Stoke's Theorem: F(x, y, z) = yi - xj + yzk, and z = ....

[shaoen01]

Use Stoke's Theorem to evaluate \(\displaystyle \int\int_S (curl F).\,dS\), where \(\displaystyle F(x,y,z) = yi - xj + yzk\) and S is part of the surface \(\displaystyle z=2(x^{2} + y^{2})for which z\leq\frac{1}{2}\) Verify your answer by direct calculation. The orientation of S is given by the outward normal vector. (i,j,k are vectors)

I managed to do the direct computation, but i do not know how to use Stoke's Theorem to apply this. I understood part of the answer, but unsure with some steps.

\(\displaystyle r(\theta) = \frac{1}{2}cos\theta i + /frac{1}{2} sin \theta j + \frac{1}{2}k\)--> I don't know how this is derived?

\(\displaystyle \int\int_S curl F.\,dS = \int_C F(r).\,dr = \int_2\pi^{0} -\frac{1}{4}(\sin^{2}\theta + \cos^{2}\theta) \,d\theta\) -->How did it come to this? (the integration for F(r) is supposed to have a circle around it but can't find how to do this).

I know
\(\displaystyle \int_C F(r) \,dr\) is the definition for line intgerals of vector fields right? Should i use Green's Theorem? I am quite confused with this question ... Any help is greatly appreciated!

 

[Unco]

First note that I readily ignore posts I see replicated on other forums (e.g. SOS), but I have not seen this question.

What does Stoke's theorem say?

So all you are left to do is evaluate a line integral along the curve given by \(\displaystyle \mbox{2(x^2 + y^2) = \frac{1}{2}}\) or \(\displaystyle \mbox{x^2 + y^2 = \frac{1}{4}}\).

How do you evaluate a line integral?

Parameterize the curve: \(\displaystyle \mbox{r = (x, y, z) = \left(\frac{1}{2}\cos{\theta}, \frac{1}{2}\sin{\theta}, \frac{1}{2}\right)}\) where theta goes from 0 to 2pi.

Note that Green's theorem applies to 2-dimensions.

 

[shaoen01]

I just like to clarify that the reason why i double-post for the other question was because i could not get the answer from this forum and also i want to look at different methods for solving the question.

I have no intention of wasting anybody's time, i am sorry if it offended you in anyway.

Parameterize the curve: \(\displaystyle \mbox{r = (x, y, z) = \left(\frac{1}{2}\cos{\theta}, \frac{1}{2}\sin{\theta}, \frac{1}{2}\right)}\) where theta goes from 0 to 2pi.

--> Why is it not \(\displaystyle (\frac{1}{4}\sin\theta,\frac{1}{4}\cos\theta, \frac{1}{4})\)?

 

[Unco]

The circle has radius 1/2, hence the x and y parameterization. It sits on the plane z=1/2, hence the z parameterization.

 

[shaoen01]

The circle has radius 1/2, hence the x and y parameterization. It sits on the plane z=1/2, hence the z parameterization.

Yeah, you're right. Thanks

 

[shaoen01]

I was looking through the model answer's working, but i am not sure why the limits for the integration is:
\(\displaystyle \int_{2\pi}^{0} F\) and not \(\displaystyle \int_0^{2\pi}F\)?

Could it because it is a outward normal vector, which means it is going in the opposite direction. So instead of \(\displaystyle \int_{c}F\), i could rewrite it as \(\displaystyle \int_{-c} F = -\int_{c} F\) based on the line integral definition.

And so instead of integrating from \(\displaystyle \int_{2\pi}^{0} F\) , i could do it using \(\displaystyle \int_0^{2\pi}-F\). I tried both ways and the answers both work out to be the same. I know the correct answer does not mean the working is correct, so if anyone has a second opinion do let me know.

 

[Unco]

The surface S is part of a parabaloid. We are told S is oriented with an outward - i.e., downward- pointing - normal vector. Draw or imagine looking down onto the parabaloid from the positive z-axis, so you see a circle in the x-y plane. Point your right thumb down. Your fingers curl in a clockwise direction. That is the direction we must evaluate the line integral in. Thus theta goes form 2pi to 0, which, as you have correctly pointed out, is equivalent to reversing the direction and multiplying by -1.

 

[shaoen01]

I am sorry, but i can't really imagine the thumb part. So the thumb is like the outward normal vector right? But i think don't get the fingers moving clockwise part? How does that show it goes from 2 pi to 0? Sorry, i can't seem to really picture it.