1. ## Stokes' Theorem

Consider the cylindrical can of radius 1 with a
differentiable board in the counter-clockwise's direction. Calculate:

$\displaystyle \int\limits_C \vec{F} .d\vec{r}$:

$\displaystyle \vec{F} (x, y, z) = ( y(x-2), x^2y, z)$

OBS.: there is the figure of the cylinder and the curve, but it just says that the curve is closed and impossible to be parametrized.

A.: 3pi
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Attempt:

Parametrization: S --> (cosv, senv, u)
n = (-cosv, sinv, 0)

$\displaystyle \nabla\vec{F} = (0, 0, 6x^2 + 6y^2)$

$\displaystyle \nabla\vec{F}\vec{r}(u,v)\vec{n} = 0$

Where is the mistake?

2. ## Re: Stokes' Theorem

Your question is unclear. In order to apply Stokes's theorem, you need a surface whose boundary is the curve $\displaystyle C$. You seem to be saying that the surface is the open cylinder but then the boundary for this is composed of two circles, not some closed curve you seem to be suggesting. A figure will help a lot.

Figure:

4. ## Re: Stokes' Theorem

OK, that makes things clearer. Using Stokes's theorem, we can write the line integral as a surface integral. Now split the surface integral up into two parts; one over the curved lateral side of the cylinder and the other over the bottom of the cylinder. Since we don't know what the lateral surface is exactly, we expect this integral to be zero. That is, check that $\displaystyle \text{ curl }\mathbf{F\cdot n}=0$ here. The bottom one you can do directly since you know the equation for this surface.

5. ## Re: Stokes' Theorem

Originally Posted by PedroMinsk
Attempt:

Parametrization: S --> (cosv, senv, u)
n = (-cosv, sinv, 0)

$\displaystyle \nabla\vec{F} = (0, 0, 6x^2 + 6y^2)$

$\displaystyle \nabla\vec{F}\vec{r}(u,v)\vec{n} = 0$

Where is the mistake?
Do you mean $\displaystyle \nabla \times \vec{F}$ here?

6. ## Re: Stokes' Theorem

Yes, it was what I meant.

I forgot to calcalute the bottom, that was the problem.

Thanks a lot for your help.