Verify Stokes' theorem for the helicoid $\Psi(r,\theta) = \Big\langle r\cos \theta, r\sin \theta, \frac{2\theta}{\pi} \Big\rangle$ oriented upwards, where $0\leq r\leq 1$, $0\leq\theta\leq\frac{\pi}{2}$, and $\mathbf{F}$ is the vector field $\mathbf{F} = \langle 3 z, 6 x^2, 3 y \rangle$.
First, compute the surface integral:

$\displaystyle\Phi = \int\limits_{\hskip 7pt\sigma}\hskip -4pt\int ({\rm curl} \hskip 2pt\mathbf{F})\cdot \mathbf{n} \hskip 3pt dS$

$\hskip 20pt =$

$\displaystyle\int$

$\displaystyle\int$

$\hskip 3pt dr\,d\theta$

$\hskip 20pt =$

Compare that computation with the line integral on the boundary of $\Psi$. From the picture, notice that boundary consists of 4 curves. Parametrize each curve by restricting the domain of $\Psi$ to an appropriate subset.

$\bf C_1$ Straight line with $\theta = 0$

$\displaystyle\Phi_1=\int_{C_1} \mathbf{F}\cdot d\mathbf{r}$

$\displaystyle\int$

$\hskip 3pt dr$

$\hskip 20pt =$

$\bf C_2$ Straight line with $\theta = \frac{\pi}{2}$

$\displaystyle\Phi_2=\int_{C_2} \mathbf{F}\cdot d\mathbf{r} =$

$\displaystyle\int$

$\hskip 3pt dr$

$\hskip 20pt =$

$\bf C_3$ Straight line with $r = 0$

$\displaystyle\Phi_3=\int_{C_3} \mathbf{F}\cdot d\mathbf{r} =$

$\displaystyle\int$

$\hskip 3pt d\theta$

$\hskip 20pt =$

$\bf C_4$ Arc with $r = 1$

$\displaystyle\Phi_4=\int_{C_4} \mathbf{F}\cdot d\mathbf{r} =$

$\displaystyle\int$

$\hskip 3pt d\theta$

$\hskip 20pt =$

Check that the sum of these integrals agrees with your answer from Stokes' theorem.