Did I do this right? (Stokes' Theorem, Flux)

[Raziel2701]

Homework Statement

Evaluate [tex]\int\int Curl F\cdot dS[/tex] where [tex]F=<z,x,y>[/tex] (NOTE: the vector in my post preview is showing me the wrong one despite me trying to correct it, the right one is F=<z,x,y>) and S is the surface [tex] z=2-\sqrt{x^2 +y^2}[/tex] above z=0.

Homework Equations

I used Stokes' Theorem, choosing to evaluate [tex]\int_C F\cdot dr[/tex] and using [tex]r(t)=<2cos(t),2sin(t),0>[/tex] as the parametric form of my Curve and after differentiating and dotting with the composed form of F and r, I got 4Pi as my answer after evaluating [tex]\int_0^{2\pi} 4cos^2(t) dt[/tex]

Did I do this correctly? Did I get the right answer? I need to know because as part of this extra credit assignment, I must do this same integral but by not using Stokes Theorem, and before I venture into trying to use the formulas for Flux, I want to know if I got the right answer first.

 

[LCKurtz]

I haven't checked your answer but I would observe that in less time than it must have taken you to post this, you could have worked the flux integral out. That might give you the same answer, in which case your question answers itself and you are done. Try it.

 

[Raziel2701]

Yes, I get 0 with the other integral. The formula I'm using is that the surface integral [tex]\int\int_S F\cdot dS[/tex] is equal to [tex]\int\int_D F\cdot n dA[/tex] where n is the normal vector of the surface.

This is the integral I evaluated:

[tex]\int\int (-\frac{zx}{\sqrt{x^2 +y^2}} -\frac{xy}{\sqrt{x^2 +y^2}} +y) r dr d\theta[/tex]

I substitute x=rcos(theta)
I substitute y=rsin(theta)
I substitute z= 2-sqrt(x^2 + y^2)

And after carrying out the substitutions I evaluate this integral:

[tex]\int_0^{2\pi}\int_0^2 -2rcos(\theta) +r^2cos(\theta) -r^2cos(\theta)sin(\theta) +r^2sin(\theta) dr d\theta[/tex] which gives me 0...

Also, the preview feature is not working properly, it doesn't display what I typed.

 

[LCKurtz]

Yes, I get 0 with the other integral. The formula I'm using is that the surface integral [tex]\int\int_S F\cdot dS[/tex] is equal to [tex]\int\int_D F\cdot n dA[/tex] where n is the normal vector of the surface.

This is the integral I evaluated:

[tex]\int\int (-\frac{zx}{\sqrt{x^2 +y^2}} -\frac{xy}{\sqrt{x^2 +y^2}} +y) r dr d\theta[/tex]

I substitute x=rcos(theta)
I substitute y=rsin(theta)
I substitute z= 2-sqrt(x^2 + y^2)

And after carrying out the substitutions I evaluate this integral:

[tex]\int_0^{2\pi}\int_0^2 -2rcos(\theta) +r^2cos(\theta) -r^2cos(\theta)sin(\theta) +r^2sin(\theta) dr d\theta[/tex] which gives me 0...

Also, the preview feature is not working properly, it doesn't display what I typed.

I get a different dS. If you let the cone be

[tex]\vec R = \langle r\cos\theta,r\sin\theta,2-r\rangle[/tex]

and calculate

[tex]d\vec S = \vec R_r \times \vec R_\theta\, drd\theta[/tex]

I get

[tex]d\vec S = \langle r\cos\theta, r\sin\theta,r\rangle dr d\theta[/tex]

which eventually leads me to an answer [itex]16\pi/3[/itex] for the surface integral. This isn't the same as the line integral, but then, if you think about it, that cone has a point and is not a smooth surface. The hypotheses of Stokes' theorem aren't satisfied.

It's late and I'm tired so you need to check it.

 

[Raziel2701]

Ah I got it. I was mixing and matching F and the normal vector. The point is that a way to evaluate the given integral from the first post is to not calculate the flux, but rather finding a substitute for dS which turns out to be the normal vector of the surface times dA, with limits of integration determined by integrating over the region D.

With that, I got the same result as I got with my line integral (4pi), which can be calculated using Stokes' Theorem because the surface is piece-wise smooth. So that solves my problem :)