Question

Find the flux integral SSs curl(Ē).d5, where F(x, y, z) = [2 cos(ny)+22 +22, 22 cos(z7/2) – sin(ny)e24, 222]T and S is the surface parametrized by F(s, t) = [(1 – 51/3) cos(t) – 4s, (1 – 51/3) sin(t), 5s]T with 0 <t< 27,0 < s < 1 and oriented so that the normal vectors point to the outside of the thorn.

Answer 1

Remember that Stokes theorem says

1/ curl(F) .dš= Fider

where $\Gamma$ is the boundary of . In this case is the following figure

Z 2

and $\Gamma$ is the unit circle in the
fic
-plane. We parametrise $\Gamma$ as $\Gamma (\theta) =(\cos\theta, \sin\theta , 0)$ for $\theta\in [0,2\pi]$ . Then,

1/ curl(F) .dš= Fider

$= \int_0^{2\pi} \vec F( \Gamma (\theta)) \cdot \Gamma '(\theta)\,\text d\theta$
$= \int_0^{2\pi} \vec F( \cos\theta, \sin\theta, 0) \cdot ( \cos\theta, - \sin\theta,0)\,\text d\theta$
$= \int_0^{2\pi} \cos\theta \left( 2 \cos( \pi \sin\theta) e^{2 \cos\theta} \right ) - \sin\theta \left( \cos^2 \theta - \pi \sin (\pi \sin\theta)e^{2\cos\theta} \right ) \,\text d\theta$

This integral seems very complicated. However, it is zero. If you split term by term,

$2\int_0^{2\pi } e^{2\cos \theta }\cos\theta \cos(\pi \sin\theta)\,\text d\theta - 2\int_0^{2\pi } \sin\theta\cos^2 \theta\,\text d\theta + \pi \int_0^{2\pi } e^{2\cos\theta} \sin(\pi \sin\theta)\,\text d\theta$

each trigonometric term does a full cycle. If you are not convinced, look at the plot of $e^{2\cos x}$

2 N -5 09 - - - 03

which is also symmetric for $[0,\pi], [\pi,2\pi]$ just as the trigonometric expressions. That is, does not change the value of the integral when accompanied of trigonometric terms.