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Verify Stokes' Therorem when the vector field is

$F(x, y, z)$ = $yi - xj + yxk$, where $k$ is constant.

and $S$ is the paraboloid $z = x^{2} + y^2$ with the circle $x^2 + y^2 = 1$, $z = 1$ as its boundary. Let $S$ be oriented with the outward normal.


I get $2\pi$ when I evaluate the flux of the curl side, but when I try to find the work around the curve, I get $-2\pi$. I'm not sure when I'm going wrong.

UPDATE

Turns out I Integrated my $x(t)$ and $y(t)$ terms instead of differentiating them.

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  • $\begingroup$ Yeah, I'm pretty sure I did it correctly. The answer is 2pi. Here is my working and how I got to -2pi. imgur.com/ANaA3ZI $\endgroup$ – braziil Sep 8 '13 at 1:56
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    $\begingroup$ When you differentiate $\cos t$ you should get $-\sin t$. $\endgroup$ – Bennett Gardiner Sep 8 '13 at 1:58
  • $\begingroup$ I'm not sure If I parametrize the Circle Curve properly. I'm always confused when it's clockwise and whether i put a negative in front of the x(t) or y(t) and what happens to the terminals :/ $\endgroup$ – braziil Sep 8 '13 at 1:58
  • $\begingroup$ Wow -_-! Failed hard there. Thanks guys haha @amzoti, Can you give me a run down on what to do when the parametrizing a circle? I don't actually understand it when its travelling clockwise. I've just been told to add a negative to x(t)? and even then, I'm not sure what I'm meant to do with the terminals. Thanks! $\endgroup$ – braziil Sep 8 '13 at 2:04
  • $\begingroup$ Does this help? jwilson.coe.uga.edu/EMAT6680Fa05/Parveen/Assignment%2010/… $\endgroup$ – Amzoti Sep 8 '13 at 2:15
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\begin{align} \oint\vec{F}\cdot{\rm d}\vec{r} &= \int_{0}^{2\pi}\left\lbrace\vphantom{\LARGE A}% \sin\left(\theta\right) \left\lbrack -\sin\left(\theta\right){\rm d}\theta\vphantom{\Large A}\right\rbrack + \left\lbrack\vphantom{\Large A}-\cos\left(\theta\right)\right\rbrack \left\lbrack\cos\left(\theta\right){\rm d}\theta\right\rbrack \right\rbrace = \color{#0000ff}{\large-2\pi} \end{align}

$$ \left(\nabla\times\vec{F}\right)_{z} = {\partial F_{y} \over \partial x} - { \partial F_{x} \over \partial y} = {\partial\left(-x\right) \over \partial x} - { \partial y \over \partial y} = -2 $$

$$ \int_{S}\nabla\times\vec{F}\cdot{\rm d}\vec{S} = \int_{S}\left(\nabla\times\vec{F}\right)_{z}\left({\rm d}\vec{S}\right)_{z} = \int_{S}\left(-2\right)\left({\rm d}\vec{S}\right)_{z} = \color{#0000ff}{\large-2\pi} $$

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    $\begingroup$ Does this use the correct normal? @lorde said the answer is $2\pi$. $\endgroup$ – Bennett Gardiner Sep 8 '13 at 2:26
  • $\begingroup$ Yeah, the answer is 2π. $\endgroup$ – braziil Sep 8 '13 at 2:32
  • $\begingroup$ @BennettGardiner ${\rm d}\vec{S} = \left\vert{\rm d}\vec{S}\right\vert\,\,\vec{\rm k}$ $\endgroup$ – Felix Marin Sep 8 '13 at 2:35

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