Solve Stokes Equation Help: Homework Statement

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SUMMARY

The discussion centers on solving the Stokes Equation for a vector field defined as \(\vec F = \cos (xyz)\hat j + (\cos (xyz) - 2x)\hat k\). The problem involves calculating the line integral \(\left| {\int {\vec Fd\vec r} } \right|\) over a curve defined by the intersection of a cylinder and a plane. Participants suggest using Stokes' Theorem, noting the normal vector to the plane and discussing parametrization strategies. The conversation highlights the equivalence of using Green's Theorem and Stokes' Theorem in this context, emphasizing the challenges posed by the sine function in the integral.

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Homework Statement


let F be vector field:
\[\vec F = \cos (xyz)\hat j + (\cos (xyz) - 2x)\hat k\]
let L be the the curve that intersects between the cylinder \[(x - 1)^2 + (y - 2)^2 = 4<br /> \] and the plane y+z=3/2
calculate:
\[\left| {\int {\vec Fd\vec r} } \right|\]

Homework Equations


in order to solve this i thought of using the stokes theorem because the normal to the plane is \[\frac{1}{{\sqrt 2 }}(0,1,1)\]
thus giving me
\oint{Fdr}=\int\int{curl(F)*n*ds}=\int\int{2/sqrt{2}*\sin(xyz)}


i tried to parametries x y and z x= rcos(t)+1 y=rsin(t)+2 z=1/2-rsin(t)

but it won't work
 
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Would x = 1 + 2 cos(t), y = 2 + 2 sin(t) and z = -1/2 - 2 sin(t) do the trick?
 


i wonder if it is allowed given we have to do a multiple integral needing 2 variables
 


Why wouldn't you just use Green's?
 


using green or stokes is the same thing green is just a private solution of stokes and if you use it you are still stuck with that sin(xyz)
 


See what Halls answered you in the other thread.
 

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