# Definition of curl

1. Dec 24, 2014

### albsanrom

Hi,

I'm trying to determine $\vec{\bigtriangledown }\times \vec{a}$ , where $\vec{a}=\vec{\omega }\times \vec{r}$, being $\vec{\omega }$ a constant vector, and $\vec{r}$ the position vector, using this definition:

$\vec{curl}(\vec{a})=\lim_{V\rightarrow 0}\frac{1}{V}\oint _{S}\vec{n}\times \vec{F}da$

where $\vec{n}$ is the unit vector perpendicular to the surface.

First of all, I know the answer is $2\vec{\omega }$. That's what I'm trying to get but I can't.

I've tried to do it with a sphere. So, $V=\frac{4}{3}\pi r^3$ , $\vec{n}=\vec{u_{r}}$ and $\vec{F}=\vec{a}$.
$\vec{\omega }$ is the angular velocity, so it's always perpendicular to $\vec{r}$, then $\vec{a}=\vec{\omega }\times \vec{r}=\omega r \vec{u_{\varphi}}$, because I have chosen $\vec{u_{\omega}}=−\vec{u_{\theta}}$. [$\theta$ is the zenith angle, and $\varphi$ is the azimuth angle.]
$da$ would be $r^2 sin\theta d\varphi d\theta$ .

So now I have:

$\vec{curl}(\vec{a})=\lim_{r\rightarrow 0}\frac{1}{\frac{4}{3}\pi r^3}\int_{0}^{\pi}\int_{0}^{2\pi} \omega r^3 sin\theta (\vec{u_{r}}\times \vec{u_\varphi)} d\varphi d\theta$

where $\vec{u_{r}} \times \vec{u_\varphi}=\vec{u_\omega}$.

The final result is $3\vec{u_\omega}$. What am I doing wrong?

Thanks in advance and Merry Christmas

2. Dec 24, 2014

### DivergentSpectrum

your definition for curl is wrong.

curl is defined as

where

is a closed LINE integral, and A is the surface area bound by the path.

edit: this is actually the absolute value of curl, while the curl vector points in the direction of the vector normal to A. Also, the line integral and enclosed surface are always oriented to give the maximum absolute value of the line integral

Last edited: Dec 24, 2014
3. Dec 28, 2014

### Cruikshank

Hello! My understanding of curl is that it is defined using a line integral around a flat area, not a surface integral around a volume. Divergence uses a surface integral around a volume. I hope this helps.

4. Jan 1, 2015

### albsanrom

My definition of curl comes from Foundations of Electromagnetic Theory (2nd edition) [John R. Reitz, Frederick J. Milford]:

"The curl of a vector, written $\mathbf{curlF}$, is defined as follows:

The curl of a vector is the limit of the ratio of the integral of its cross product with the outward drawn normal, over a closed surface, to the volume enclosed by the surface as the volume goes to zero. That is,

$\mathbf{curlF}=\lim_{V\rightarrow 0}\frac{1}{V}\oint_{S}\mathbf{n}\times \mathbf{F}da$

(...) This definition is convenient for finding the explicit form of the curl in various coordinate systems; however, for other purposes a different but equivalent definition is more useful. This definition is:

The component of $\mathbf{curlF}$ in the direction of the unit vector $\mathbf{a}$ is the limit of a line integral per unit area, as the enclosed area goes to zero, this area being perpendicular to $\mathbf{a}$. That is,

$\mathbf{a}\cdot\mathbf{curlF}=\lim_{S\rightarrow 0}\frac{1}{S}\oint_{C}\mathbf{F}\cdot d\mathbf{l}$ "

Following that, they proceed to show why the two definitions are equivalent.

I don't understand why I can't use the first definition as I have used it to do the calculations... and don't know how to apply the second one.

5. Jan 2, 2015

### Cruikshank

Hello again! I've spent a while working on this. You are correct that those definitions are equivalent; I was rusty on the details.
omega vector x r vector = omega r sin theta phi unit vector. You forgot the sin theta; omega vector and r vector are not perpendicular.
Also remember that the unit vector in the theta direction is a variable that has to be integrated as well, you can't pull it out of the integral.
As for applying the second one, take a disk in the x-y plane, integrate around it, divide by pi r^2 and you get the z component in about
four lines. I assume you would integrate over disks in the other planes, or rectangles or whatever is convenient on those axes, to get
that the other components are zero.
Also, cylindrical coordinates might give you fewer headaches than spherical, doing it the volume way.
I hope this helps. Good luck!