Calculating the flux of the curl of $F=z\hat{i}+x\hat{j}+y\hat{k}$ with Stokes $C=${$(x,y,z)|x^2+y^2+z^2=1, z\geq 0, y\geq 0$}
Given is the vectorfield $F=z\hat{i}+x\hat{j}+y\hat{k}$.
Calculate the flux of the curl of $F$ through $C$, with a normal with a positive $z$-component. Calculate it with Stokes's theorem, but also without.
So I calculated the curl of $F$, which is $\hat{i}+\hat{j}+\hat{k}$.
Without Stokes's Theorem I calculated $F \bullet \hat{N} dS= 2x+2y+2z dA$. I used polar coordinates to calculate the integral, and found that the flux of the curl of $F$ through $C$ was $\frac{\pi}{2}$.
But when using Stokes I got stuck. $\oint F \bullet dr= \int \int \mathrm{curl} F \bullet \hat{N} dS$. We want to know $\int \int \mathrm{curl} F \bullet \hat{N} dS$, so we want to calculate $\oint F \bullet dr$. So we have to find a parametrisation. But I don't know which one.
 A: Your first calculation is off. We do indeed have $\nabla \times \textbf{F} = \textbf{i}+\textbf{j}+\textbf{k}$ but when we parametrize this "quarter-sphere" as 
$$\textbf{X}(\phi,\theta)=(\cos \theta \sin \phi , \sin \theta \sin \phi , \cos \phi), \phi \in [0,\pi/2], \theta \in [0,\pi]$$
we have normal vector $\textbf{N}(\phi,\theta) = (\sin^2 \phi \cos \theta, \sin^2 \phi \sin \theta, \cos \phi \sin \phi)$ so evaluating the integral gives us:
$$ 
\begin{align}
\int_{0}^{\pi} \int_{0}^{\pi/2} \left[\sin^2 \phi (\cos \theta + \sin \theta) + \cos \phi \sin \phi \right]\, d\phi \, d\theta &= \frac{1}{4} \int_{0}^{\pi} \left(\pi \sin \theta + \pi \cos \theta + 2 \right)\, d\theta = \pi.
\end{align}
$$
Now, to use Stoke's theorem, we need a closed boundary so we can parametrize the boundary piecewise as $\textbf{x}_1 \cup \textbf{x}_2$ where
$$ 
\textbf{x}_1(\theta) = (\cos \theta, \sin \theta, 0) \quad \theta \in [0,\pi] \\
\textbf{x}_2(\theta) = (\cos \theta, 0, \sin \theta) \quad \theta \in [\pi,0].
$$
Evaluating a piecewise line integral gives
$$ 
\begin{align}
\int_{0}^{\pi} (0,\cos \theta,\sin \theta) \cdot (-\sin \theta, \cos \theta, 0) \, d\theta \, + 
\\ \int_{\pi}^{0} (\sin \theta, \cos \theta, 0) \cdot (-\sin \theta, 0, 0) \, d\theta &= \int_{0}^{\pi} \cos^2 \theta + \sin^2 \theta \, d\theta \\
&= \pi.
\end{align}
$$
