Calculate surface integral $\iint_S \nabla \times \mathbf{F} \bullet \mathbf{n} \; dS$ with stokes, when $$\mathbf{F}=\left\langle\frac{5y(z-1)}{6},xz, 6e^{xy}\cos{z}\right\rangle$$ and $S$ is surface $x^2+y^2+z^2=4$, $z \geq0$, which's positive side is the above side.
My work so far.. $$S=\{r\in[0,2], \phi \in [0,2\pi]\}$$ $$\begin{align} \int_0^{2\pi}\int_0^2 ((\frac{\overbrace{z}^{=0}+5}{6})\mathbf{k}\bullet \mathbf{k}) r \; d rd \phi &= \frac 56 \times \text{area of circle with radius of 2} \\ &=\frac 56\times4\pi = \frac {10}{3} \pi \end{align}$$
Hint: The flux of $\text{curl}\,\mathbf F$ is the same across any oriented surface with the same boundary curve.