Solve Stokes Equation Help: Homework Statement

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Homework Help Overview

The problem involves evaluating a line integral of a vector field using Stokes' theorem. The vector field is defined as \(\vec F = \cos (xyz)\hat j + (\cos (xyz) - 2x)\hat k\), and the curve of integration is described by the intersection of a cylinder and a plane.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster considers using Stokes' theorem and attempts to parameterize the curve but encounters difficulties. Other participants suggest alternative parameterizations and question the appropriateness of using multiple integrals. There is also a mention of Green's theorem as a potential alternative approach.

Discussion Status

The discussion is ongoing with various approaches being explored. Some participants are questioning the validity of certain parameterizations and the use of different theorems, indicating a productive exchange of ideas without reaching a consensus.

Contextual Notes

There are constraints related to the requirement of using multiple variables for integration, and the discussion includes considerations of the appropriateness of different theorems for the problem at hand.

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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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