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Do I understand Stokes' theorem? (Calc 3)

  1. Jun 5, 2010 #1
    Here's where I try to explain Stokes' theorem in my own words and you tell me if I'm right / what I need to clarify on.

    Essentially, it's a method to compute a line integral around a closed curve in three dimensions, with a given vector field F, without having to parametrize this field and solve it how one would normally solve a line integral. This closed curve needs to be a boundary of some surface S in 3 dimensions. Stokes' theorem states that the line integral one needs to compute is equal to double integral with respect to area of the curl of F dotted with the vector field n that is normal to the plane in which your curve that you wish to evaluate the line integral of lies. This means that the area on which you evaluate your curl of F dotted with n lies on some 2-dimensional plane. This makes n whatever vector (i, j, k) that lies orthogonal to that plane. Then you simply dot the 2 vectors and solve for the area with respect to whatever plane your surface resides in.


    As a bonus, here's a problem:


    1. The problem statement, all variables and given/known data
    Verify Stokes' Theorem for the given vector field F, surface S, and curve C.


    2. Relevant equations
    Vector field F = (z-y)i + (x+z)j - (x+y)k
    Surface S: z= 9 - x2 - y2 (paraboloid)
    Curve C: The circle x2 + y2 = 9 (oriented counterclockwise as seen from above)

    curl of F
    Stokes' Theorem: outlined above (clearly, I hope)

    3. The attempt at a solution

    This prompt I assumed wants me to solve both the line integral of the curve I got and if it indeed equals curlF dotted with n over the area (in this case, a circle in the xy plane with radius of 3).

    For the line integral:

    =(z-y)dx + (x+z)dy - (x+y) dz
    since z = 0 (xy plane)
    =-y(dx) + x(dy)
    Since the surface is a circle w/ radius of 3, we can parametrize x and y as:
    x = 3cos(t)
    y = 3sin(t)
    dx/dt = -3sin(t)
    dy/dt = 3cos(t)
    Allowing us to rewrite the line integral:
    = -(3sin(t))(-3sin(t))+3cos(t)(3cos(t) dt
    = 9(sin2(t)) + 9(cos2(t)) dt
    = 9 (sin2(t)) + cos2(t)) dt
    = 9
    swithing to polar for convenience yields
    9 * integral from 2pi to zero dt
    = 18 [tex]\pi[/tex]

    For the other method:

    The surface I have to work with is the same one I used with the line integral - the circle in the xy plane with a radius of 3. My curl of F is -2i + 2j. Since my normal is in the direction -i - j + k, dotting these two leads me to a number: 2. Then I set up the double integral as I normally would for a circle with a radius 3 (again, in polar): the outer integral is from 2[tex]\pi[/tex] to 0, the inner from 3 to zero, and the inside is r dr d[tex]\theta[/tex]. Integrating here and multiplying by my curl F dotted with n yields 18[tex]\pi[/tex].

    Sorry for my lack of clarity with tex, I couldn't get it to work at all w/ integrals.

    Also, if anyone feels like it, could you explain why, if k is positive, i and j are negative (and vice-versa) when it comes to the normal vector n? Also, if someone wants to say how the curl of F helps to calculate line integrals, feel free.

    Thanks in advance.
     
  2. jcsd
  3. Jun 5, 2010 #2
    Yes.

    The other way to describe it is that it's a way to compute a surface integral (on an open surface) by only evaluating a line integral of a related function around the edge.
     
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