Let $\partial S$ be the closed, piece wise smooth curve that travels from $(0,0,0)$, $(2,0,4)$, $(3,2,6)$, $(1,2,2)$ and then back to $(0,0,0)$ in precisely that order. Let $S$ be the planar region with boundary $\partial S$ (this region is necessarily contained in the plane $z = 2x$). Use Stokes' theorem to calculate the line integral $$ \oint_{\partial S} \mathbf{F} \cdot \mathrm{d}r$$ where $\mathbf{F}$ is the vectorfield $\mathbf{F}(x,y,z) = z (\cos x) \cdot \mathbf{i} + x^2 y z \cdot \mathbf{j} + yz \cdot \mathbf{k}$.
Simply using Stokes we can parameterize the surface as $$ \mathbf{r}(u,v) = (2u + v, 2v, 4u + 2v) \ , \qquad (u,v) \in [0,1]^2 $$ Some very tedious calculations give then $$ \oint_{\partial S} \mathbf{F} \cdot \mathrm{d}r \stackrel{\text{Stokes'}}{=} \iint_{S} \mathrm{curl}\,\mathbf{F} \,\mathrm{d}S = \iint_{[0,1]^2} \mathrm{curl}\,\mathbf{F}(\mathbf{r}(u,v)) \cdot \left( \frac{\partial \mathrm{r}}{\partial u}\times\frac{\partial \mathrm{r}}{\partial v}\right) \mathrm{d}(u,v) = 52 $$ I was hoping there was a quicker way to do this. Plotting the function in the $xz$-plane it looks like this.
Is there any reason why I can not use Stokes backwards and instead calculate the line integral over this simpler curve? I should be able to take the projection of S, down into the $xz$-plane as is still a piecewise-smooth boundary to $S$. Any other simpler way would also be appreciated.