1. ## Using Stoke's Theorem..

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

2. 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.

3. 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!

4. 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

5. 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.

6. 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

7. 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.