Find Work Done in Contour Integration of Vector Field: F=ix^2 +j2xy

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Homework Help Overview

The discussion revolves around calculating the work done by a vector field, specifically F = ix^2 + j2xy, as a particle moves counterclockwise around a rectangular path. Participants are exploring the conditions under which the work done in closed line integrals equals zero.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants discuss the result of a line integral yielding 96 and question the conditions that would lead to a zero result. They explore the implications of the curl of the vector field being zero and the concept of conservative fields.

Discussion Status

There is an ongoing exploration of the conditions required for a vector field to be conservative, including the necessity of the curl being zero and the implications of the field's domain. Some participants have provided insights into the requirements for applying Green's Theorem, while others raise questions about the continuity of the vector function.

Contextual Notes

Participants note that the vector function must be continuous and possess continuous partial derivatives within the region enclosed by the loop, and they discuss the implications of these conditions in relation to the specific vector field presented.

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A particle moves in a counter clockwise direction around a rectangle having the vertices (0, 0), (6, 0), (0, 4), and (6, 4) under the influence of the vector field:
F= ix^2 +j2xyOur objective is to find the work done by after one complete circuit. The work done is given by the close loop line integral.

And here's my question: the result of this operation is 96. well, which conditions provide closed line integrals result is 0? ..sorry for my bad english

thanks in advance
 
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ssrtac said:
And here's my question: the result of this operation is 96. well, which conditions provide closed line integrals result is 0? ..sorry for my bad english

thanks in advance

The curl of the vector field must be zero.
 
Yes, any time the field is conservative. If the field has a potential function (like gravity, etc) then coming back to the starting point requires no net work on the field.
 
Redbelly98 said:
The curl of the vector field must be zero.

This is actually not the only requirement. For this to be the only requirement, the field's domain of definition must be simply connected. For instance, consider the vector field

[tex]\mathbf{F} = -\frac{y}{x^2+y^2} \mathbf{i} + \frac{x}{x^2+y^2} \mathbf{j}.[/tex]​

Its curl is indeed [tex]0[/tex], but if you find the work done when going around a circle centered at [tex](0, 0)[/tex], you will see that it is not [tex]0[/tex]. So, even though [tex]\nabla \times \mathbf{F} = 0[/tex], it is not a conservative field. For vector fields where the domain of definition is multiply-connected, the condition [tex]\nabla \times \mathbf{F} = 0[/tex] is necessary but not sufficient.

Of course, for the simple functions ssrtac is asking, this is enough.
 
Yes, I should have said the curl must exist, and equal zero, everywhere on and within the loop.
 
The vector function must be continuous and have continuous partial derivatives on the region enclosed by the loop.
 
Curl said:
The vector function must be continuous and have continuous partial derivatives on the region enclosed by the loop.
But this is true of the function given in Post #1, is it not? Yet it is nonconservative.
 
I mean that the function must be continuous etc in order to apply Green's Theorem
 

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