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Problem

Compute the integrals $$I=\iint_\Sigma \nabla\times\mathbf F\cdot d\,\bf\Sigma$$

And $$J=\oint_{\partial\Sigma}\mathbf F\cdot d\bf r$$

For $F=(x^2y,3x^3z,yz^3)$, and $$\Sigma:\begin{cases}x^2+y^2=1\\-1\le z\le 1 \end{cases}$$

I don't quite understand which kind of curve could be the boundary of such a cylinder, so I was not able to calculate the line integral, how can I do it?

For the surface integral, I parametrized the surface:: \begin{align}\Sigma_1(r,t)&=(\cos t, \sin t,r)\\ \end{align}

So $I=\iint_{\Sigma_i}\nabla\times F\, \mathrm{d}\bf\Sigma_i$.

I just like to get a verification on the first one.

We have $\nabla\times F= (z^3-3x^3,0,x^2(9z-1))$.

$\Sigma_{1r}\times\Sigma_{1t}=(-\cos t,-\sin t,0)$, and $\nabla\times F(\Sigma_1(r,t))=(r^3-3\cos^3t,0,\cos^2t(9r-1))$, then

$$ I_1=\iint_D (\nabla\times F)\cdot dS= \int_0^{2\pi}\int_{-1}^1 (-r^3\cos t-3\cos ^4t)drdt $$

Then I just calculate that. Is my procedure correct so far? I know Stokes' theorem says that $I=J$, but we're asked to calculate these two to just 'verify' it.

What would $\oint_{\partial \Sigma} F\cdot \,dr$ be here? Do I have to pick different orientations for each of the circles on the top and bottom of the cylinder?

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2 Answers 2

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The mistake you seem to be making is that you misunderstood what $\Sigma$ is. It is the cilinder without the two 'caps' at the ends (it is what you would get is you took a piece of A4 paper and folded two sides together). This means that the boundary of $\partial \Omega$ are the two circles at the ends of the cilinder. While checking the rest of your calculation I found the parameter $t$ in the $z$-component of the parametrisation of the 'long' part of the cilinder a bit odd. You can call it that off course, but it seems better to call it z to avoid confusion. The rest of your approach seems to be correct.

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  • $\begingroup$ Thank you a lot! Yes, you're correct, I rushed through the question and misunderstood the surface. Could you explain how Stokes' theorem applies for surfaces like this one (where the boundary are two disjoint closed curves)? $\endgroup$ Feb 4, 2016 at 23:47
  • $\begingroup$ Actually I was mistaken, it will still hold true that $I = J$. In general Stokes' theorem states that $$ \int_{S} \nabla \times \vec{F} \cdot d\vec{S} = \int_{\partial S} \vec{F} \cdot d\vec{l} $$ Which in more general form reads (for those who are familiar with manifold theory) $$ \int_M d\alpha = \int_{\partial M} \iota^* \alpha $$ Where $M$ is a manifold (with boundary), $\alpha$ is a form on $\alpha$ and $\iota: \partial M \to M$ the natural inclusion. Note that in this case the form has to have compact support. You are probably unfamiliar with this but it might wet your appetite ;). $\endgroup$
    – Anonymous
    Feb 6, 2016 at 11:20
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To begin with, $\Sigma$ is a $2$-surface in $\mathbb{R^3}$. The parameter domain of $\Sigma$ is a rectangle $D$ in $\mathbb{R^2}$, $0\le \theta \le 2\pi$, $-1\le z \le 1$. As you have already noted, with this characterisation, one may write $$\Sigma (\theta, z)=\left(\cos \theta, \sin \theta, z\right).$$ By definition (Refer to PMA, Rudin Chapter $\boldsymbol{10}$) the positively oriented boundary of $\Sigma$ may be written as $$\partial \Sigma (\theta, z)=\Sigma\left(\partial D\right)$$ for a choice of a positively oriented boundary of $D$. It is easy to see that such a boundary of $D$ is simply the perimeter of $D$ traversed in the plane $\mathbb{R^2}$ so that the points belonging to $D$ are always to the "left" of the direction of traversal. Representing the coordinates in $\mathbb{R^2}$ as $(\theta, z)$, we have for the boundary of $D$, $\partial D=[0,1]$ to $[0,-1] + [0,-1]$ to $[2\pi,-1] + [2\pi, -1]$ to $[2\pi, 1]+[2\pi, 1]$ to $[0,1]$. Indeed the notation for representing the boundary can be improved but, I hope you get the idea. Then $\partial \Sigma (\theta, z) = \Gamma_1+\Gamma_2+\Gamma_3+\Gamma_4$ where each of the $\Gamma_i$ is a curve in $\mathbb{R^3}$ obtained from the above mentioned definition of $\partial \Sigma (\theta, z)$. Explicitly $$\Gamma_1=(1,0,z) \space \space \space\space\space z \space\space from\space\space1\space to\space-1$$ $$\Gamma_2=(\cos \theta,\sin \theta,-1) \space \space \space\space\space \theta \space\space from \space\space 0 \space to\space 2\pi$$ $$\Gamma_3=(1,0,z) \space \space \space\space\space z \space\space from\space\space-1\space to\space 1$$ $$\Gamma_4=(\cos \theta,\sin \theta ,1) \space \space \space\space\space \theta \space\space from\space\space 2\pi \space to\space 0.$$ With such a choice of boundary, one can easily see that the sense in which the circles at the top and the bottom of the cyclinder are traversed are opposite. Looking from the top in the negative $z$ direction, the upper circle is traversed in clockwise sense and the lower one in the anticlockwise sense. Now $$J=\oint_{\partial\Sigma}\mathbf F\cdot d\bf r = \sum_{i=1}^4 \int_{\Gamma_i}\boldsymbol{F.}d\boldsymbol{r}$$ It is fairly obvious that the line integrals along $\Gamma_1$ and $\Gamma_3$ cancel each other. In the parametric form, the field is given by $$\boldsymbol{F}=\left(\cos^2\theta\sin\theta, 3z\cos^3\theta, z^3\sin\theta\right)$$ and $\boldsymbol{dr}$ along the two circles is given by $$\boldsymbol{dr}= \left(-\sin\theta d\theta, \cos\theta d\theta,0\right).$$ Now, one has $$J=\int_{\Gamma_2}\boldsymbol{F.}d\boldsymbol{r} + \int_{\Gamma_4}\boldsymbol{F.}d\boldsymbol{r}$$ $$\implies J = \int_0^{2\pi}d\theta \left[-\cos^2\theta\sin^2\theta+3(-1)\cos^4\theta\right]+\int_{2\pi}^0 d\theta \left[-\cos^2\theta\sin^2\theta+3(+1)\cos^4\theta\right] $$ $$\implies J = -6 \int_0^{2\pi}\cos^4\theta d\theta.$$ I will leave it here so you can easily verify that indeed $I=J$.

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