0
$\begingroup$

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.

  • 0
    @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

1 Answers 1

2

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.

  • 0
    @BillCook: Nice, thank you!2011-11-28