What do the symbols in the line integral equation mean?

[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?

 

[micromass]

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.

 

[nayfie]

Firstly, thanks for the reply.

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 :(

 

[HallsofIvy]

"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.

 

[CallMeShady]

Not sure about the second part but the "ds" basically means the derivative of variable "s." This vairable could have been any other letters, such as x - in that case, the derivate of "x" would be stated as "dx."

 

[lugita15]

Not sure about the second part but the "ds" basically means the derivative of variable "s." This vairable could have been any other letters, such as x - in that case, the derivate of "x" would be stated as "dx."

You mean differential, not derivative. Derivatives are things like dy/dx, and dx and dy are called differentials.