# Vector line integral notation.

1. Mar 10, 2013

### Craptola

Hey, I'm studying for a physics degree and have a general curiosity about vector calculus. Having learned about surface and line integrals for scalar functions in multivariable calculus I've been having some issues translating them into vector calculus. Though conceptually I haven't had much trouble yet I find myself struggling to interpret some notation.

My main concern concerns have been with $\vec{dl}$ (I've sometimes seen it written $\vec{dr}$) and $\vec{ds}$. I've encountered $dl$ as a scalar when doing line integrals but not as a vector. After much searching I was able to discover that the vector $\vec{ds}$ is equal to $ds\mathbf{\hat{n}}$ where n is the unit vector normal to the surface. But I've still not been able to find such a definition for $\vec{dl}$. I would appreciate if anyone could shed some light on what this actually is.

2. Mar 10, 2013

### vela

Staff Emeritus
Hmm, can you give an example of a line integral where you didn't have something like $d\vec{l}$ or $d\vec{r}$?

There's nothing too mysterious going on here: $d\vec{r} = dx\,\hat{i} + dy\,\hat{j} + dz\,\hat{k}$. If you have a vector field $\vec{F} = F_x\,\hat{i} + F_y\,\hat{j} + F_z\,\hat{k}$, then you get $\vec{F}\cdot d\vec{r} = F_x\,dx + F_y\,dy + F_z\,dz$

3. Mar 10, 2013

### LCKurtz

Suppose you have a force field $\vec F(x,y,z)$ and a curve parameterized by $\vec r(t) =\langle x(t),y(t),z(t)\rangle,\ a\le t \le b$. Since $\frac {d\vec r}{dt}$ is parallel to the curve, if you want to calculate the work done by the force moving along the curve you would calculate the integral$$W=\int_a^b \vec F(x(t),y(t),z(t))\cdot \frac{d\vec r}{dt}\, dt =\int_a^b \vec F(x(t),y(t),z(t))\cdot \langle \frac{dx}{dt},\frac {dy}{dt}\frac{dz}{dt}\rangle\, dt$$
This is sometimes written in the differential form, using $\frac{d\vec r}{dt}dt=d\vec r$ and $\frac{dx}{dt}dt = dx$ etc, as $\int_C\vec F \cdot d\vec r$. You can think of $d\vec r = \langle dx, dy,dz\rangle$. Whatever notation you use, remember that it means $\int_a^b\vec F \cdot \frac {d\vec r}{dt}\, dt$.
For surface integrals, it is a good idea to use capital "$S$" as in $d\vec S = \hat n\cdot dS$ so as not to confuse arc length notation with surface area notation.

4. Mar 10, 2013

### Craptola

One of the example questions I have is.
Evaluate $\oint \vec{a}\cdot \vec{dl}$ around the circle $x^{2} +y^{2}=b^{2}$ for $\vec{a}=\frac{\vec{r}}{r^{3}}$. Where r has its usual meaning is spherical polars.

It's nothing complicated I was just never taught what $\vec{dl}$ actually is when represented as a vector, and can't find it anywhere in my notes.

My first instinct was to assume that since $\vec{ds}$ is just $ds\mathbf{\hat{n}}$ then $\vec{dl}$ would just be a scalar line element (which in polars I figured would be $rd\phi$) multiplied by a vector normal to the contour, ie $\mathbf{\hat{r}}$ but after I did the line integral I tried to verify it using stokes theorem only to find that the curl of a is zero. I'm almost certain this is because of the assumption I made on the definition of the vector dl.

5. Mar 10, 2013

### vela

Staff Emeritus
You can think of it similarly to $d\vec{S}$ except that $d\vec{l}$ is tangent, not normal, to the contour and has magnitude ds. It's a little piece of the contour.