Vectore area element?

1. May 6, 2009

-EquinoX-

1. The problem statement, all variables and given/known data

Let $$\vec{F} = xy\vec{i} + yz\vec{j} + xz\vec{k}$$ and C is the boundary of S, the surface z = 9 - x2 for 0 ≤ x ≤ 3 and -6 ≤ y ≤ 6, oriented upward. Use Stokes' Theorem to find [tex] \int\limits_c \vec{F} \cdot d\vec{r}.

2. Relevant equations

3. The attempt at a solution

well I've found the curlF at least.. I don't know what I should do now

2. May 6, 2009

gabbagabbahey

Re: stoke

Find an expression for the vector area element of S (it should have an x-component and a z-component)and integrate over the surface.

3. May 6, 2009

-EquinoX-

Re: stoke

vectore area element??

4. May 6, 2009

gabbagabbahey

Re: stoke

Yes, it's the product of the infinitesimal area element (usually denoted $dS$ or $da$) with the unit normal to the surface ($\vec{da}=\hat{n}da$). Have you not heard that term before?

For example, the outward vector area element for a spherical shell of radius $R$ is $\vec{da} =R^2 \sin \theta d \theta d \phi \hat{r}$, where $\theta$ is the polar angle, $\phi$ is the azimuthal angle, and $\hat{r}$ (sometimes written $\hat{e}_r$) is the radial unit vector.

Different authors use different notations.

Last edited: May 6, 2009
5. May 6, 2009

-EquinoX-

Re: stoke

yes, I've heard of it.. I now need to find the normal vector first... how can I do that...

6. May 6, 2009

gabbagabbahey

Re: stoke

Start by drawing a sketch of the surface, you should see that the outward normal to the surface is the same as the outward normal of the curve z=9-x^2. Parameterize that curve (I suggest using x=t) and find the tangent and normal in the usual ways.