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Stokes' theorem

  1. Jul 31, 2008 #1
    1. The problem statement, all variables and given/known data

    I got stuck using the Stokes' theorem, the problem is at the bottom of the pic.
    I found the Curl of F, and also the normal of the Triangle. As you can see, I ended up with an area integer with 3 variables, how do I solve this? did I do it right?

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

  2. jcsd
  3. Jul 31, 2008 #2


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    The area over which you now need to integrate is the interior of the triangle with vertices A= (0,0,1), B= (2,0,9), and C= (2,2,11). The vector from A to B is <2, 0, 8> and from A to C is <2, 2, 10>. A normal to that plane is given by the cross product: <-16, -2, 4> or, dividing by 2 to simplify, <8, -1, 2>. The equation of the plane is 8x- y+ 2(z-1)= 0 or 8x- y+ 2z= 2 which we can finally write as z= 1- 4x+ (1/2)y. Writing that as a vector equation, using x and y as parameters, [itex]\vec{r}(x,y)= x\vec{i}+ y\vec{j}+ (1- 4x+ y/2)\vec{i}[/itex]. Then [itex]\vec{r}_x= \vec{i}- 4\vec{k}[/itex] and [itex]\vec{r}_y= \vec{j}+ 1/2\vec{k}[/itex]. The "fundamental vector product for the surface is the cross product of those, [itex]2\vec{i}- (1/2)\vec{j}+ \vec{k}[/itex]. You need to integrate the dot product of that with curl F with respect to x and y, over the area in the xy-plane with vertices (0,0), (2,0), and (2,2), the projection of the points given to the xy-plane.
  4. Jul 31, 2008 #3

    Thanks a lot HallsofIvy.

    I think the first cross product should be (-16, -4, 4), right?
  5. Jul 31, 2008 #4
    Another thing.
    I marked everything in the pic.
    Shouldn't I normalize the normal vector?
    Did I write the boundaries right?

    10x again.

    Attached Files:

  6. Jul 31, 2008 #5


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    Whether you should normalize the normal vector depends upon exactly how you are calculating the differential of area. I prefer to use the "fundamental vector product" as above. That is both normal to the plane and its length is the correct differential of area. That you do NOT want to normalize.
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