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Math Help - Stokes Theorem - double integral

  1. #1
    Member mybrohshi5's Avatar
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    Cool Stokes Theorem - double integral

    Use Stokes’ Theorem to evaluate \int_c^.F \cdot dr

    C: is the boundary of the portion of  z = x^2 + y^2 below z = 4, oriented downward,

     F = <x^2, y^4 - x, z^2sin z>

    Can someone please check over my work thanks!

    First I found N -->  N = \frac{<2x, 2y, -1>}{\sqrt{4x^2 + 4y^2 + 1^2}}

    Then i found  \bigtriangledown \times F = <0, 0, -1>

    from using the formula  \bigtriangledown \times F = <F_{3y} - F_{2z}, -F_{3x} + F_{1z}, F_{2x} - F_{1y}>

    Now  \bigtriangledown \times F \cdot N = \frac{1}{\sqrt{4x^2 + 4y^2 + 1^2}}

    Then  dS = {\sqrt{4x^2 + 4y^2 + 1}}dA

    So putting it all together....

     \int\int_R^. \frac{1}{\sqrt{4x^2 + 4y^2 + 1^2}} \cdot \sqrt{4x^2 + 4y^2 + 1}

    The square roots cancel and reduce to....

     \int\int_R^. 1 dA

    This double integral of 1 dA is equal to the Area of the Region

    And using  z = x^2 + y^2 and  z = 4 we come up with  r^2 = 4 .... r = 2

     A(R) = \pi r^2 = \pi(2^2) = 4\pi

    Does  4\pi look correct for the answer?

    Thanks for looking
    Last edited by mybrohshi5; December 8th 2010 at 01:14 PM. Reason: Corrected with N oriented downward
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  2. #2
    MHF Contributor

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    I've always disliked the notation \vec{n}dS because used that unthinkingly results in doing two square roots which, just as you see here, cancel!

    Instead think of d\vec{S} as the "vector differential of surface area" and calculate it directly. We can write the surface z= x^2+ y^2 in a vector equation with x and y as parameter: \vec{r}(x,y)= x\vec{i}+ y\vec{j}+ z\vec{k}= x\vec{i}+ y\vec{j}+ (x^2+ y^2)\vec{k}. The derivatives with respect to x and y give two tangent vectors, \vec{i}+ 2x\vec{k} and \vec{j}+ 2y\vec{k}. The cross product of those gives the "vector differential of surface area": 2x\vec{i}+ 2y\vec{j}- \vec{k} where I have chosen the order of multiplication to get a negative value for \vec{k}, "oriented downward".

    You are correct that \nabla\times \vec{F}= -\vec{k} so that \nabla\vec{F}\cdot\vec{n} dS= \nabla\vec{F}\cdot d\vec{S}[tex]= (0(2x)+ 0(2y)- 1(-1))dxdy= dxdy.

    That gives 4\pi, not -4\pi.

    Your normal vector, \frac{-2x\vec{i}-2y\vec{j}+ \vec{k}}{\sqrt{4x^2+ 4y^2+ 1}} was oriented upward, not downward.
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