]> 23.4 Multiple Integrals and Area Integrals

## 23.4 Multiple Integrals and Area Integrals

When given an integral defined over area, the standard procedure for reducing it to ordinary integral it is to make it into a multiple integral.

A multiple integral is an expression involving two or more ordinary integrals, in which the limits of integration on one may depend on the values of the variables of integration of the next, so that the integrals may have to be performed in a specific order.

To reduce an area integral to a pair of multiple integrals you must perform the following steps:

1. Choose a set of variables that is convenient to you. This can include rectangular variables or polar coordinates or anything else. You choose them to make either the integrand, or the limits of integration that will be required (preferably both) as simple as possible; and choose an order in which you intend to define the integration in these variables. When you have no clue you can start with the obvious variables $x$ and $y$ .

2. Express the limits of integration appropriately so that the area you are integrating over is represented by the points within your limits of integration.

3. Express the element of area $d A$ in terms of your variables $u$ and $v$ and $d u$ and $d v$ .

4. Express the integrand in terms of your variables.

5. Integrate each of the ordinary integrals you have produced, if necessary in the appropriate order.

Example: we wish to integrate the function $exp ⁡ ( − r 2 )$ over the entire $x y$ plane.

One way to do this is with ordinary $x , y$ coordinates.

The limits of integration are then from $− ∞$ to $∞$ in each variable.

(This is an improper integral as the limits are infinite, and we have really defined integrals only over finite regions. However the integrand here is so well behaved for large values of any variables here that we can ignore this fact and imagine that we cut off the integrals at huge finite values. This will not change the answer significantly.)

The element of area $d A$ is $d x d y$ in rectangular coordinates.

The integrand can be expressed as $exp ⁡ ( − x 2 − y 2 )$ .

Our task then is to evaluate the product of two factors each one of which is the integral over all finite values of $exp ⁡ ( − x 2 )$ integrated $d x$ .

The answer is the square of $∫ exp ( − x 2 ) d x$ , where the limits of integration are $− R$ and $R$ for very large $R$ .

We can also perform this integration in polar coordinates.

Here $r$ goes from 0 to infinity, and $θ$ goes from 0 to $2 π$ .

The area element is the product of the line elements $d r$ and $r d θ$ , so that we have $d A = r d r d θ$ .

The integrand is $exp ⁡ ( − r 2 )$ .

Here the $θ$ integration is trivial and yields the value $2 π − 0$ or just $2 π$ .

We are left with the integral from 0 to infinity of $r * exp ⁡ ( − r 2 ) d r$ . If we set $u = r 2$ , we have $d u = 2 r d r$ , and the integrand $d r$ becomes $exp ⁡ ( − u ) d u 2$ and the limits of integration on $u$ are still 0 and infinity. The value of this integral is therefore $1 2$ , and the product of the two integrals here is just $π$ .

Notice that we have just proven that the integral of $exp ( − x 2 )$ over all $x$ values is $( π ) 1 / 2$ .

Of the steps required to evaluate integrals of this kind, the second and third merit further discussion.

The third, expressing the area element in your coordinate system, will be discussed in section 24.2 of the next chapter.

The second, finding the correct limits on your variables of integration in order to match the area you want to integrate over, is discussed in detail in section 31.5 .

The applet below allows you to choose limits and see what the resulting region in the plane looks like. A good way to get a feeling for this is to set up limits integrating first over one variable and then another, and then try to match it by integrating in the opposite order. You can even try to match a region given in rectangular coordinates by specifying polar variables or vice versa. Generally people screw this up the first time they try to do it, so practicing with it is probably worthwhile.