# Line/Curve Integral Question

1. May 2, 2012

### nayfie

Hey guys. I've just had a few lectures on line integrals. My lecturer has told me the following:

$\int_{C}^{} f(s) \text{d}s = \int_{a}^{b} f(\gamma(t))\|\gamma'(t)\| \text{d}t$

Unfortunately he hasn't explained this topic very well. I understand what's going on with a line integral but have these few questions:

- What is the 'ds' on the left hand side?
- How do we determine what gamma is?

2. May 2, 2012

### micromass

Staff Emeritus
The "ds" is just a notation to indicate the variable towards we integrate. You can change it by dt or du or whatever (although ds seems like the canonical choice).

The $\gamma$ is a parametrization of the curve. This is usually given, although not always.

For example, if C is a circle, then

$\gamma:[0,2\pi[\rightarrow \mathbb{R}^2:t\rightarrow (\cos(t),\sin(t))$

This is one of the many possible parametrizations of the circle.
Note that the integral does depend on the parametrization used: for example, if we decide to run C in the other direction, then the integral can turn up to be negative. Thus C should always come with some kind of orientation if we want to find a right parametrization.

3. May 2, 2012

### nayfie

Can you describe what you mean by a parametrization? In your example you've used a range, but in an example we did in class the lecturer used the function of the curve.

I'm confused :(

4. May 2, 2012

### HallsofIvy

Staff Emeritus
"ds" is the "differential of arc length". A curve, being one dimensional, can be expressed in terms of one variable. For example, a curve in 3 dimensional space, given an ordinary xyz-coordinate system, can be written x= f(t), y= g(t), z= h(t). If you like, you think of t as "time" and those three equations give the position of an object moving along that curve.

If we have two points, say $(x_0, y_0, z_0)$ and $(x_1, y_1, z_1)$, then the straight line distance between the points is $\sqrt{(x_1-x_0)^2+ (y_1-y_0)^2+ (z_1-z_0)^2}$, from the Pythagorean theorem. For a curve, do the usual reduction to "differentials" or derivatives that you have surely seen in Calculus to get the "differential of arc length", $ds= \sqrt{dx^2+ dy^2+ dx^2}$ or, in terms of the parameter, t, $dx= \sqrt{(dx/dt)^2+ (dy/dt)^2+ (dz/dt)^2}dt$.

IF the curve, in two dimensions, is given as a function, y= f(x), you can use x itself as parameter. Formally, that would be "x= t, y= f(t)" so that the differential of arc length would be $ds= \sqrt{(dx/dt)^2+ (dy/dt)^2}dt= \sqrt{1+ (df/dt)^2}dt$ or, less formally, $ds= \sqrt{(1+ (dy/dx)^2}dx$.

Last edited: May 2, 2012
5. May 2, 2012