# Vectore area element?

• -EquinoX-
In summary, the problem involves finding the vector area element of the surface S and using it to integrate over the surface in order to apply Stokes' Theorem. The vector area element is the product of the infinitesimal area element and the unit normal to the surface. The outward normal to the surface can be found by parameterizing the curve z=9-x^2 and finding the tangent and normal.

## Homework Statement

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}.

## The Attempt at a Solution

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

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.

vectore area element??

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.

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

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.