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Stokes's theorem should be applied to vector field $$\vec F (x,y,z) = 3z\vec i + 5x\vec j + 2y\vec k$$ where area $\sigma$ is part of paraboloid $z=9-x^2-y^2$ with $z \ge 0$ oriented up.

$C$ is positive oriented circle $x^2 + y^2 = 9$ that is boundary of area $\sigma$ in the $xy$-plane.

Edit: This is the text of my assignment and it is not clear to me how can setup this problem as integral and use Stokes's theorem to solve it.

Any help is appreciated.

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    what specifically is the question?2011-11-27
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    @10smi To apply Stoke's theorem, you need to first find out which integral you are computing, and clearly state it in the question. And then, do not forget to show what _you_ have tried.2011-11-27
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    This is the text of my assignment and it is not clear to me how can setup this problem as integral and use Stoke's theorem to solve it.2011-11-27
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    His name was "Stokes", not "Stoke"; so it should be "Stokes's Theorem". (Proper names that end in "s" take "'s" as possessives, except those that end in an accented "eez" sound, such as Xerxes, and for traditional reasons, "Jesus".)2011-11-27
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    @ArturoMagidin: Or Stokes' theorem.2011-11-27
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    @Matt: A common spelling, though technically considered incorrect according to all sources I consulted when I was writing a paper on Gauss's Lemma (see the parenthetical rule). Of course, repeat it often enough, and it will become "standard". (Did you know "delight" is spelled that way by mistake? When they started standardizing spelling, they thought it belonged with the same words that were derived from the german suffix "licht", but it doesn't; it should be "delite") Certainly, "Stoke's" is as wrong as using "it's" for the possessive.2011-11-27
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    @ArturoMagidin: I wrote "Stokes'", not "Stoke's". And no, I did not know all that. I agree with your last sentence.2011-11-27
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    @Matt: Sorry if I wasn't clear: I know you didn't write "Stoke's"; that's what it was 'corrected' to later, and what was written in many of the comments, but none by you. "Stokes' " seems to be very common (much like "Gauss '").2011-11-27

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I had a go. If you want to apply Stokes' theorem you first need to decide whether it's easier to compute $$ \int \int F \cdot dS$$ in terms of $$ \oint F \cdot d\vec{r}$$ or the other way around.

Here $dS$ is the normal vector of your surface and $\vec{r}$ is the parametrisation of $C$, the boundary curve of your surface. In your homework, I decided that computing $ \oint F \cdot d\vec{r}$ was easier.

First you parameterise $C$ as follows: $$ \vec{r}(t) = \Big ( \begin{array}{c} 3 \cos t \\ 3 \sin t \\ 0 \end{array} \Big )$$

How did I get this parameterisation? The boundary circle is in the $xy$ plane, so $z = 0$. The other two are just the standard parameterisation of a circle of radius $3$.

Now we're good to go:

$$\begin{align} \int \int F \cdot dS = \oint_C F \cdot d \vec{r} = \int_0^{2\pi} F(r(t)) \cdot \vec{r} dt = \int_0^{2 \pi} \Big ( \begin{array}{c} 0 \\ 15 \cos t \\ 6 \sin t \end{array} \Big ) \cdot \Big ( \begin{array}{c} 3 \cos t \\ 3 \sin t \\ 0 \end{array} \Big ) dt = 45 \int_0^{2 \pi} \cos t \sin t dt = 0 \end{align}$$

Where I used integration by parts in the last step. Hope this helps.

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    Hey @Matt just a minor $\LaTeX$ thing...not important, but I thought you might want to know...double integrals are "\iint" which displays $\iint$ and triple ints are "\iiint" which displays $\iiint$.2011-11-27
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    @BillCook: Nice, thank you!2011-11-28