Evaluating both sides of Stoke's Theorem for a square surface How does one set up, and evaluate stokes theorem for a case such as this.
$$\iint_S \operatorname{curl}\mathbf F \cdot \mathbf n \,dA = \oint_C \mathbf F \cdot \mathbf r' ds$$
I'm considering the function...
$$\mathbf F(x,y,z) = \langle z^2,5x,y^2 \rangle$$
and the square surface, S: $~~0 \le x \le 1,~~ 0 \le y \le 1,~~ z = 1$
I need to evaluate both sides of the equation. It seems like it should be simple enough, but I'm just unsure how to set up integrals. For the right side, I did $$\operatorname{curl} \textbf{F} = \langle 2y,2z,5 \rangle$$
Where do I go from here? How do I set up my bounds of integration, find n etc.? What about the right hand side? Any help would be greatly appreciated!
 A: Your surface is a square on the plane $z = 1$, this can be parametrized simply as
$$ \boldsymbol{\sigma} (u,v) = \langle u, v, 1 \rangle, \quad 0 \le u \le 1, \quad 0 \le v \le 1 $$
Then the LHS is
$$ \int_{v=0}^{v=1} \int_{u=0}^{u=1} (\nabla \times \textbf{F}) \cdot \left( \frac{\partial\boldsymbol{\sigma}}{\partial u} \times \frac{\partial \boldsymbol{\sigma}}{\partial v} \right) \,du\,dv $$
The RHS is the path integral over the boundaries of this surface, which consists of 4 lines segments, connecting the 4 points $(0,0,1),(1,0,1),(0,1,1),(1,1,1)$. So all you need to do is parametrize theese 4 lines. For example
$$ C_1: \boldsymbol{r}_1(t) = \langle t, 0, 1 \rangle, \quad 0 \le t \le 1 $$
And
$$ \int_{C_1} \textbf{F}\cdot d\textbf{r}_1 = \int_{t=0}^{t=1} \textbf{F}\cdot \textbf{r}_1' \, dt $$
Then the total RHS would be
$$ \oint_c \textbf{F}\cdot d\textbf{r} = \int_{C_1} \textbf{F}\cdot d\textbf{r}_1 + \int_{C_2} \textbf{F}\cdot d\textbf{r}_2 + \int_{C_3} \textbf{F}\cdot d\textbf{r}_3 + \int_{C_4} \textbf{F}\cdot d\textbf{r}_4 $$
