Calculating Work: Contour Integration vs Stokes Theorem

[jeff1evesque]

Homework Statement

Given the attached picture, Calculate the work required to go around the contour shown if the force is \vec{F} = y\hat{x} - x^{2} \hat{y}First by Contour integration:
Work = \oint \vec{F} \bullet \vec{dl} = \int_{a}^{b} (\vec{F} \bullet \hat{x})dx + \int_{b}^{c} (\vec{F} \bullet \hat{y})dy - \int_{c}^{d} (\vec{F} \bullet \hat{x})dx - \int_{d}^{a} (\vec{F} \bullet \hat{y})dy

= \int_{1}^{2} (2) dx + \int_{2}^{3} (4)dy - \int_{2}^{1} (3)dx - \int_{3}^{2} (1)dy = 0

I think the answer above is suppose to be -4 not 0.Now by Stokes Theorem:
When this was done with "Stokes theorem", we get the following:
Work = \oint \vec{F} \bullet \vec{dl} \equiv \int \int (\nabla \times \hat{F})\bullet \vec{ds} = - \int \int (2x + 1)( \hat{z} \bullet \hat{z})ds = \int_{2}^{3} \int_{1}^{2} (2x + 1)dxdy = -4
Note: ( \hat{z} \bullet \hat{z}) = 1 since ( \hat{z} \bullet \vec{ds}) = ds

Question:
When I look at this, Stokes method seems apparently correct, but the method before it seems wrong. Can someone help me find my error?Thanks,JL

 

[E_M_C]

Your attachment is still pending approval, but if we're speaking of the work done around a closed path, it's certainly not zero. This is due to the fact that the force field is not conservative. While the force is position dependent (as is required by a conservative force), it's curl, \nabla \times F, is not the zero vector.

 

[HallsofIvy]

Homework Statement

Given the attached picture, Calculate the work required to go around the contour shown if the force is \vec{F} = y\hat{x} - x^{2} \hat{y}

The contour is the square with corners at (1, 2), (2, 2), (2, 3), and (1, 3).

First by Contour integration:
Work = \oint \vec{F} \bullet \vec{dl} = \int_{a}^{b} (\vec{F} \bullet \hat{x})dx + \int_{b}^{c} (\vec{F} \bullet \hat{y})dy - \int_{c}^{d} (\vec{F} \bullet \hat{x})dx - \int_{d}^{a} (\vec{F} \bullet \hat{y})dy

Why do you have minus signs on the last two integrals? Assuming that your contour goes from point a to point b, then to point c, then to point d, then back to point a, these should all be "+". (Although the integrals themselves might be 0.)

= \int_{1}^{2} (2) dx + \int_{2}^{3} (4)dy - \int_{2}^{1} (3)dx - \int_{3}^{2} (1)dy = 0

On the first leg, with x= t, y= 2, the integral is \int_1^2 2 dx as you say. On the second leg, with x= 2, y= t, the integral is \int_2^3 (-4) dt. Did you forget the "-" on "-x^2\vec{j}"? On the third leg, with x= t, y= 3, the integral is \int_2^1 3dt= -\int_1^2 3dt. You have the wrong sign. On the fourth leg, with x= 1, y= t, the integral is \int_3^2(-1) dt= \int_2^3 dt. That is what you have, but, I think, because of two canceling sign errors!. Evaluating, the integral is 2- 4+ (-3)+ (1)= -4.

I think the answer above is suppose to be -4 not 0.


Now by Stokes Theorem:
When this was done with "Stokes theorem", we get the following:
Work = \oint \vec{F} \bullet \vec{dl} \equiv \int \int (\nabla \times \hat{F})\bullet \vec{ds} = - \int \int (2x + 1)( \hat{z} \bullet \hat{z})ds = \int_{2}^{3} \int_{1}^{2} (2x + 1)dxdy = -4
Note: ( \hat{z} \bullet \hat{z}) = 1 since ( \hat{z} \bullet \vec{ds}) = ds

Question:
When I look at this, Stokes method seems apparently correct, but the method before it seems wrong. Can someone help me find my error?


Thanks,


JL

 

[jeff1evesque]

Thanks Halls.