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Math Help - Contradictory Stokes Theorem verification problem/ even and odd functions

  1. #1
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    Contradictory Stokes Theorem verification problem/ even and odd functions

    Problem- Let F = <0, -z, 1> and S be the part of the unit sphere where z > \frac{1}{2}. Evaluate \iint_{\textit{S}} FdS directly as a surface integral, then verify that curl(A) = F where A = <0, x, xz> and evaluate the surface again using Stokes' Theorem.

    The easy part of this problem was verifying that curl(A) = F and then parameterizing the boundary of the spherical cap as c = < \frac { \sqrt { 3 }  }{ 2 } cos t, \frac { \sqrt { 3 }  }{ 2 } sin t, \frac { 1 }{ 2 } > for 0<t< 2\pi and then evaluating the integral \oint _{ \partial S }^{  }{ \overrightarrow { A }  } \cdot \quad d\overrightarrow { S } =\quad \int _{ 0 }^{ 2\pi  } <0, \frac { \sqrt { 3 }  }{ 2 } cos t, \frac { \sqrt { 3 }  }{ 4 } cos t> <- \frac { \sqrt { 3 }  }{ 2 } sin t, \frac { \sqrt { 3 } }{ 2 } cos t, 0> = \frac { 3\pi  }{ 4 }

    The trouble came in verifying via \iint_{\textit{S}} FdS which I attempted by using G( \theta,  \phi ) = ( sin\theta sin\phi ,\quad cos\theta sin\phi ,\quad cos\phi ) with a normal vector \overrightarrow { n } (\theta ,\phi )=\quad ({ R }^{ 2 }sin\phi )<cos\theta sin\phi ,\quad sin\theta sin\phi ,\quad cos\phi )\quad . I took the dot product of F which I equated in terms of G( \theta,  \phi ) as <0, -cos \phi, 1>, with the normal vector to get my integral of \int _{ 0 }^{ 2\pi  }{ \int _{ -\frac { \pi  }{ 3 }  }^{ \frac { \pi  }{ 3 }  }{ -sin\theta sin^{ 2 }\phi cos\phi \quad +\quad sin\phi cos\phi \quad d\phi \quad d\theta  }  } .

    I tried evaluating this, however I ran into several problems. If I split the integrand and had two separate integrals, then as is, I calculated both to be 0 and checked this using Wolfram Alpha. Furthermore, \int _{ 0 }^{ 2\pi  }{ \int _{ -\frac { \pi  }{ 3 }  }^{ \frac { \pi  }{ 3 }  }{ -sin\theta sin^{ 2 }\phi cos\phi \quad d\phi \quad d\theta  }  } could be split into the product of two single variable integrals, and the integral for \theta will always be zero, because of the symmetry of the bounds and how sine is an odd function. Also, -sin\theta sin^{ 2 }\phi cos\phi is an odd function because sin^{ 2 }\phi is even, cos\phi is even, and sin\theta is odd, so when evaluated from -\frac { \pi  }{ 3 } to \frac { \pi  }{ 3 } it will end up as 0. This is no surprise.

    The big punch to the face comes with the other integral, \int _{ 0 }^{ 2\pi }{ \int _{ -\frac { \pi }{ 3 } }^{ \frac { \pi }{ 3 } }{ \quad sin\phi cos\phi \quad d\phi \quad d\theta } }. The integrand should be odd according to the logic of sine being odd and cosine being even therefore their product is odd, however, when I turned the integral into 2\int _{ 0 }^{ \pi  }{ \int _{ 0 }^{ \frac { \pi  }{ 3 }  }{ \quad sin\phi cos\phi \quad d\phi \quad d\theta  }  } , the magic answer of \frac { 3\pi }{ 4 } showed up. Seeing that I got the right answer by "cheating" on an odd integrand this way, I tried splitting the first integrand of \int _{ 0 }^{ 2\pi  }{ \int _{ -\frac { \pi  }{ 3 }  }^{ \frac { \pi  }{ 3 }  }{ -sin\theta sin^{ 2 }\phi cos\phi \quad d\phi \quad d\theta  }  } but this had a non-zero answer.

    Is there some way to rationalize all this even and odd function reasoning that I used to solve this problem? I feel like splitting \int _{ 0 }^{ 2\pi }{ \int _{ -\frac { \pi }{ 3 } }^{ \frac { \pi }{ 3 } }{ \quad sin\phi cos\phi \quad d\phi \quad d\theta } } into 2\int _{ 0 }^{ \pi  }{ \int _{ 0 }^{ \frac { \pi  }{ 3 }  }{ \quad sin\phi cos\phi \quad d\phi \quad d\theta  }  } and getting the right answer and then leaving the other half of the flux integral as-is just to validate my answer from calculating the circulation integral was a little slight-handed.

    Thanks for any input
    Anthony
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  2. #2
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    Re: Contradictory Stokes Theorem verification problem/ even and odd functions

    Let  x=\sin{\theta}\cos{\phi}, y=\sin{\theta}\sin{\phi}, z=\cos{\theta},
    d\sigma = \sin{\theta}d\theta\wedge d\phi
    \int \langle F, n \rangle d\sigma=\int (-yz+z) d\sigma
    =\int_0^{\frac\pi{3}} \int_0^{2\pi} \cos{\theta}(1-\sin{\theta}\sin{\phi})\sin{\theta}d\theta\wedge d\phi
    =\int_0^{\frac\pi{3}} \int_0^{2\pi} \cos{\theta}\sin{\theta}d\theta\wedge d\phi-\int_0^{\frac\pi{3}} \int_0^{2\pi} \cos{\theta}\sin{\theta}\sin{\phi}\sin{\theta} d\theta \wedge d\phi
    =2\pi\int_0^{\frac\pi{3}} \cos{\theta}\sin{\theta}d\theta-\int_0^{\frac\pi{3}} \cos{\theta}\sin^2{\theta} d\theta  \int_0^{2\pi}  \sin{\phi} d\phi
    =\pi\int_0^{\frac\pi{3}} \sin{2\theta}d\theta-\int_0^{\frac\pi{3}} \cos{\theta}\sin^2{\theta} d\theta \cdot 0
    =\frac{3\pi}{4}
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