Work Done in Circle - Why Not Zero? (Calc 3)

[Weather Freak]

Homework Statement

Find the work done by the force field F(x,y) = (x+2y^2)j as an object moves once counterclockwise about the circle (x-2^2+y^2=1 by first evaluating the integral directly then use Green's Theorem and evaluate the double integral.

Homework Equations

W = F*D*cos(theta)

The Attempt at a Solution

My problem is not in getting the answer. I haven't gotten the final answer yet, but that's not the issue at hand.

My Calc teacher told me that it is NOT zero, and I am wondering why. If you move around in a circle, there is no displacement, so therefore, work should be equal to zero. So what is it that makes this case different from everything I did in Physics class?

 

[0rthodontist]

If you move around in a circle through a force field that is a gradient of a scalar function, the work done by the force field is zero. Gravity is a conservative force for this reason--gravity is the gradient of gravitational potential. But if the force field is not the gradient of a function, the work done can be nonzero.

 

[quasar987]

The force fields in physics (gravitational & coulombian) are conservatives. This means that they have the special caracteristic that if the net displacement is 0, then so is the total work. But not all vector fields have this property.

 

[HallsofIvy]

Homework Statement

Find the work done by the force field F(x,y) = (x+2y^2)j as an object moves once counterclockwise about the circle (x-2^2+y^2=1 by first evaluating the integral directly then use Green's Theorem and evaluate the double integral.

Homework Equations

W = F*D*cos(theta)

The Attempt at a Solution

My problem is not in getting the answer. I haven't gotten the final answer yet, but that's not the issue at hand.

My Calc teacher told me that it is NOT zero, and I am wondering why. If you move around in a circle, there is no displacement, so therefore, work should be equal to zero. So what is it that makes this case different from everything I did in Physics class?

Surely you were not taught that the work done in moving away from and then back to an initial point is always 0? That is true only for a conservative force field. This field is not conservative.

 

[Weather Freak]

Okay, I think I get it now, thanks! One more thing though... if the work ends up being zero, does that necessarily imply that the vector field is conservative, or is it only true the other way around?

 

[quasar987]

If the work is 0 over ANY closed loop, then the field is conservative. But if it is 0 over one particular loop, it does not mean that it is conservative. It could be non-zero around some other loop.