1. ## 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 $\iint_{\textit{S}}$ F•dS 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 $\iint_{\textit{S}}$ F•dS 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 ). 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

2. ## Re: Contradictory Stokes Theorem verification problem/ even and odd functions

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