Finding a Plane with Zero Circulation for a Given Vector Field

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Homework Help Overview

The problem involves finding a plane through the origin for a vector field defined by its curl, specifically curl f = (1,2,5) in R^3. The objective is to determine the equation of a plane such that the line integral of the vector field over any closed curve in that plane equals zero.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the curl of the vector field and the normal vector to the plane. There is an attempt to apply Stokes' theorem and algebraic manipulation to derive the plane's equation. Questions arise regarding the correctness of the proposed equations for the plane.

Discussion Status

The discussion includes attempts to derive the equation of the plane, with some participants questioning the initial conclusions and suggesting alternative equations. There is no explicit consensus on the correct plane equation, but multiple interpretations are being explored.

Contextual Notes

Participants are working under the constraint of ensuring that the line integral condition holds for any closed curve in the specified plane. There is an ongoing examination of the implications of the curl's direction and its relationship to the plane's normal vector.

bfr
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Homework Statement



Suppose that f is a vector field such that curl f=(1,2,5) at every point in R^3. Find an equation of a plane through the origin with the property that \oint_{C}f dot dX = 0 for any closed curve C lying in the plane.

Homework Equations



http://img187.imageshack.us/img187/291/1fdf437d8e18a23191b63dfnj8.png

The Attempt at a Solution



With Stokes' theorem and a bit of algebra I get: \int\int ( 1,2,5) dot \nablag dy dx) = 0 . So, 1*dx+2*dy+3*dz=0; let dx=1; let dy=1; dz=-1. The resulting plane is x+y-z=0. Is this right?
 
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You want curl(F)=(1,2,5) to be normal to the plane, right? I don't think that gives you x+y-z=0.
 
Er, oops :eek:.

Then I guess x+2y+5z=0 would simply be the answer?
 
bfr said:
Er, oops :eek:.

Then I guess x+2y+5z=0 would simply be the answer?

Seems so to me.
 

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