# Stokes' theorem application to vector field

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? – analysisj Nov 27 '11 at 20:38
@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. – Sasha Nov 27 '11 at 21:20
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. – 1osmi Nov 27 '11 at 21:54
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".) – Arturo Magidin Nov 27 '11 at 22:17
@ArturoMagidin: Or Stokes' theorem. – Rudy the Reindeer Nov 27 '11 at 22:23

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 $\iiint$. – Bill Cook Nov 27 '11 at 23:10
@BillCook: Nice, thank you! – Rudy the Reindeer Nov 28 '11 at 6:12