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Thread: Stokes' Theorem

  1. #1
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    Stokes' Theorem

    $\displaystyle \int\limits_C \vec{F} . dr $

    C:
    $\displaystyle x^2 + y^2 + z^2 = a^2$ intersection $\displaystyle y + z = a$

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

    _____________________________________

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

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

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

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

    $\displaystyle \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:

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

    The solution is $\displaystyle -3A$ being $\displaystyle A$ the area enclosed by the circle $\displaystyle x^2+y^2+z^2=a^2,\;y+z=a$ .
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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 = $\displaystyle \frac{\pi a^2}{\sqrt{2}}.(-3)$ = $\displaystyle \frac{-3\sqrt{2} \pi a^2}{2}$.

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

    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 $\displaystyle A$ the area enclosed by the circle $\displaystyle x^2+y^2+z^2=a^2,\;y+z=a$ . Then, $\displaystyle \iint_{S}\textrm{rot}\vec{F}\cdot \vec{n}\;dS=(-3/\sqrt {2})A$ ( notice that we forgot to write $\displaystyle \vec{n}=(1/\sqrt{2})(0,1,1)$ ). The area enclosed by the circle is $\displaystyle A=\pi a^2/2$ so, the solution certainly is $\displaystyle \dfrac{-3\sqrt{2}\pi a^2}{4}$ .
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