# Condition for a force to be conservative

I get that a force dependent on path is non-conservative or a force which does non-zero work in completing a cycle. But how do you prove that mathematically.

Let's assume I have a force F= 2yi + x^2j. Now how do I conclude that it's work done is path dependent?

My attempt Work done=F.dr

(2yi + x^2j).(dxi + dyj)= 2y.dx + x^2dy

Now what do I integrate it to? Let's assume they take 2 paths to reach (10,10) 1.(0,10) to (10,10) 2. (10,0) to (10,10). Now if i get different work done, they are non-conservative. But again, how do I integrate it? Is there a way to tell it's non-conservative without all this?

• Yes. Given that your field is continuously differentiable, if its curl is zero then it's conservative. – Omar Nagib Mar 24 '16 at 19:38

You need to use Stoke's theorem. $$\oint_{C} \mathbf{F} \cdot d\mathbf{r} = \int_S (\nabla \times \mathbf{F}) \cdot \mathbf{\hat{n}} dS$$ Where $\nabla \times \mathbf{F}$ is equal to $$({\partial F_z \over \partial y} - {\partial F_y \over \partial z}) \mathbf{\hat{x}} + ({\partial F_x \over \partial z} - {\partial F_z \over \partial x}) \mathbf{\hat{y}} + ({\partial F_y \over \partial x} - {\partial F_x \over \partial y}) \mathbf{\hat{z}}$$ For the line integral of the force to vanish on every closed path, its curl ($\nabla \times \mathbf{F}$) must be zero everywhere, too.
If the force is generated by a potential, $F=-\nabla\Phi$, then its curl has to vanish, since $\nabla\times(\nabla\Phi)=0$. You can check the curl of your force field. Remember that the curl of a vector field at a point, according to its definition, is proportional to the line integral of the field along an infinitesimal loop around such point.
• Consider a vector field $\mathbf{F}$ and an infinitesimal loop $C$ along which you want to integrate $F$. The loop $C$ encloses a region $S$ with area $A$ and unit normal $\hat{\mathbf{n}}$. This line integral is $\oint_C\mathbf{F}\cdot d\mathbf{s}$. As you shrink the loop, both the line integral and the area $A$ go to zero, however, the ratio does not. This ratio is precisely the curl, which is defined as $(\nabla\times\mathbf{F})\cdot\hat{\mathbf{n}}\equiv\frac{1}{A}\oint_C\mathbf{F}\cdot d\mathbf{s}$. This is what erenust just wrote. – Pedro Aguilar Mar 25 '16 at 19:21