In Lecture 2 we talked about one type of line integral, the work integral. In this lecture we cover two more line integrals: the Circulation integral and the flux integral.

where

This integral is exactly the same as the Work integral from

The velocity of a fluid perpendicular to a path is **F**(dot)**n**,
where **n** is the unit normal to the path C.

The flux integral is

Recall from before that **T** = ( dx/dt **i** +
dy/dt **j** ) / ||**v**|| = **v** / ||**v**||

`
n = T x k = ( ( dx/dt i + dy/dt j ) x
k ) / ||v||`

= ( dy/dt i - dx/dt j ) / ||v||

Alternatively, **n** can be found by inspection. For example, Given a
semicircle as below, we can easily pick out the normal vectors
indicated.

An alternative method of computing the flux integral is as follows:

Pictorially the flux can be looked at in the following way:

With fluid leaving the area enclosed by the path, the flux is > 0.

With fluid entering the area enclosed, the flux < 0

If fluid flow is uniform, as much fluid enters the area as exits, and the
flux is 0.

Example:

Find the Circulation and Flux for **F** = x**i** + y**j** and C is
a semicircle:

We can do the integral in two parts - over C1, the
circular portion of the path, and over C2, the flat
part of the curve along the x-axis.

Paramatrizing the Curves:

C1: **r**1(t) = a cos t **i**
+ a sin t **j**; 0<t<Pi

C2: **r**2(t) = at**i**;
-1<t<1

On C1: **T**1 = **v**(t) / ||**v**(t)|| = - a sin t**i** + a cos t
**j** / ||**v**||

**n**1 = a cos t **i** + a sin t **j** /
||**v**||

On C2: **T**2 = **i**

**n**2 = -**j**

Note that the circulation was zero, as expected for a radial field - there
is no flow around the path, and the flux was positive, as expected for a
field expanding outward from the area enclosed by the path.

Lecture written by Lawrence C. Weintraub on Wednesday, September 13, 1995

Edited by Aric Shafran on Thursday, September 14, 1995

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