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Math Help - Stokes' Theorem

  1. #1
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    Stokes' Theorem

    \int\limits_C \vec{F} . dr

    C:
    x^2 + y^2 + z^2 = a^2 intersection y + z = a

    \vec{F} = (3y +z, x + 4y, 2x + y)

    _____________________________________

    \nabla_X\vec {F} = (1, -1, -2)

    \vec{r(u,v)} = (u, v, a-v)

    \vec{n} = (0, 1, 1)

    I made the intersection between the sphere and the plane and got:

    \frac{x^2}{a^2} + \frac{(y-a)^2}{\frac{a^2}{2}} = 1

    I know the area of this ellipse, but what is the area of this ellipse projected on the xy plane?

    If I knew this could substitute an got the value:

    \iint\limits_D -3 dudv = -3A(D)
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Stokes' Theorem

    The solution is -3A being A the area enclosed by the circle x^2+y^2+z^2=a^2,\;y+z=a .
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  3. #3
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    Re: Stokes' Theorem

    I think you meant 'sphere'.

    But the problem is that the intersection between the sphere and the plane give us the ellipse that I wrote in the first post.

    The area of this ellipse multiply by -3 is = \frac{\pi a^2}{\sqrt{2}}.(-3) = \frac{-3\sqrt{2} \pi a^2}{2}.

    But the answer is \frac{-3\sqrt{2} \pi a^2}{4}.

    Shouldn't I project this area on the xy plane?
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: Stokes' Theorem

    Quote Originally Posted by PedroMinsk View Post
    I think you meant 'sphere'.
    No, I exactly meant by A the area enclosed by the circle x^2+y^2+z^2=a^2,\;y+z=a . Then, \iint_{S}\textrm{rot}\vec{F}\cdot \vec{n}\;dS=(-3/\sqrt {2})A ( notice that we forgot to write \vec{n}=(1/\sqrt{2})(0,1,1) ). The area enclosed by the circle is A=\pi a^2/2 so, the solution certainly is \dfrac{-3\sqrt{2}\pi a^2}{4} .
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