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

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The discussion centers on calculating the work done by a particle moving in a vector field defined by F=ix^2 + j2xy around a rectangular path, which results in 96. It emphasizes that for closed line integrals to yield a result of zero, the curl of the vector field must be zero, indicating a conservative field. However, this condition alone is insufficient if the field's domain is multiply-connected, as illustrated by a counterexample where the curl is zero but the work done is not. Continuous partial derivatives and continuity of the vector function within the loop are also necessary for applying Green's Theorem. Overall, the conversation highlights the complexities of vector fields and the conditions required for conservative behavior.
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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

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

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