## Intro

The concept of work is key in the study of mechanics and electromagnetism. Work is the amount of energy we need to put in, in order move an object along a certain path.

But how to know how much work we need to use? Line integrals, is the answer.

When bungee jumping off a cliff, we lose energy. This is because the field of gravity is doing work on *us*, pulling us in the direction of the field.

As we, all high on adrenaline and probably nauseous, are pulled back up, someone has to use energy to get us back to the platform. The amount of energy needed is the same as the energy we lost when jumping.

This energy can be calculated as a line integral, where we sum up the pulling force along the path we follow, as we are dangling from the elastic band slowly approaching the platform again after the jump.

This example comes from mechanics, but the notion of line integrals pops up in all of physics!

For example, when the folks at NASA plan the trajectory of a satellite, they need to consider work, in order to have it stay in orbit around the Earth.

## Concept

The work done when a particle travels through a vector field is calculated using a line integral. For example, the work done by you when you are climbing a hill is calculated with a line integral, since you are traveling through the Earth's gravitational field.

Another example is when you push a piece of iron towards a magnet. Here, the vector field is the magnetic field generated by the magnet and you can literally feel the work being done.

## Math

Math always have a lot of notation in order to write equation in a beautiful and efficient manner. Below we will introduce some new notation.

Where is the vector field and is a small change in the line . We can also write this integral as

Here and are the vectors magnitude at a point , and are the small changes of the vector components in . Another imporanant symbol that also look really cool is the following

This symbol simply means that our curve is a closed loop. For example the curve below is closed but the curve is not closed

## Line integrals over scalar fields

We use line integrals to determine the *total amount* of quantity along a curve, if we know the *density* at each point of the curve.

You know those video games, where you're running along a path collecting golden coins? That is what we are talking about.

We want to know how much gold we get, as we run from to along the path.

Summing all the coins along the path up is analogous to taking the line integral of a coin density function along the path. (We may imagine that the air along the path is full of little shiny coins, forming a golden cloud of varying density...)

Actually, we have already seen lots of line integrals:

This is a line integral: the curve along which we integrate is just a straight line from to .

We can think of the function as a line density associated with every point on the -axis. Then, the integral is the total quantity of that something is the density of.

Now, we want to do line integrals along curves in . This can be done by using what we know about curve parametrization. Let the curve be parameterized by:

We need to be able to express a small segment of the curve , in terms of and . It's necessary because we have expressed the curve in terms of . Thus, in order to be allowed to integrate, the length element need to live in the same coordinate space as the integrand.

We'll do the derivation of the expression for for a curve in . In , becomes the same as the result we'll get now, it's just a bit more heavy to derive.

First, let a small segment along the line be denoted as . Then, using Pythagoras theorem we get

Dividing by , we get

Thus, the curve length is

where .

This is a special case of line integrals along the curve , with the function value at each point is just . Say we assign to each point on the curve a value depending on where on the curve we are:

Then, multiplying with a tiny line segment and summing it all up, we get the line integral of along :

Note that the value of the line integral is independent of parametrization. If we wish, we may change to another coordinate system; we'd still get the same value of the integral.

### Example 1

Calculate the following line integral

were and is a line from to .

Let's start by parameterizing the line as

(Note that the integral is the same regardless of the parametrization). This gives:

so the integral becomes

### Example 2

Calculate the length of the curve parametrized by

*Solution:* The length of the curve is given by the integral

is given by

Thus,

Now, we can calculate our integral

## Line integrals over vector fields

If we let a rubber duck move freely at the surface of a bath tub, described by a vector field, it will follow the direction of the field in each point.

As Aristotle famously argued, nothing on earth moves without a mover, and in this case the mover is the field. But how much work is done by the field while moving the duck?

For those not so familiar with physics, work in this sense is a physical quantity measured in joules that is related to change in energy, and is given by the total force applied over a distance moved. To answer the question, we turn to line integrals.

Line integrals over vector fields aggregate the field strength in the direction of the given path

Only the field's component along the path we integrate over contributes, and if it is opposite in direction, the contribution there is negative. This won't happen in nature without some outside interaction, such as us moving the duck around, and the field then does negative work.

If the path we integrate a vector field over is a closed loop, meaning the start and end points are the same, we get a special case called *the circulation of around *, denoted by:

The process of parametrizing the curve and finding the differential is the same for vector fields as it is for line integrals over scalar fields.

However, since the function is vector valued, and we are interested in its component along the path of integration in each point, the integrand takes the form of a dot product between the vector field and the unit tangent vector of the curve:

Hence, the line integral of over the curve starting at and ending at is given by:

### Example

Let the force asserted on the duck by the vector field at each point on the surface, measured in meters, be given in Newtons as

How much work is done by the field as the duck is moved from to along the path given by ?

*Solution:*

The curve in the interval can be split into two pieces, parametrized individually as:

is then given by:

Further, the vector field in terms of becomes:

Then, the line integral evaluates to:

Hence, the answer comes out to around joule.

## Surface integrals

Remember how integrating over the -axis actually was our first line integral? The double integral as we know it is actually a surface integral, over a flat domain in the -plane:

As in the case of the line integral, we can see the integrand of the double integral as a function assigning some *surface density* to each point on the surface.

Surface density is amount per area. So the integral becomes the total quantity, as we multiply at each point with a tiny area , and sum all these quantities up by integration.

Let be the surface density, then the integral over the surface is the total amount.

In order to compute a surface integral over a surface with more personality than that flat fellow in the -plane, we need surface parametrization.

Recall that a surface being a -dimensional object, we can describe it using two coordinates :

Now, we need to write the surface element in terms of and , to have and the area element in the same coordinate system.

We go about it like this: let be the small surface element arising from taking a small step in the direction, and a small step in the direction, from the point .

Using the surface parametrization , the two tangent vectors and are parallel to the tangent plane at the point .

As the vectors are not parallel to *each other*, we can approximate as the area of the parallelogram with sides and .

Now, letting all these things become infinitesimal, we actually get an equivalence:

has the same area as the parallelogram.

That parallelogram area can be computed by taking the absolute value of the cross product between the vectors creating it. This is the geometrical interpretation of the cross product. Thus, we get:

So that's our expression for . If we want to calculate the area of a surface delimited by some domain in the -plane, find its corresponding domain in the -plane. Then:

This can be regarded as the special case where the aforementioned surface density is just . If the surface density varies according to some function

then we multiply the surface elements by this function, getting:

As always, the integral is independent of which parametrization we use.

## Surface orientation

Say we have a parameterized surface:

We call its positive side the side towards which its normal vector

points. The way we parameterize thus decides the orientation of the surface.

If we change the order of parametrization, the orientation changes as well.

We normally choose to parameterize closed surfaces so that the normal points out from it.

The orientation of a surface induces an orientation of all of its boundaries.

The orientation of the boundary is given by the right hand rule: point the thumb of your right hand in the direction of the normal.

Then, the orientation of the boundary curve is defined as the direction of your fingers.

## Flux integrals

Flux integrals are for calculating how much flux goes through a surface.

Flux is all over. For example, the water flowing through a fire hose has a flux. The surface in this case is the cross-section of the hose.

This highlights an important point: The surface we are talking about is *imaginary*. It is not a wall or filter blocking the flow. It's just the 'area' through which we'd like to know the flow.

The flux integral of a vector field looks like this:

The geometrical interpretation of this dot product is the projection of the field onto the normal vector of the surface .

If we imagine dividing into a component parallel to the surface and one perpendicular to it, the perpendicular component will be the one that flows through the surface. And that is the same as we get when projecting onto the normal vector.

This is the reason for why we have the dot product in the integral.

Note that . If the surface normal is not given, we can find it by taking the cross product of the partial derivative vectors with respect to the parameters.

If the surface is parameterized by

then, the normal vector is

### Example

Water is flowing at a constant rate through a pipe with a cross sectional area of . The water flow (length per unit time) is described by the vector field

and the wall of the pipe is given by

What amount of water is flowing through the pipe?

The flow of water through the pipe is the volume of water that is flowing through a cross section per second and is given by the following flux integral

The pipe can be thought of as a cylinder, and the cross section as a circle. The cross section can be parameterized using cylindrical coordinates in the following way

The normal vector of the cross section is . The integral becomes

When we evaluate the dot product in the integral before integration, we get a scalar value as a result. Thus, it doesn't matter that we actually wrote the vector field and normal vector using the Cartesian , and -coordinates, and not cylindrical coordinates.