# How to check if a 2 dimensional vector field is irrotational (curl=0)?

I need to check if a this vector field is irrotational and conservative:

$$F=\langle e^{x \cos y} \cos y,-x e^{x \cos y}\sin{y} \rangle$$

curl of F should be => $$\dfrac{dQ}{dx} - \dfrac{dP}{y}=0$$ (where 'd' = partial derivative).

With some calculation, I found out that the partials are equal :

$$-e^{x \cos y}\sin y[1+x \cos y]$$

But here is my confusion, does that mean that the vector field is irratational? or conservative? My doubt arises from the fact that the definition of curl I know is only for 3 dimensional vector field.

• Your field is the gradient of $e^{x \cos y}$, confirming it is conservative – kmm Jan 14 at 14:26
• yes, but how do I check if it is irrotational ? – NPLS Jan 14 at 14:27

You are right that the usual notion of curl is only defined for 3 dimensions. To prove a two-dimensional vector field is conservative, you can use Green's theorem which says that the following identity holds for a region $$D$$ bounded by a curve $$C$$

$$\oint (P\,dx+Q\,dy)=\iint _{D}\left({\frac {\partial Q}{\partial x}}-{\frac {\partial P}{\partial y}}\right)\,dx\,dy$$

A conservative vector field is a vector field which can be expressed as the gradient of a scalar function. Remember that a conservative vector field has the property that the result of a line integral is path independent, implying the line integral over a simple closed path is always zero (there are some subtleties involving non-simply connected spaces). Hence the left integral will be zero if and only if the vector field $$\vec{F} = P \hat{e}_x + Q \hat{e}_y$$ is conservative. So to check if a field is conservative, we can simply check if the integrand of integral on the right hand side is zero. This expression is the two-dimensional analog of curl.

$$curl \vec{F} = \frac {\partial Q}{\partial x}-\frac {\partial P}{\partial y}$$

This is the formula you wrote down. In two-dimensions, the curl is a scalar, not a vector field itself. To get a completely general description of the concept of curl, you need differential forms, but this a relatively complex matter.

Another way to see it is that your expression is the z-component of the 3-dimensional curl for a vector field which does not depend on the z-direction.

• So what I can say about the irrotationality of my vector field? Does that mean that conservativeness implies irrotationality? (in the domain of definition) – NPLS Jan 15 at 8:18
• It's irrotational if this two-dimensional analog of the curl vanishes. A conservative field will always be irrotational. To see this, consider a gradient $\vec{F} = (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})$ . The 2D curl will be $\frac{\partial }{\partial x}\frac{\partial f}{\partial y} - \frac{\partial }{\partial y}\frac{\partial f}{\partial x}$ which will always be zero as partial derivatives commute. So conservativeness implies irrotationality – kmm Jan 15 at 23:10