Why is the work done non-zero even though it's along a closed path?

Question

So, in this problem I just solved there is a force field given by $\mathbf{F} = -x \hat{\mathbf{j}}$ and I need to compute the work done on a particle along a circular path of radius $R$, centred at the origin, clockwise direction. I checked that the correct answer is $W = \pi R^2$.

So, what I'm confused about is that this is a closed path, since it's a circle and it goes from $0$ to $2\pi$, but the work done is non-zero. I thought closed-path line integrals would always be zero, although I also know that the force must be conservative for the closed-path integral to actually be zero. What's wrong with my thinking? Why is the integral $$ \oint -R\cos\theta\rm{d}\theta = \pi R^2 $$ non-zero?

Answer 1

You're thinking that an integral over a closed path for any field $\bf F$ will always evaluate to zero. This is incorrect.

However, and by definition, for a force $\bf F$ to be conservative, the line integral $$\oint \bf F\cdot dl= 0$$ If this line integral is not zero, then the force cannot be conservative.

You can also verify that the force is non-conservative by computing the curl$^1$. That is, $$\nabla\times {\bf F}=\nabla\times (0,-x,0)= \begin{vmatrix} \hat i & \hat j & \hat k\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ 0 & -x & 0 \end{vmatrix} =-\hat k$$

Any field with a nonzero curl is not conservative. This force is not conservative, hence why you do not get a zero value for the integral.

$^1$ Note that from Stoke's theorem, $$\bf\iint_S(\nabla\times F)\cdot dS=\oint_l F\cdot dl$$ where $S$ is the surface enclosed by the closed path $l$.

Answer 2

I also know that the force must be conservative for the closed-path integral to actually be zero

And since the integral is non-zero... that must not be a conservative force. Closed paths over such forces may be non-zero.

Answer 3

The correct integral is $\int_{o}^{2\pi} \vec F \cdot d\vec r$. Here $d\vec r = R d\theta \hat l$ where $\hat l$ is unit vector in increasing $\theta$ counterclockwise direction, and $\vec F = -R cos\theta \hat j$. To evaluate the dot product for the integrand use $\hat l = - \hat l sin\theta + \hat j cos\theta$, where $-\hat j \cdot \hat l = 0$ and $-\hat j \cdot \hat j = -1$. Then, the integral is $\int_{o}^{2\pi} +R^2cos^2\theta d\theta$ which evaluates to $+\pi R^2$.

The force is non-conservative since the work is non-zero over the closed path. Also, for a non-conservative force $\nabla \times \vec F \ne 0$.)