SBurchette
  • SBurchette
Stokes' Theorem I'm trying to evaluate the line integral around the curve obtained by intersecting the cylinder x^2+z^2=1 with the plane y=3. The vector field is F=<3xz, e^(xz), 2xy> and the curve is oriented counterclockwise when viewed from the right.
Mathematics
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schrodinger
  • schrodinger
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SBurchette
  • SBurchette
I parameterized the surface needed to use stokes' thm as r(u,v)= where 0 <=u<=1 and 0<=v<=2*pi. The curl of F and the cross product of the partials of r(u,v) are quite long, but I got an answer of 6*pi which is far from the answer in the solution key.
amistre64
  • amistre64
F(r) . r' right?
SBurchette
  • SBurchette
For the line integral it is, but we are supposed to use Stokes' theorem

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amistre64
  • amistre64
i can never keep the names straight
amistre64
  • amistre64
what is stokes thrm by chance?
slaaibak
  • slaaibak
Flux is F.n
slaaibak
  • slaaibak
The line integral in space is equal to the flux of the surface
SBurchette
  • SBurchette
\[\int\limits_{C} F.dr = \int\limits \int\limits_{S} curl(F).dS\]
slaaibak
  • slaaibak
what was your normal vector? <0,y,0? ?
amistre64
  • amistre64
oy, its way to late for me to be doing anything remotely sane with that :/
anonymous
  • anonymous
If the plane intersects the cylinder you know the cross section's shadow on the x-y plane is a circle right? So you need only to integrate around the circle? And since it's a surface integral r if FIXED at r=1 not to be integrated over. So integrate over z and phi (the polar angle).
anonymous
  • anonymous
I believe.
anonymous
  • anonymous
Especially with y=3, its not even slanted so it will DEFINITELY only give you a circle of constant radius.
SBurchette
  • SBurchette
The formula I have been following states that flux integral over a paramterized surface is \[\int\limits \int\limits_{D} F(r).(r_u \times r_v) dA\] This, in effect, eliminates having to compute the normal directly.
anonymous
  • anonymous
And since you're only integrating with scalars you don't need to worry about integrating over a "position" vector which would NOT give you a circle if the plain was slanted. And you still compute the normal directly, you just don't normalize it.
anonymous
  • anonymous
And I guess it would be the x-z plane not x-y.
SBurchette
  • SBurchette
Integrating along the curve would likely be more simple, but we are supposed to use Stokes' theorem instead of directly computing the line integral
slaaibak
  • slaaibak
but the normal is quite obvious though, isn't it? It just points in the y-direction. so <0,1,0> is fine
SBurchette
  • SBurchette
Yes, I meant we don't compute the nomalized normal, the wording gets tricky ;)
anonymous
  • anonymous
And I misread, since you want to calculate the line integral you only need to integrate over phi. So: |dw:1354845022728:dw| So: \[\Phi(\phi)=(- \cos(\phi),3,\sin(\phi))\] That is your parameterization for that circle. So \[0 \le \phi \le 2 \pi\] And then you have: \[\oint \vec{F} \cdot d \vec{l}=\int\limits_0^{2 \pi} \vec{F}(\vec{\Phi}(\phi)) \cdot \vec{\Phi}'(\phi)d \phi\]
slaaibak
  • slaaibak
wait, I did this a long while ago... but if you just use xyz coordinates, and calculate the curl of the vector field, then dot it with <0,1,0> and then use parametrizations, it's not really tedious at all? since you're only dealing with the j component?
SBurchette
  • SBurchette
That's what we need if we were doing the line integral directly, but to use stokes thm, we have to make it into a flux integral. and @slaaibak I didn't have too much trouble actually doing it, but the solution was incorrect and I can;t figure out where I went wrong
anonymous
  • anonymous
A flux integral of the curl of F.
SBurchette
  • SBurchette
I just realized that I was checking the wrong section of the manual... so the answer I got actually was correct. Thanks for the help guys, I apologize for my oversight,.

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