the partial derivatives

**arbolis** · August 3rd 2009, 01:01 PM

Originally Posted by calciihelp

I'm thinking you'd somehow have to use the integral of region S of Curl F multiply by dS..not sure if i'm on the right track..if anyone else can provide more help, please do!

Yes, sorry.
The boundaries are good I think. $\int _0^{\frac{\pi}{4}} \int _0^{2} F dA$ should read $\int _0^{\frac{\pi}{4}} \int _0^{2} curl F \hat n dS$ I believe.

**Calculus26** · August 3rd 2009, 01:08 PM

Yes, sorry.
The boundaries are good I think.

should read

I believe.

Not quite int (curlF*nds) = int (curlF*NdA) once you have the region R in the plane and integration limits dS is replaced with dA and n with N
recall n is a unit vector n = N/|N| dS = |N|dA

**arbolis** · August 3rd 2009, 01:18 PM

Originally Posted by Calculus26

Not quite int (curlF*nds) = int (curlF*NdA) once you have the region R in the plane and integration limits dS is replaced with dA and n with N
recall n is a unit vector n = N/|N| dS = |N|dA

I get it, thanks a lot.

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