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Thread: 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 > $\displaystyle \frac{1}{2}$. Evaluate $\displaystyle \iint_{\textit{S}}$ 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 = <$\displaystyle \frac { \sqrt { 3 } }{ 2 } $cos t, $\displaystyle \frac { \sqrt { 3 } }{ 2 } $sin t, $\displaystyle \frac { 1 }{ 2 } $> for 0<t<$\displaystyle 2\pi $ and then evaluating the integral $\displaystyle \oint _{ \partial S }^{ }{ \overrightarrow { A } } \cdot \quad d\overrightarrow { S } =\quad \int _{ 0 }^{ 2\pi }$ <0, $\displaystyle \frac { \sqrt { 3 } }{ 2 } $cos t, $\displaystyle \frac { \sqrt { 3 } }{ 4 } $ cos t> <-$\displaystyle \frac { \sqrt { 3 } }{ 2 } $sin t, $\displaystyle \frac { \sqrt { 3 } }{ 2 } $cos t, 0> = $\displaystyle \frac { 3\pi }{ 4 } $

    The trouble came in verifying via $\displaystyle \iint_{\textit{S}}$ FdS which I attempted by using G($\displaystyle \theta$,$\displaystyle \phi $) = ($\displaystyle sin\theta sin\phi ,\quad cos\theta sin\phi ,\quad cos\phi $) with a normal vector $\displaystyle \overrightarrow { n } (\theta ,\phi )=\quad ({ R }^{ 2 }sin\phi )<cos\theta sin\phi ,\quad sin\theta sin\phi ,\quad cos\phi )\quad $. I took the dot product of F which I equated in terms of G($\displaystyle \theta$,$\displaystyle \phi $) as <0, -cos$\displaystyle \phi$, 1>, with the normal vector to get my integral of $\displaystyle \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, $\displaystyle \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 $\displaystyle \theta$ will always be zero, because of the symmetry of the bounds and how sine is an odd function. Also, $\displaystyle -sin\theta sin^{ 2 }\phi cos\phi $ is an odd function because $\displaystyle sin^{ 2 }\phi $ is even, $\displaystyle cos\phi $ is even, and $\displaystyle sin\theta $ is odd, so when evaluated from $\displaystyle -\frac { \pi }{ 3 } $ to $\displaystyle \frac { \pi }{ 3 } $ it will end up as 0. This is no surprise.

    The big punch to the face comes with the other integral, $\displaystyle \int _{ 0 }^{ 2\pi }{ \int _{ -\frac { \pi }{ 3 } }^{ \frac { \pi }{ 3 } }{ \quad sin\phi cos\phi \quad d\phi \quad d\theta } }$. 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 $\displaystyle 2\int _{ 0 }^{ \pi }{ \int _{ 0 }^{ \frac { \pi }{ 3 } }{ \quad sin\phi cos\phi \quad d\phi \quad d\theta } } $, the magic answer of $\displaystyle \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 $\displaystyle \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 $\displaystyle \int _{ 0 }^{ 2\pi }{ \int _{ -\frac { \pi }{ 3 } }^{ \frac { \pi }{ 3 } }{ \quad sin\phi cos\phi \quad d\phi \quad d\theta } }$ into $\displaystyle 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 $\displaystyle x=\sin{\theta}\cos{\phi}, y=\sin{\theta}\sin{\phi}, z=\cos{\theta}$,
    $\displaystyle d\sigma = \sin{\theta}d\theta\wedge d\phi$
    $\displaystyle \int \langle F, n \rangle d\sigma=\int (-yz+z) d\sigma$
    $\displaystyle =\int_0^{\frac\pi{3}} \int_0^{2\pi} \cos{\theta}(1-\sin{\theta}\sin{\phi})\sin{\theta}d\theta\wedge d\phi$
    $\displaystyle =\int_0^{\frac\pi{3}} \int_0^{2\pi} \cos{\theta}\sin{\theta}d\theta\wedge d\phi-\int_0^{\frac\pi{3}} \int_0^{2\pi} \cos{\theta}\sin{\theta}\sin{\phi}\sin{\theta} d\theta \wedge d\phi$
    $\displaystyle =2\pi\int_0^{\frac\pi{3}} \cos{\theta}\sin{\theta}d\theta-\int_0^{\frac\pi{3}} \cos{\theta}\sin^2{\theta} d\theta \int_0^{2\pi} \sin{\phi} d\phi$
    $\displaystyle =\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 $
    $\displaystyle =\frac{3\pi}{4}$
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