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Math Help - Circulation

  1. #1
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    Circulation

    Given the vector field F(x,y,z) = (x-y)i +(y-z)j + (z-x)k and the curve \alpha which is the intersection of surfaces:

    S1 = \{ (x,y,z) \in R^3; x+y+z=1 \} and S2 = \{ (x,y,z) \in R^3; x^2+y^2=1 \}

    Find the circulation of F arrond \alpha
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  2. #2
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    Quote Originally Posted by Apprentice123 View Post
    Given the vector field F(x,y,z) = (x-y)i +(y-z)j + (z-x)k and the curve \alpha which is the intersection of surfaces:

    S1 = \{ (x,y,z) \in R^3; x+y+z=1 \} and S2 = \{ (x,y,z) \in R^3; x^2+y^2=1 \}

    Find the circulation of F arrond \alpha
    Paramterize alpha by

    x=\cos(\theta),y=\sin(\theta),z=1-\cos(\theta)-\sin(\theta)

    \vec r(\theta)= \cos(\theta) \vec i +\sin(\theta) \vec j + [1-\cos(\theta)-\sin(\theta)] \vec k

    r'(\theta) = [-\sin(\theta) \vec i +\cos(\theta) \vec j + [\sin(\theta)-\cos(\theta)] \vec k]d\theta

    From here sub in dot product ect and integrate from 0 to 2 Pi.

    You can also use Stokes theorem

    \iint_S (\nabla \times \vec F)\cdot d\vec S=\oint_{\partial S} \vec F \cdot d\vec r

    \nabla \times F = \vec i + \vec j +\vec k

    d\vec S = (\vec i +\vec j +\vec k )dA

    \iint_{\mbox{ unit circle}}3dA=3\pi
    Last edited by TheEmptySet; May 2nd 2009 at 02:50 PM. Reason: I forgot to normalize the unit vector
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    Paramterize alpha by

    x=\cos(\theta),y=\sin(\theta),z=1-\cos(\theta)-\sin(\theta)

    \vec r(\theta)= \cos(\theta) \vec i +\sin(\theta) \vec j + [1-\cos(\theta)-\sin(\theta)] \vec k

    r'(\theta) = [-\sin(\theta) \vec i +\cos(\theta) \vec j + [\sin(\theta)-\cos(\theta)] \vec k]d\theta

    From here sub in dot product ect and integrate from 0 to 2 Pi.

    You can also use Stokes theorem

    \iint_S (\nabla \times \vec F)\cdot d\vec S=\oint_{\partial S} \vec F \cdot d\vec r

    \nabla \times F = \vec i + \vec j +\vec k

    d\vec S = (\vec i +\vec j +\vec k )\sqrt{3}dA

    \iint_{\mbox{ unit circle}}3\sqrt{3}dA=3\pi \sqrt{3}

    I find:
    \int_0^{2 \pi} (-2cost + 3cos^2t + sint)dt = 3 \pi + \frac{3}{4}
    What is my mistake?
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  4. #4
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    Quote Originally Posted by Apprentice123 View Post
    I find:
    \int_0^{2 \pi} (-2cost + 3cos^2t + sint)dt = 3 \pi + \frac{3}{4}

    What is my mistake?
    the integrand is correct

    \int_0^{2 \pi} (-2cost + 3cos^2t + sint)dt =

    \int_{0}^{2\pi}-2\cos(t)+\frac{3}{2}+\frac{3}{2}\cos(2t)+\sin(t)dt  =-2\sin(t)+\frac{3t}{2}+\frac{3}{4}\sin(2t)-\cos(t) \bigg|_{0}^{2\pi}=3\pi
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  5. #5
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    Quote Originally Posted by TheEmptySet View Post
    the integrand is correct

    \int_0^{2 \pi} (-2cost + 3cos^2t + sint)dt =

    \int_{0}^{2\pi}-2\cos(t)+\frac{3}{2}+\frac{3}{2}\cos(2t)+\sin(t)dt  =-2\sin(t)+\frac{3t}{2}+\frac{3}{4}\sin(2t)-\cos(t) \bigg|_{0}^{2\pi}=3\pi
    Thank you
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