Calculate $\int\int \nabla \times\vec{F} \cdot d\vec{s}$ for $\vec{F} = (y, 3xy, x^2z)$ for the surface - $z=\frac{1}{\sqrt{3}}\sqrt{x^2+y^2}$ for - $0\le z \lt 1.$
(Attempt) -
I can use Stoke's Theorem and use the fact that the surface is "open" when z=1, therefore -
$\int_C\vec{F} \cdot d\vec{r}$ = $\int_C(y,3xy,x^2z) \cdot dr $ = (apply $z=1$) = $\int_C (y,3xy,x^2) \cdot (0,0,dxdy)$
Now we use again the fact that $z=1$ on the top of the surface, so - $x^2 +y^2 = 3$, now we can transform to cylindrical coordinates ($\rho = 3$, \phi will run from $0$ to $2\pi$) -
$\int_C (y,3xy,x^2) \cdot (0,0,dxdy)$ = $\int_C (y,3xy,x^2) \cdot (0,0,\rho d\phi) = \int\limits_0^{2\pi} x^2\rho d\phi = \int\limits_0^{2\pi} \rho^2cos^2(\phi)\rho d\phi$ =
27$\int\limits_0^{2\pi} cos^2(\phi) d\phi = 27\pi$
Might this be true?
As you've already seen, we want to make use of Stoke's Theorem. We have the surface $S = \{(x,y) \in \mathbb{R}^2 \mid \sqrt{x^2 + y^2} = \sqrt{3} z \text{ for some } 0 \le z < 1\}$. The boundary of the surface $S$ is the curve $C$ given by $C = \partial S = \{(x,y) \in \mathbb{R}^2 \mid \sqrt{x^2 + y^2} = \sqrt{3}\}$ (it's not too hard to see graphically that the boundary is at $z = 1$, but if you're having trouble with that let me know).
Now, we can parametrize the curve $C$ as $\vec{r}(t) = \sqrt{3}(\cos(t), \sin(t), 1),\ 0 \le t \le 2\pi$ (i.e., as the circle of radius $\sqrt{3}$ parallel to the $xy$ plane and at $z = 1$). Notice that the curve has already been parametrized with a positive orientation, and that we have $x = \cos(t), y = \sin(t)$, and $z = 1$.
Now, we have $$\begin{eqnarray} \int\int \nabla \times\vec{F} \cdot d\vec{s} &=& \oint_C \vec{F} \cdot d\vec{r}\\ &=& \int_0^{2\pi} \vec{F} \cdot \frac{d\vec{r}}{dt} \cdot dt\\ &=& \int_0^{2\pi} (\sin(t), 3\cos(t)\sin(t), \cos^2(t)) \cdot \sqrt{3}(-\sin(t), \cos(t), 0)\ dt\\ &=& \int_0^{2\pi} \sqrt{3}(-\sin^2(t) + 3\cos^2(t)\sin(t))\ dt\\ &=& \int_0^{2\pi} \sqrt{3}\left(\frac{\cos(2t) - 1}{2} + 3\cos^2(t)\sin(t)\right)\ dt\\ &=& \sqrt{3} \left[\frac{\sin(2t) - t}{2} - \cos^3(t) \right|_0^{2\pi}\\ &=& \sqrt{3}(-\pi) \end{eqnarray}$$ Barring some calculation error, this should be the way to go.