I want to calculate $\iint_{S}curl\vec Fd\vec S$ where $F = \begin{pmatrix} xyz \\ x \\ e^{xy}\cos (z)\end{pmatrix}$ and $S$ is the half-sphere $x^2+y^2+z^2 =1$ where $z \geq 0$ with a normal facing outwards.
This is the solution:
From Stokes theorem we know that $\iint_{S}curl\vec Fd\vec S = \int_{\Gamma}Fd\vec r$ where $\Gamma$ is the boundary of $S$.
In our case, $\Gamma$ is the circle $x^2+y^2 = 1$ since the boundary is at $z=0$. writing it in vector form $r(t) = \begin{pmatrix}\cos (t) \\ \sin (t) \\ 0\end{pmatrix}$ where $ 0 < t < 2\pi$ and so $dr(t) = \begin{pmatrix}-\sin (t) \\ \cos (t) \\ 0\end{pmatrix}$, and our function is $F(x(t),y(t),z=0) = \begin{pmatrix} 0 \\ \cos (t) \\ e^{\sin (t) \cos (t)}\end{pmatrix}$
and so $\iint_{S}curl\vec Fd\vec S = \int_{\Gamma}Fd\vec r = \int_{0}^{2\pi}\begin{pmatrix} 0 \\ \cos (t) \\ e^{\sin (t) \cos (t)}\end{pmatrix}*\begin{pmatrix} -\sin (t) \\ \cos(t) \\ 0\end{pmatrix} dt = \int_{0}^{2\pi}\cos ^2(t) dt = \pi$
This is all well and good, I just don't understand where the "normal facing outwards" comes into play. How would the answer be any different if the question stated "normal facing inwards"?