Stoke's Theorem in Component Form

Let
$\mathbf{F} = F_1 \mathbf{i} +F_2 \mathbf{j} + F_3 \mathbf{k}$
.
$d \mathbf{r} = dx \mathbf{i} + dy \mathbf{j} + dz \mathbf{k}$

Hence
$\mathbf{F} \cdot d \mathbf{r} = F_1 dx + F_2 dy + F_3 dz$
We can take
$\mathbf{n} = cos \alpha \mathbf{i} + cos \beta \mathbf{j} + cos \gamma \mathbf{k}$

$\alpha , \: \beta , \: \gamma$
are the angles the normal makes with the coordinate axes.
\begin{aligned} \mathbf{\nabla} \times \mathbf{F} &= \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ F_1 & F_2 & F_3 \end{array} \right| \\ &= (\frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial y}) \mathbf{i} + (\frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}) \mathbf{j} + (\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}) \mathbf{k} \end{aligned}

Then
$(\mathbf{\nabla} \times \mathbf{F}) \cdot \mathbf{n} = (\frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial y}) cos \alpha + (\frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}) cos \beta + (\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}) cos \gamma$

We can write Stoke's Theorem as
$\oint_C F_1 dx + F_2 dy + F_3 dz = \int \int_S (\frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial y}) cos \alpha + (\frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}) cos \beta + (\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}) cos \gamma ) dS$