Stokes's theorem says $$\oint \mathbf{F}.d\mathbf{r}=\iint_{D}^{-} \nabla \times \mathbf{F}\cdot \mathbf{n} \, dA$$
Evaluate using the RHS of Stokes' theorem the following problem.
for the hemisphere $$S:x^{2}+y^{2}+z^{2}=9,$$ $$z\geq 0,$$ its bounding circle $$C:x^{2}+y^{2}=9,$$ $$z=0$$ and the field $$\mathbf{F}=y\mathbf{i}-x\mathbf{j}$$
I have no idea how to find the outer normal vector.
I have tried grad(F)/Magnitude of grad (F) but its incorrect
Stokes theorem is basically relating the flux through a surface with a closed path around the surface. Intuitively it says that a vector field, the total flux of the surface or flow through the closed path, must be equal to the dot product of the vector field along the path.
$\iint \nabla \times \mathbf {\vec F}\cdot \mathbf n dA = \iint \nabla \times \mathbf{\vec F}(\mathbf G(u,v))\cdot (\frac{\partial{\mathbf G}}{\partial u}\times \frac{\partial {\mathbf G}}{\partial v})dudv$
One such parameterization $\mathbf G$ is $\mathbf G(\phi, \theta)$ in spherical coordinates. $\mathbf F(\phi, \theta)=\langle 3\sin\phi \sin \theta, -3\sin\phi \cos\theta, 0 \rangle $ this comes from cartesian coordinates in terms of spherical.
$\iint \nabla \times \mathbf {\vec F}\cdot \mathbf n dA = \iint \nabla \times \langle 3\sin\phi \sin \theta, -3\sin\phi \cos\theta, 0 \rangle \cdot \langle 3\cos\phi \sin\theta, 3\sin\phi \cos\theta, -3\sin\phi \rangle \times\langle -3\sin \phi \sin\theta, 3\sin\phi \cos\theta, 0 \rangle d\phi d\theta$
It should be strait forward how to integrate this, but you do need to consider the limit's of integration to fit what was asked. You should arrive at the same conclusion when you compute the line integral along the path and further it should be a little easier.