1. Stoke's Theorem

$\displaystyle F = ( y - z + 2)i +(yz +4)j -xzk$, where S is the surface of the cube $\displaystyle x = 0, y = 0, z = 0, x = 2, y = 2, z = 2$, above the xy plane

So i have computed $\displaystyle \nabla x F = -yi + (-1+z)j -k$ now i dont know what to do next

2. If you insist on using Stoke's theorem here, you need to calculate the curve integral of the six squares that make up the sides of the cube. It would be a lot easier to use the divergence theorem here since dV is a cube.

3. Stokes thm is

$\displaystyle \iint_{S} \nabla \times F \cdot {\bf n} dS = \int_C {\bf F} \cdot d{\bf r}$

Are you trying to find the LHS?

4. Originally Posted by Danny
Stokes thm is

$\displaystyle \iint_{S} \nabla \times F \cdot {\bf n} dS = \int_C {\bf F} \cdot d{\bf r}$

Are you trying to find the LHS?
Yes

5. Well, you want to compute $\displaystyle \iint_s \nabla \times F \cdot {\bf n}dS$ for 5 sides of the cube (the bottom is open). Things will cancel though

$\displaystyle S_1: \;\;x = 2 \;\;{\bf n} = <1,0,0> \;\;\;\;\;\nabla \times F \cdot {\bf n} = -y$

$\displaystyle S_2: \;\;x = 0 \;\;{\bf n} = <-1,0,0> \;\;\nabla \times F \cdot {\bf n} = y$

$\displaystyle S_3: \;\;y = 2 \;\;{\bf n} = <0,1,0> \;\;\;\;\;\nabla \times F \cdot {\bf n} = z-1$

$\displaystyle S_4: \;\;y = 9 \;\;{\bf n} = <0,-1,0> \;\;\nabla \times F \cdot {\bf n} = -(z-1)$

$\displaystyle S_5: \;\;z = 2 \;\;{\bf n} = <0,0,1> \;\;\;\;\;\nabla \times F \cdot {\bf n} = -1$

Now the surface integrals

$\displaystyle S_1: -\int_0^2\int_0^2 y dy dz$

$\displaystyle S_2: \int_0^2\int_0^2 y dy dz$

$\displaystyle S_3: \int_0^2\int_0^2 (z-1) dx dz$

$\displaystyle S_4: - \int_0^2\int_0^2 (z-1) dx dz$

$\displaystyle S_5: \int_0^2\int_0^2 -1 dx dy$
Now the first 4 add together give zero (you don't even have to integrate) so your final answer is

$\displaystyle \iint_s \nabla \times F \cdot {\bf n} = -4$