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

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?

• Please define x and j; is x the horizontal distance and j the unit vector in Cartesian vertical direction? Apr 6 at 1:26
• @JohnDarby yes, that's the unit vector in the y direction.
– Dewd
Apr 6 at 1:27
• It simply tells that $F$ is non-conservative because work done to be zero over a closed loop only follows for conservative force. Apr 6 at 9:43
• "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" - this is self-contradictory. If the closed path line integral is only zero for conservative forces, it of course cannot always be zero. Apr 6 at 12:45
• If you want an example of a non-zero closed path line integral commonly used in physics, consider Ampere's Law in integral form. The closed path line integral around a loop is proportional to the current passing through the loop. Apr 6 at 20:55

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 $$l$$.

• So... can a line integral over a closed path only be written as ∮ if it is zero? and since in this case it is non-zero, I have to use the ordinary integral notation ∫? This means that I have to know in advance, before actually solving it, whether the integral is zero or non-zero in order to choose the proper notation for it... makes no sense to me
– Dewd
Apr 6 at 2:17
• I almost have no idea what you mean. But, if you are computing an integral over a closed loop, you can use the $\oint$ notation. Whether or not it evaluates to zero depends on whether or not the field is conservative. Apr 6 at 2:19
• @Dewd The circle on the integral represents a closed path; it's not a $0$ on the integral Apr 6 at 2:46
• @Dewd are you confused by the notation $\oint$? It means you take the integral over a closed path, i.e. a path where the initial and final points coincide. It has nothing to do with the value of the integral, which may be zero or non-zero, depending on what you are integrating. Apr 6 at 20:12

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.

• maybe this is more of a notation problem, but I've learned that all closed-path line integrals are always zero - which conflicts with the statement you quoted.
– Dewd
Apr 6 at 2:09
• "all closed-path line integrals" are always zero is not true. Apr 6 at 2:19
• @Dewd: The integral of 1 around the unit circle is equal to $\pi$, which is definitely not zero. Apr 6 at 3:30

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 $$\nabla \times \vec F \ne 0$$.)

• so it's not a closed-path integral (∮)? why not?
– Dewd
Apr 6 at 2:01
• It is a closed path integral but the force is non conservative. See updated answer. Apr 6 at 2:08
• @Dewd This may all seem confusing at the moment, but as you progress through your studies, things will become clearer. Don't pay too much attention to notation. The important thing to understand is that not all integrals over closed paths need to be zero. But when they are, the field in question is conservative. Apr 6 at 2:56
• @josephh it's clearer to me now, thank you both for your help!
– Dewd
Apr 6 at 3:08
• @Dewd The notation $\oint$ is essentially a helpful hint that the path is closed. If you have a path $C$, you may write the integral of force $\vec F$ along this path as $\int_C\vec F\cdot d\vec l$. If the path $C$ happens to be a closed loop, the previous notation is still valid, but it is conventional to give a hint to the reader by using the notation $\oint_C\vec F\cdot d\vec l$. (However if the limits of integration are specified explicitly, as in $\int_0^{2\pi}$, then the $\oint$ notation is not very common, although still possible.) Apr 6 at 20:22