1 answer

Find the flux integral SSs curl(Ē).d5, where F(x, y, z) = [2 cos(ny)+22 +22, 22 cos(z7/2)...

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 su

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.

Answers

Remember that Stokes theorem says

\iint_S \text{curl}(\vec F) \cdot \text d\vec S = \int_{\Gamma} \vec F\cdot\text d\Gamma

where \Gamma is the boundary of S . In this case S 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,

\iint_S \text{curl}(\vec F) \cdot \text d\vec S = \int_{\Gamma} \vec F\cdot\text d\Gamma
= \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}

1597008998217_image.png

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.

.

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