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.
Best Answer
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.