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Thread: 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 $\displaystyle {\Psi(r,\theta) = (r\cos \theta, r\sin \theta, \theta)}$ where $\displaystyle {(r,\theta)}$ lies in the rectangle $\displaystyle {[0,1] \times [0,\pi/2]}$, and $\displaystyle {\mathbf{F}}$ is the vector field $\displaystyle {\mathbf{F} = (5 z, 3 x, 7 y)}$.

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

    Next compute the line integral on that part of the boundary from $\displaystyle {(1,0,0)}$ to $\displaystyle {(0,1,\pi/2)}$.
    $\displaystyle {\int_C \mathbf{F}\cdot d\mathbf{s} = \int_a^b g(\theta)\,d\theta}$, where
    $\displaystyle {a =}$ _, $\displaystyle {b =}$ _,
    and $\displaystyle {g(\theta) =}$ ____ (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 $\displaystyle {f(r, \theta) = }$ ____. I tried finding:

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

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

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

    Which is what I thought $\displaystyle {f(r, \theta) = }$ should be.. but my webwork says that it is incorrect. Can anyone find where I am going wrong?
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  2. #2
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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 $\displaystyle {\Psi(r,\theta) = (r\cos \theta, r\sin \theta, \theta)}$ where $\displaystyle {(r,\theta)}$ lies in the rectangle $\displaystyle {[0,1] \times [0,\pi/2]}$, and $\displaystyle {\mathbf{F}}$ is the vector field $\displaystyle {\mathbf{F} = (5 z, 3 x, 7 y)}$.

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

    Next compute the line integral on that part of the boundary from $\displaystyle {(1,0,0)}$ to $\displaystyle {(0,1,\pi/2)}$.
    $\displaystyle {\int_C \mathbf{F}\cdot d\mathbf{s} = \int_a^b g(\theta)\,d\theta}$, where
    $\displaystyle {a =}$ _, $\displaystyle {b =}$ _,
    and $\displaystyle {g(\theta) =}$ ____ (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 $\displaystyle {f(r, \theta) = }$ ____. I tried finding:

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

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

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

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

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

    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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