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About stoke's thm.

  1. Nov 20, 2012 #1
    1. The problem statement, all variables and given/known data
    if my curve is a ellipse intersect by a cylinder ,$$ (x^2+y^2=a^2 )$$ and plane $$ ax+by+cz=d$$ , and the $$ curl.F=<0,0,f(x,y)>$$
    and the question is about find the line integral of $$ \oint F\cdot dr $$
    then I apply stoke's thm. for the $$ S_{1}$$ surface which is the projection of the ellipse on x-y plane
    $$(x^2+y^2≤a^2,z=0) $$+$$ S_{2}$$ the surface which is the surface area of cylinder from the ellipse to circle. since $$ \int \int_{D} curl.F \cdot dS_{2} $$ is 0 ,
    so what I need to do is compute $$ \int \int_{D} curl.F \cdot dS_{1} = \int \int_{x^2+y^2≤a^2} f(x,y) dxdy $$
    2. Relevant equations

    3. The attempt at a solution
    just want to confirm whether the approach is correct. sorry for typing ugly, and poor english
  2. jcsd
  3. Nov 21, 2012 #2


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    Presumably, those are not the same ##a##.

    You haven't told us what F is. Do you have an F whose curl is that form? (And you would normally write it as curl F without the dot.)

    It isn't the way I would have analyzed the problem, but if I understand what you are doing I think the answer is yes. And I would work the last integral in polar coordinates. Don't forget about the correct orientation.
  4. Nov 22, 2012 #3
    ya , not the same ##a## ,and I mean the $$curl.F$$ is only contain $$k$$
    such as $$F=<-y^3,x^3,z^3> $$ , then $$curl.F = <0,0,3x^2+3y^2>,$$
    can you tell me about your approach for these questions?
  5. Nov 22, 2012 #4


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    You have $$
    \int_C\vec F\cdot d\vec r = \iint_S\nabla \times \vec F\cdot d\vec S
    =\iint_S \langle 0,0,f(x,y)\rangle\cdot d\vec S$$where ##S## is the portion of the plane ##ax+by+cz = d##. I would then parameterize the surface using$$
    x=x,\, y = y,\, z =\frac{d-ax-by} c$$so$$
    \vec R(x,y) = \langle x,y, \frac{d-ax-by} c\rangle$$and use$$
    \iint_S \langle 0,0,f(x,y)\rangle\cdot d\vec S=
    \pm\iint_{(x,y)}\langle 0,0,f(x,y)\rangle\cdot\vec R_x\times \vec R_y\, dydx$$
    That last integral is over the circle in the ##xy## plane and the sign is chosen to agree with the orientation around the curve using the right hand rule. You would probably want to change that last integral to polar coordinates to evaluate it because of the circle.

    You don't have to think about the lateral surface of the cylinder and this method would work whether or not your curl had zeroes in the first two components.
  6. Nov 22, 2012 #5
    thank you guy!!
    you are so helpful
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