# Section13.1Iterated Integrals and Area¶ permalink

In Chapter 12 we found that it was useful to differentiate functions of several variables with respect to one variable, while treating all the other variables as constants or coefficients. We can integrate functions of several variables in a similar way. For instance, if we are told that $f_x(x,y) = 2xy\text{,}$ we can treat $y$ as staying constant and integrate to obtain $f(x,y)\text{:}$ \begin{align*} f(x,y) \amp = \int f_x(x,y)\ dx\\ \amp = \int 2xy\ dx\\ \amp = x^2y + C. \end{align*}

Make a careful note about the constant of integration, $C\text{.}$ This “constant” is something with a derivative of $0$ with respect to $x\text{,}$ so it could be any expression that contains only constants and functions of $y\text{.}$ For instance, if $f(x,y) = x^2y+ \sin(y) + y^3 + 17\text{,}$ then $f_x(x,y) = 2xy\text{.}$ To signify that $C$ is actually a function of $y\text{,}$ we write: \begin{equation*} f(x,y) = \int f_x(x,y)\ dx = x^2y+C(y). \end{equation*}

Using this process we can even evaluate definite integrals.

##### Example13.1.1Integrating functions of more than one variable

Evaluate the integral $\ds \int_1^{2y} 2xy\ dx\text{.}$

Solution

We can also integrate with respect to $y\text{.}$ In general, \begin{equation*} \int_{h_1(y)}^{h_2(y)} f_x(x,y)\ dx = f(x,y)\Big|_{h_1(y)}^{h_2(y)} = f\big(h_2(y),y\big)-f\big(h_1(y),y\big), \end{equation*} and \begin{equation*} \int_{g_1(x)}^{g_2(x)} f_y(x,y)\ dy = f(x,y)\Big|_{g_1(x)}^{g_2(x)} = f\big(x,g_2(x)\big)-f\big(x,g_1(x)\big). \end{equation*}

Note that when integrating with respect to $x\text{,}$ the bounds are functions of $y$ (of the form $x=h_1(y)$ and $x=h_2(y)$) and the final result is also a function of $y\text{.}$ When integrating with respect to $y\text{,}$ the bounds are functions of $x$ (of the form $y=g_1(x)$ and $y=g_2(x)$) and the final result is a function of $x\text{.}$ Another example will help us understand this.

##### Example13.1.2Integrating functions of more than one variable

Evaluate $\ds \int_1^x\big(5x^3y^{-3}+6y^2\big)\ dy\text{.}$

Solution

In the previous example, we integrated a function with respect to $y$ and ended up with a function of $x\text{.}$ We can integrate this as well. This process is known as iterated integration, or multiple integration.

##### Example13.1.3Integrating an integral

Evaluate $\ds \int_1^2\left(\int_1^x\big(5x^3y^{-3}+6y^2\big)\ dy\right)\ dx\text{.}$

Solution

The previous example showed how we could perform something called an iterated integral; we do not yet know why we would be interested in doing so nor what the result, such as the number $89/8\text{,}$ means. Before we investigate these questions, we offer some definitions.

##### Definition13.1.4Iterated Integration

Iterated integration is the process of repeatedly integrating the results of previous integrations. Integrating one integral is denoted as follows.

Let $a\text{,}$ $b\text{,}$ $c$ and $d$ be numbers and let $g_1(x)\text{,}$ $g_2(x)\text{,}$ $h_1(y)$ and $h_2(y)$ be functions of $x$ and $y\text{,}$ respectively. Then:

1. $\ds \int_c^d\int_{h_1(y)}^{h_2(y)} f(x,y)\ dx\ dy = \int_c^d\left(\int_{h_1(y)}^{h_2(y)} f(x,y)\ dx\right) dy\text{.}$

2. $\ds \int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)\ dy\ dx = \int_a^b\left(\int_{g_1(x)}^{g_2(x)} f(x,y)\ dy\right) dx\text{.}$

Again make note of the bounds of these iterated integrals.

With $\ds \int_c^d\int_{h_1(y)}^{h_2(y)} f(x,y)\ dx\ dy\text{,}$ $x$ varies from $h_1(y)$ to $h_2(y)\text{,}$ whereas $y$ varies from $c$ to $d\text{.}$ That is, the bounds of $x$ are curves, the curves $x=h_1(y)$ and $x=h_2(y)\text{,}$ whereas the bounds of $y$ are constants, $y=c$ and $y=d\text{.}$ It is useful to remember that when setting up and evaluating such iterated integrals, we integrate “from curve to curve, then from point to point.”

We now begin to investigate why we are interested in iterated integrals and what they mean.

# Subsection13.1.1Area of a plane region

Consider the plane region $R$ bounded by $a\leq x\leq b$ and $g_1(x)\leq y\leq g_2(x)\text{,}$ shown in Figure 13.1.5. We learned in Section 7.1 that the area of $R$ is given by \begin{equation*} \int_a^b \big(g_2(x)-g_1(x)\big)\ dx. \end{equation*}

We can view the expression $\big(g_2(x)-g_1(x)\big)$ as \begin{equation*} \big(g_2(x)-g_1(x)\big) = \int_{g_1(x)}^{g_2(x)} 1\ dy =\int_{g_1(x)}^{g_2(x)} \ dy, \end{equation*} meaning we can express the area of $R$ as an iterated integral: \begin{equation*} \text{ area of } R = \int_a^b \big(g_2(x)-g_1(x)\big)\ dx = \int_a^b\left(\int_{g_1(x)}^{g_2(x)} \ dy\right) dx =\int_a^b\int_{g_1(x)}^{g_2(x)} \ dy\ dx. \end{equation*}

In short: a certain iterated integral can be viewed as giving the area of a plane region.

A region $R$ could also be defined by $c\leq y\leq d$ and $h_1(y)\leq x\leq h_2(y)\text{,}$ as shown in Figure 13.1.6. Using a process similar to that above, we have \begin{equation*} \text{ the area of } R = \int_c^d\int_{h_1(y)}^{h_2(y)} \ dx\ dy. \end{equation*}

We state this formally in a theorem.

The following examples should help us understand this theorem.

##### Example13.1.8Area of a rectangle

Find the area $A$ of the rectangle with corners $(-1,1)$ and $(3,3)\text{,}$ as shown in Figure 13.1.9.

Solution
##### Example13.1.10Area of a triangle

Find the area $A$ of the triangle with vertices at $(1,1)\text{,}$ $(3,1)$ and $(5,5)\text{,}$ as shown in Figure 13.1.11.

Solution
##### Example13.1.12Area of a plane region

Find the area of the region enclosed by $y=2x$ and $y=x^2\text{,}$ as shown in Figure 13.1.13.

Solution

Changing Order of Integration

In each of the previous examples, we have been given a region $R$ and found the bounds needed to find the area of $R$ using both orders of integration. We integrated using both orders of integration to demonstrate their equality.

We now approach the skill of describing a region using both orders of integration from a different perspective. Instead of starting with a region and creating iterated integrals, we will start with an iterated integral and rewrite it in the other integration order. To do so, we'll need to understand the region over which we are integrating.

The simplest of all cases is when both integrals are bound by constants. The region described by these bounds is a rectangle (see Example 13.1.8), and so: \begin{equation*} \int_a^b\int_c^d 1\ dy\ dx = \int_c^d\int_a^b1\ dx\ dy. \end{equation*}

When the inner integral's bounds are not constants, it is generally very useful to sketch the bounds to determine what the region we are integrating over looks like. From the sketch we can then rewrite the integral with the other order of integration.

Examples will help us develop this skill.

##### Example13.1.14Changing the order of integration

Rewrite the iterated integral $\ds \int_0^6\int_0^{x/3} 1\ dy\ dx$ with the order of integration $dx\ dy\text{.}$

Solution
##### Example13.1.16Changing the order of integration

Change the order of integration of $\ds\int_0^4\int_{y^2/4}^{(y+4)/2}1\ dx\ dy\text{.}$

Solution

This section has introduced a new concept, the iterated integral. We developed one application for iterated integration: area between curves. However, this is not new, for we already know how to find areas bounded by curves.

In the next section we apply iterated integration to solve problems we currently do not know how to handle. The real" goal of this section was not to learn a new way of computing area. Rather, our goal was to learn how to define a region in the plane using the bounds of an iterated integral. That skill is very important in the following sections.

# Subsection13.1.2Exercises

Terms and Concepts

In the following exercises, evaluate the integral and subsequent iterated integral.

In the following exercises, a graph of a planar region $R$ is given. Give the iterated integrals, with both orders of integration $dy\ dx$ and $dx\ dy\text{,}$ that give the area of $R\text{.}$ Evaluate one of the iterated integrals to find the area.