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Vectore area element?

  • Thread starter -EquinoX-
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  • #1
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Homework Statement



Let [tex] \vec{F} = xy\vec{i} + yz\vec{j} + xz\vec{k} [/tex] 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}.

Homework Equations





The Attempt at a Solution



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

Answers and Replies

  • #2
gabbagabbahey
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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
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vectore area element??
 
  • #4
gabbagabbahey
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Yes, it's the product of the infinitesimal area element (usually denoted [itex]dS[/itex] or [itex]da[/itex]) with the unit normal to the surface ([itex]\vec{da}=\hat{n}da[/itex]). Have you not heard that term before?

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

Different authors use different notations.
 
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  • #5
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yes, I've heard of it.. I now need to find the normal vector first... how can I do that...
 
  • #6
gabbagabbahey
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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.
 

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