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SBurchette
 3 years ago
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.
SBurchette
 3 years ago
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.

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SBurchette
 3 years ago
Best ResponseYou've already chosen the best response.0I parameterized the surface needed to use stokes' thm as r(u,v)=<u*cos(v), 3, u*sin(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.

SBurchette
 3 years ago
Best ResponseYou've already chosen the best response.0For the line integral it is, but we are supposed to use Stokes' theorem

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0i can never keep the names straight

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0what is stokes thrm by chance?

slaaibak
 3 years ago
Best ResponseYou've already chosen the best response.1The line integral in space is equal to the flux of the surface

SBurchette
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{C} F.dr = \int\limits \int\limits_{S} curl(F).dS\]

slaaibak
 3 years ago
Best ResponseYou've already chosen the best response.1what was your normal vector? <0,y,0? ?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0oy, its way to late for me to be doing anything remotely sane with that :/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If the plane intersects the cylinder you know the cross section's shadow on the xy 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
 3 years ago
Best ResponseYou've already chosen the best response.0Especially with y=3, its not even slanted so it will DEFINITELY only give you a circle of constant radius.

SBurchette
 3 years ago
Best ResponseYou've already chosen the best response.0The 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
 3 years ago
Best ResponseYou've already chosen the best response.0And 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
 3 years ago
Best ResponseYou've already chosen the best response.0And I guess it would be the xz plane not xy.

SBurchette
 3 years ago
Best ResponseYou've already chosen the best response.0Integrating along the curve would likely be more simple, but we are supposed to use Stokes' theorem instead of directly computing the line integral

slaaibak
 3 years ago
Best ResponseYou've already chosen the best response.1but the normal is quite obvious though, isn't it? It just points in the ydirection. so <0,1,0> is fine

SBurchette
 3 years ago
Best ResponseYou've already chosen the best response.0Yes, I meant we don't compute the nomalized normal, the wording gets tricky ;)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And 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
 3 years ago
Best ResponseYou've already chosen the best response.1wait, 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
 3 years ago
Best ResponseYou've already chosen the best response.0That'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
 3 years ago
Best ResponseYou've already chosen the best response.0A flux integral of the curl of F.

SBurchette
 3 years ago
Best ResponseYou've already chosen the best response.0I 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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