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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.

enter image description here

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.

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  • $\begingroup$ Unlike your map of the $z=2x$ plane to the $u$-$v$ plane, this map is singular. $\endgroup$ – amd Apr 19 '18 at 23:11
  • $\begingroup$ The other way is to simply take the required line integrals, which involve parameterizing each leg of the journey, substituting that data into F, taking the required scalar products then summing the results....which takes longer than calculating the flux of the curl. $\endgroup$ – Triatticus Apr 19 '18 at 23:12

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