Evaluating a double integral over a hemisphere Evaluate \begin{align*} \iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{\hat{N}} \ dS, \end{align*} where $S$ is the hemisphere $x^2 + y^2 + z^2 = a^2, z \geq 0$ with outward normal, and $\mathbf{F} = 3y \hat{i} - 2xz \hat{j} + (x^2 - y^2) \hat{k}$.
I'm not sure how to do this. We first compute the curl: \begin{align*} \nabla \times \mathbf{F} = (2x-2y) \hat{i} - 2x \hat{j} + (-2z-3) \hat{k} \end{align*} I know that Stoke's theorem says that \begin{align*} \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{\hat{N}} dS, \end{align*} but that doesn't seem to help me here.
 A: Evaluate \begin{align*} \iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{\hat{N}} \ dS, \end{align*} where $S$ is the hemisphere $x^2 + y^2 + z^2 = a^2, z \geq 0$ with outward normal, and $\mathbf{F} = 3y \hat{i} - 2xz \hat{j} + (x^2 - y^2) \hat{k}$.
Stoke's theorem :
 \begin{align*} \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot \mathbf{\hat{N}} dS, \end{align*} 
The edge of the hemisphere is the circle:
$$x^2 + y^2= a^2 \text{ and } z=0$$
This has to be parameterized. A possible parameterization of this curve is:
$$ \forall t \in [0,2 \pi]: \textbf{r}(t)= a.cos(t) \hat{i}+a.sin(t) \hat{j}+0\hat{k}$$
For every t we get a point on C.
$\textbf{dr}$ is an infinitesimal length in the direction of the curve. Just by calculating the differential of $\textbf{r}(t)$, we get:
$$\textbf{dr} = \left(-a.sin(t) \hat{i}+a.cos(t) \hat{j}+0\hat{k} \right) dt$$
We substitute the parameterization of the curve in $\textbf{F}$:
$$\mathbf{F} = 3a.sin(t) \hat{i} - 2a.cos(t).0 \hat{j} + ({a.cos(t)}^{2} - {a.sin(t)}^{2}) \hat{k}$$
$$\mathbf{F} = 3a.sin(t) \hat{i} - 0 \hat{j} + ({a.cos(t)}^{2} - {a.sin(t)}^{2}) \hat{k}$$
Calculating $\mathbf{F} \cdot \textbf{dr}$:
$$\mathbf{F} \cdot \textbf{dr} = \left( 3a.sin(t) \hat{i} - 0 \hat{j} + ({a.cos(t)}^{2} - {a.sin(t)}^{2}) \hat{k} \right) \cdot \left(-a.sin(t) \hat{i}+a.cos(t) \hat{j}+0\hat{k} \right) dt$$
$$\mathbf{F} \cdot \textbf{dr} =-3{a.sin(t)}^{2}dt$$
So al that remains is to calculate the integral:
$$\int_{0}^{2 \pi}{-3{a.sin(t)}^{2}dt}$$
