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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 => $\dfrac{dQ}{dx} - \dfrac{dP}{dy}=0$ (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.

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  • $\begingroup$ Your field is the gradient of $e^{x \cos y}$, confirming it is conservative $\endgroup$
    – Kasper
    Jan 14, 2019 at 14:26
  • $\begingroup$ yes, but how do I check if it is irrotational ? $\endgroup$
    – NPLS
    Jan 14, 2019 at 14:27

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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.

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  • $\begingroup$ So what I can say about the irrotationality of my vector field? Does that mean that conservativeness implies irrotationality? (in the domain of definition) $\endgroup$
    – NPLS
    Jan 15, 2019 at 8:18
  • $\begingroup$ 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 $\endgroup$
    – Kasper
    Jan 15, 2019 at 23:10

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