## Intro

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## Koncept

Do you remember single variable integrals? They measured the area under the curve. We had something one-dimensional (a curve), and we got something two-dimensional (an area).

Now we'll up our game. Double integrals are computed pretty much like normal integrals, but they represent the volume under the graph.

This should seem quite natural. I mean, we're taking into account another dimension, so we might as well expect something three-dimensional to pop out.

## Summering

When calculating double integrals we do exactly like regular integrals but first we do it along one axis and then the other, in practice this amounts to regarding the first integral as brackets.

## Double integral

Do you remember the intuitive idea behind calculating the following definite integral?

We can imagine this as calculating the area enclosed by the function's graph, the -axis, and two vertical lines placed at and .

Now for a function of two variables , the graph becomes a surface, and our domain do not lie on a single axis, but in the -plane.

Therefore, integrating yields the volume enclosed by the surface , the -plane, and some defined region in the -plane. Since we have two variables to integrate with respect to, it takes the form of a *double integral*:

The differential refers to a tiny area inside the region , for which multiplying by produces a narrow bar that is part of the enclosed volume. When adding up all such bars, we get an approximation for the total volume.

When painting a fence, it does not matter whether you stroke vertically or horizontally, the paint will cover the entire fence when finished. Similarly, we can add the bars row-wise or column-wise first.

However, just like stroking along the slats rather than across them is generally easier, it will prove extremely useful when it comes to solving double integrals that we can choose to integrate in the - or -direction first depending on the situation.

A double integral consist of two entangled integrals, and we can choose which one to evaluate first

We can think of the base of each bar as a rectangle with sides and . As we make these sides approach zero, we get the differentials and , and we express the area as . In such case, the previously approximate volume becomes perfect, and is given by the double integral

When we evaluate this integral, we need to dissect the region into conditions on and . Here, it is to our advantage that we can choose to sum row-wise or column-wise first as we like.

This corresponds to evaluating them with respect to or first. The only thing we need to beware of is that the inner limits of integration has to be expressed as functions of the outer variable.

The outer limits are simply start and end values, just like in regular integration.

### Example

Calculate the following integral

Note in the picture that the line restricting the region can be expressed either as giving an upper bound of , or as providing a lower bound of . The way we choose to view it depend on which integral we put as the inner one.

In this case, it will be a lot easier to solve if we change the direction of integration as follows

Using the variable substitution we get

which we can evaluate to

### Integrability on a region

Integrability on a region.

A function is said to be integrable on a region if

is closed and bounded,

The boundary of consist of finitely many curves of finite lengths, and

is continuous on

### Properties of double integrals

Properties of single integrals generally hold for double integrals as well, such as the fact that we can split the region of integration into two sub-regions and perform the integral piece-wise:

where

Another property of double integrals is that:

where is the area of the domain . This is because the volume it computes is a cylinder with base and height .

## Improper integrals

Just like in the case of single variable calculus, improper double integrals can arise if either the integration domain is unbounded, or the integrand is unbounded on the boundary of or inside the integration domain.

What is new here is that we have the two iterated integrals to evaluate. In opposition to double integrals that are not generalized, we may *not* just flip the order of integration. Doing that can give rise to conflicting results.

In improper double integrals, we may not change the integration order at our whim.

We also have to be careful with integrands that change sign inside the domain of integration. We'll see how to deal with this soon. For now, we'll just assume the integrand is positive on the whole integration domain.

A generalized integral can either *converge* to a value, or be divergent.

Suppose the inner double integral, say, with respect to y, converges for *any* fixed . Then, the double integral is convergent if and only if the outer integral with respect to is convergent.

If the integral is not convergent, we say it is divergent.

We can calculate generalized integrals analytically, as we will see in the two examples at the end of this note.

If only interested in convergence, we can also use comparison to determine if an integral converges or not. You may recognize this from single variable calculus.

Calling the domain of integration , if the following criterion is satisfied

then the integral

will converge.

The number can be any number for the criterion to work.

### Integrands that change sign

The rules and methods we have mentioned so far about generalized integrals only work if the integrand has the same sign on the whole domain. The reason for this is related to the infinity notion, which has to be treated with care. Negative and positive infinities may seem to cancel each other out, but that's not how it works.

If the integrand is negative on parts of the integration domain, we therefore split it into two:

and

Both of these functions are positive, and

The integral of is convergent if and only if the integrals of *and* converge. If the integral of converges, it has the value

### Example 1

Evaluate the integral

where is described by

The domain of is depicted in the figure below

The integral is improper since we are integrating over an unbounded domain. We will begin by integrating with respect to

The figure below shows the appearence of the integrand.

### Example 2

Evaluate the integral

where is described by

The figure below depicts the domain

The integral is improper since the integrand is not defined when . We start by integrating with respect to

We can see that this integral diverges for all , as is not defined. Thus, the integral is divergent. The figure below shows us the appearance of the integrand.

## Integration with polar coordinates

When motivating the arbitrary choice of inner and outer integrals in a double integral, we used the analogy of painting a fence with vertical or horizontal strokes depending on the particular fence.

Now for some fences, the most efficient way to paint them will be to throw away the paint brush and pick up the roller instead. In terms of integration, the roller corresponds to using *polar coordinates* as the variables.

Faced with a double integral whose region is bounded by a circle centered around the origin, the limits of integration expressed in terms of and are rather complicated, while in terms of and they become trivial.

simply ranges from to the radius of the circle, and from to .

What we need to be aware of when dealing with polar coordinates is that has a factor of built into it. It arises from the arc length , which we take to be the side of an approximately rectangular area , with as its other side:

The double integral in polar coordinates is then:

### Example 1

We will calculate

using Cartesian coordinates and polar coordinates to demonstrate the benefits of changing variables. We will start with Cartesian coordinates.

The bounds of integration will be

We choose to integrate with respect to first

Now let's calculate this integral using polar coordinates.

Let

This gives the following bounds of integrations

The integral then becomes

The takeaway here is that if the domain where we integrate over is axially symmetric (described by a circle for example), then we should integrate using polar coordinates.

### Example 2

The following is an important result in mathematics and it can be shown by using integration with polar coordinates:

Let be the improper integral, then

Thus

## Change of variables in integration

In this lecture note, we will return to the Jacobian matrix and its determinant to see how these concepts can be employed in practice to help us perform integration.

As it turns out, integrating using polar coordinates is just a special case of a *change of variables in integration*.

Such a change is always connected to a Jacobian matrix, regardless of whether we use it to make the substitution or not.

In the case of changing to polar coordinates, it looks as follows:

When we perform an integration with polar coordinates, picks up a factor of . This was shown intuitively using the formula for the area with one side being an arc with length .

Alternatively, we can motivate the emergence of this factor by recalling that the determinant of a matrix is the scaling factor of a linear transformation.

The Jacobian determinant for the transformation between coordinate systems (which is when going from Cartesian to polar coordinates) is what accounts for the extra factor in the integral.

### Example 1

Here, we shall integrate the following expression

Where is the region that is defined by

To integrate the above it is much easier using polar coordinates. First we remind our self that

we also know that:

Then we find that and then .

Next we must calculate the determinant of the Jacobi matrix,

Next we note that

and therefore we find that

### Changing to any variables

If the coordinate system we want to change our variables to are not the polar coordinates, but have some other variables and , the scaling factor will not be the same.

Instead we must find the Jacobian determinant of the particular Jacobian matrix corresponding to that variable change:

Let us see what this means in practice:

### Example 2

Find the area of the region in the figure below.

The area of is given by the integral

The problem will be easier if we switch variables to

The bounds of integration is given by the figure. Observe that

The integral is then given by

where

tells us how much the area is changing in this coordinate transformation. Notice how we haven't expressed and as functions of and , so it will be kinda tricky to calculate .

However, we can employ the following diabolic trick to calculate .

Going from the -coordinate system, we will scale our area by a factor of . Going back from the -coordinate system to the , we will scale our area by a factor of .

But since we returned to were we started, our area actually hasn't changed! This means that

Thus, we can calculate through .

Finally, we can calculate the integral