# $\nabla \times \bf{u} \neq 0$ but $\oint_{c} \bf{u} \cdot \textit{d}r \textit{=0}$?

Consider the vector field $$\vec{u}=(xy^2,x^2y,xyz^2)$$

The curl of the vector field is $$\nabla \times\vec{u}=(xz^2,-yz^2,0)$$ Consider the line integral of $$\vec{u}$$ around the ellipse $$C$$ $$x^2+4y^2=1, z=-1$$.

With $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a=1, b=\frac{1}{2}$$, the parameterisation gives $$\vec{r}=(x,y,z)=(cos\theta,\frac{1}{2}sin\theta,-1)$$ $$d\vec{r}=(-sin\theta,\frac{1}{2}cos\theta,0))$$

$$\vec{u}=(\frac{1}{4}cos\theta sin^2\theta,\frac{1}{2}cos^2\theta sin\theta,\frac{1}{2}sin\theta cos\theta)$$ $$\oint_{C} \vec{u} \cdot d\vec{r}=\frac{1}{4}\int^{2\pi}_0sin\theta cos\theta(cos^2\theta -sin^2\theta)d\theta=0$$ (I got the same result by working in cartesian cooridnates without parameterizing the curve)

But this does not make sense because $$\nabla\times \vec{u}=\lim_{\delta S \to 0}\frac{1}{\delta S}\oint_{\delta C}\vec{u} \cdot d\vec{r}$$ so if a vector field is conservative, its curl should be zero.

Can someone please explain where my conceptual errors lie?

• The left hand side of your last equation is a vector while the right hand side is a scalar. Jun 9, 2020 at 10:54

The path you have chosen is special in a way, for this vector field. The path is cylindrically symmetric. The fact that $$z$$ is constant in your path means that $$u_z$$ becomes irrelevant to the integral. What remains is $$(xy^2,yx^2)$$. This is odd in $$x,y$$. Thus if the path is symmetric with $$(x,y)\to(-x,-y)$$ (as is the same for cylindrically symmetric path), the contribution from the positive half of the path is cancelled by the negative half exactly.

But for a vector field to be conservative, the path integral must be independent for any path chosen. So if you choose a path that’s not cylindrically symmetric then you’ll get a non-zero value.

Can someone please explain where my conceptual errors lie?

OP's conceptual error lies in the preceding equation:

$$\nabla\times \vec{u}=\lim_{\delta S \to 0}\frac{1}{\delta S}\oint_{\delta C}\vec{u} \cdot d\vec{r} \qquad \color{red}{\text{Wrong!}}$$

The left hand side is a vector while the integral on the right hand side is a scalar.

The correct equation should be: $$( ∇ × \vec{u} ) (\color{green}{ p} ) \color{blue}{⋅ \vec n } = \lim _{δ S → 0} \frac{1}{ δ S } ∮_{δC} ⁡ \vec{u} ⋅ d \vec{r}$$

Note, that:

• We find this way a projection of curl at the specific point $$p$$ onto the normal vector $$\vec n$$. So if a curve lies in a plane $$z=\mathrm{const}$$ we can only obtain $$z$$-component of the curl vector.

• The limiting procedure means that in order to find curl through calculation of line integrals we must use families of curves contracting around a given point.

So one line integral around one ellipse of finite size is not enough to make conclusions about value of curl (or even its components) in any specific point.