# Use Stokes's on Line Integral to Show Path Independence

1. Feb 26, 2013

### sikrut

Use Stokes's theorem to show that line integral of $\vec{F}(\vec{r})$ over an curve $L$, given by $\int_L \vec{F}(\vec{r}) d\vec{r}$, depends only on the start and endpoint of $L$, but not on the trajectory of $L$ between those two points.

Hint: Consider two different curves, $L$ and $M$, say, which share a common start and endpoint. Construct from them a closed curve $C$ to which Stokes's theorem can be applied.

$$\vec{F}(\vec{r}) \equiv [F_x(\vec{r}), F_y(\vec{r}), F_z(\vec{r})] = [2z^2, 3z^2, (4x+6y)z] \rightarrow where \rightarrow \vec{r} \equiv [x,y,z].$$

2. Feb 26, 2013

### hapefish

First, you need to remember that Stoke's theorem says that $$\iint\limits_S curl( \vec{F} ) d \vec{S} = \oint\limits_C \vec{F} d \vec{r}$$ You need to compute $curl( \vec{F})$ and show that it is the zero vector. This would mean that any closed line integral will be zero. This makes the integral over the path $C$ zero, which forces the two line integrals $\int\limits_L \vec{F} d\vec{r}$ and $\int\limits_M \vec{F} d \vec{r}$ to be equal.

Good Luck!

3. Feb 26, 2013

### rude man

It's not necessarily the curl of F that needs to be zero, but the integration over the circumscribed area of the curl dotted into an element of normal area.

4. Feb 26, 2013

### hapefish

this is true, but I think if the curl is nonzero at some point then you could find a surface that makes the surface integral nonzero (you would need some sort of continuity assumption to prove this). Since this problem is talking about an arbitrary curve, L, I believe the only possibility is to show that the curl itself is zero.

I could imagine, however, a problem stated "for an arbitrary path on the surface, S, show that..." in which the curl isn't zero, but the same property holds.

5. Feb 26, 2013

### rude man

Consider a single-turn copper coil with a mixture of ∂B/∂t of both polarities thruout its area. Curl of the E field = -∂B/∂t could be nonzero everywhere within its planar area but ∫curl E*dA = 0 is entirely possible, so that the circulation of E (the emf) around the coil would be zero too.

Of course, what you say is entirely correct for a lamellar field like gravity or electrostatic E field.