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Math Help - Stokes' theorem question

  1. #1
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    Stokes' theorem question

    I have the following homework problem:

    In part, we compute both integrals from Stokes' theorem for the helicoid {\Psi(r,\theta) = (r\cos \theta, r\sin \theta, \theta)} where {(r,\theta)} lies in the rectangle {[0,1] \times [0,\pi/2]}, and {\mathbf{F}} is the vector field {\mathbf{F} = (5 z, 3 x, 7 y)}.

    First, compute the surface integral:
    {\iint_M (\nabla \times \mathbf{F})\cdot d\mathbf{S}= \int_a^b\int_c^d f(r, \theta) dr\,d\theta}, where
    {a = } _, {b = } _, {c = } _, {d =} _, and
    {f(r, \theta) = } ____ (use "t" for theta).
    Finally, the value of the surface integral is ____.

    Next compute the line integral on that part of the boundary from {(1,0,0)} to {(0,1,\pi/2)}.
    {\int_C \mathbf{F}\cdot d\mathbf{s} = \int_a^b g(\theta)\,d\theta}, where
    {a =} _, {b =} _,
    and {g(\theta) =} ____ (use "t" for theta).

    ----

    So far I have a=0, b=pi/2, c=0, d=1, and those are right. But, I'm stuck at {f(r, \theta) = } ____. I tried finding:

    {\Psi_r = (\cos \theta, \sin \theta, 0)}
    {\Psi_\theta = (-r\sin \theta, r\cos \theta, 1)}
    {\Psi_r \times \Psi_\theta = (\sin \theta, - \cos \theta, r)}

    Then I plugged {\Psi} into {\mathbf{F} = (5 z, 3 x, 7 y)} to get:
     (5 \theta, 3r \cos \theta, 7r \sin \theta)

    I dotted that with the cross product and found:
     5 \theta \sin \theta - 3r \cos^2 \theta + 7 r^2 \sin \theta

    Which is what I thought {f(r, \theta) = } should be.. but my webwork says that it is incorrect. Can anyone find where I am going wrong?
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  2. #2
    MHF Contributor

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    [QUOTE=Poison;522315]I have the following homework problem:

    In part, we compute both integrals from Stokes' theorem for the helicoid {\Psi(r,\theta) = (r\cos \theta, r\sin \theta, \theta)} where {(r,\theta)} lies in the rectangle {[0,1] \times [0,\pi/2]}, and {\mathbf{F}} is the vector field {\mathbf{F} = (5 z, 3 x, 7 y)}.

    First, compute the surface integral:
    {\iint_M (\nabla \times \mathbf{F})\cdot d\mathbf{S}= \int_a^b\int_c^d f(r, \theta) dr\,d\theta}, where
    {a = } _, {b = } _, {c = } _, {d =} _, and
    {f(r, \theta) = } ____ (use "t" for theta).
    Finally, the value of the surface integral is ____.

    Next compute the line integral on that part of the boundary from {(1,0,0)} to {(0,1,\pi/2)}.
    {\int_C \mathbf{F}\cdot d\mathbf{s} = \int_a^b g(\theta)\,d\theta}, where
    {a =} _, {b =} _,
    and {g(\theta) =} ____ (use "t" for theta).

    ----

    So far I have a=0, b=pi/2, c=0, d=1, and those are right. But, I'm stuck at {f(r, \theta) = } ____. I tried finding:

    {\Psi_r = (\cos \theta, \sin \theta, 0)}
    {\Psi_\theta = (-r\sin \theta, r\cos \theta, 1)}
    {\Psi_r \times \Psi_\theta = (\sin \theta, - \cos \theta, r)}[quote]
    This is wrong. \Psi_r \times \Psi_\theta = \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ \cos(\theta) & \sin(\theta) & 1 \\ -sin(\theta) & cos(\theta) & 0 \end{array}\right|= -cos(\theta)\vec{i}- sin(\theta)\vec{j}+ r\vec{k}

    Then I plugged {\Psi} into {\mathbf{F} = (5 z, 3 x, 7 y)} to get:
     (5 \theta, 3r \cos \theta, 7r \sin \theta)

    I dotted that with the cross product and found:
     5 \theta \sin \theta - 3r \cos^2 \theta + 7 r^2 \sin \theta

    Which is what I thought {f(r, \theta) = } should be.. but my webwork says that it is incorrect. Can anyone find where I am going wrong?
    Last edited by Jester; June 2nd 2010 at 05:11 AM. Reason: fixed latex
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  3. #3
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    Jun 2010
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    Your Latex code isn't showing up correctly for me, but I'm assuming that you crossed:
    (cos \theta, sin\theta, 1) \times (-sin \theta, cos\theta, 0)

    To get:
    = -cos(\theta)\vec{i}- sin(\theta)\vec{j}+ r\vec{k}

    But im not sure where the two vectors you crossed came from, or why neither of them has r but your answer does. I think it may be a problem with me trying to read the latex code.
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