In Chapter 13 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 ${f}_{x}(x,y)=2xy$, we can treat $y$ as staying constant and integrate to obtain $f(x,y)$:
$f(x,y)$ | $={\displaystyle \int {f}_{x}(x,y)\mathrm{d}x}$ | ||
$={\displaystyle \int 2xy\mathrm{d}x}$ | |||
$={x}^{2}y+C.$ |
Make a careful note about the constant of integration, $C$. This “constant” is something with a derivative of $0$ with respect to $x$, so it could be any expression that contains only constants and functions of $y$. For instance, if $f(x,y)={x}^{2}y+\mathrm{sin}y+{y}^{3}+17$, then ${f}_{x}(x,y)=2xy$. To signify that $C$ is actually a function of $y$, we write:
$$f(x,y)=\int {f}_{x}(x,y)\mathrm{d}x={x}^{2}y+C(y).$$ |
Using this process we can even evaluate definite integrals.
Evaluate the integral ${\int}_{1}^{2y}}2xy\mathrm{d}x$.
SolutionWe find the indefinite integral as before, then apply the Fundamental Theorem of Calculus to evaluate the definite integral:
${\int}_{1}^{2y}}2xy\mathrm{d}x$ | $={{x}^{2}y|}_{1}^{2y}$ | ||
$={(2y)}^{2}y-{(1)}^{2}y$ | |||
$=4{y}^{3}-y.$ |
We can also integrate with respect to $y$. In general,
$${\int}_{{h}_{1}(y)}^{{h}_{2}(y)}{f}_{x}(x,y)\mathrm{d}x={f(x,y)|}_{{h}_{1}(y)}^{{h}_{2}(y)}=f({h}_{2}(y),y)-f({h}_{1}(y),y),$$ |
and
$${\int}_{{g}_{1}(x)}^{{g}_{2}(x)}{f}_{y}(x,y)\mathrm{d}y={f(x,y)|}_{{g}_{1}(x)}^{{g}_{2}(x)}=f(x,{g}_{2}(x))-f(x,{g}_{1}(x)).$$ |
Note that when integrating by $x$, the bounds do not depend on $x$, and the result is no longer a function of $x$. When integrating by $y$, the bounds do not depend on $y$, and the result is no longer a function of $y$. Another example will help us understand this.
Evaluate ${\int}_{1}^{x}}\left(5{x}^{3}{y}^{-3}+6{y}^{2}\right)\mathrm{d}y$.
SolutionWe consider $x$ as staying constant and integrate with respect to $y$:
${\int}_{1}^{x}}\left(5{x}^{3}{y}^{-3}+6{y}^{2}\right)\mathrm{d}y$ | $={\left({\displaystyle \frac{5{x}^{3}{y}^{-2}}{-2}}+{\displaystyle \frac{6{y}^{3}}{3}}\right)|}_{1}^{x}$ | ||
$=\left(-{\displaystyle \frac{5}{2}}{x}^{3}{x}^{-2}+2{x}^{3}\right)-\left(-{\displaystyle \frac{5}{2}}{x}^{3}+2\right)$ | |||
$={\displaystyle \frac{9}{2}}{x}^{3}-{\displaystyle \frac{5}{2}}x-2.$ |
Note how the bounds of the integral are from $y=1$ to $y=x$ and that the final answer is a function of $x$.
In the previous example, we integrated a function with respect to $y$ and ended up with a function of $x$. We can integrate this as well. This process is known as iterated integration, or multiple integration.
Evaluate ${\int}_{1}^{2}}\left({\displaystyle {\int}_{1}^{x}}\left(5{x}^{3}{y}^{-3}+6{y}^{2}\right)\mathrm{d}y\right)\mathrm{d}x$.
SolutionWe follow a standard “order of operations” and perform the operations inside parentheses first (which is the integral evaluated in Example 14.1.2.)
${\int}_{1}^{2}}\left({\displaystyle {\int}_{1}^{x}}\left(5{x}^{3}{y}^{-3}+6{y}^{2}\right)\mathrm{d}y\right)\mathrm{d}x$ | $={\displaystyle {\int}_{1}^{2}}\left({\left[{\displaystyle \frac{5{x}^{3}{y}^{-2}}{-2}}+{\displaystyle \frac{6{y}^{3}}{3}}\right]|}_{1}^{x}\right)\mathrm{d}x$ | ||
$={\displaystyle {\int}_{1}^{2}}\left({\displaystyle \frac{9}{2}}{x}^{3}-{\displaystyle \frac{5}{2}}x-2\right)\mathrm{d}x$ | |||
$={\left({\displaystyle \frac{9}{8}}{x}^{4}-{\displaystyle \frac{5}{4}}{x}^{2}-2x\right)|}_{1}^{2}$ | |||
$={\displaystyle \frac{89}{8}}.$ |
Note how the bounds of $x$ were $x=1$ to $x=2$ and the final result was a number.
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 $89/8$, means. Before we investigate these questions, we offer some definitions.
Iterated integration is the process of repeatedly integrating the results of previous integrations. Integrating one integral is denoted as follows.
Let $a$, $b$, $c$ and $d$ be numbers and let ${g}_{1}(x)$, ${g}_{2}(x)$, ${h}_{1}(y)$ and ${h}_{2}(y)$ be functions of $x$ and $y$, respectively. Then:
${\int}_{c}^{d}}{\displaystyle {\int}_{{h}_{1}(y)}^{{h}_{2}(y)}}f(x,y)\mathrm{d}x\mathrm{d}y={\displaystyle {\int}_{c}^{d}}\left({\displaystyle {\int}_{{h}_{1}(y)}^{{h}_{2}(y)}}f(x,y)\mathrm{d}x\right)\mathrm{d}y.$
${\int}_{a}^{b}}{\displaystyle {\int}_{{g}_{1}(x)}^{{g}_{2}(x)}}f(x,y)\mathrm{d}y\mathrm{d}x={\displaystyle {\int}_{a}^{b}}\left({\displaystyle {\int}_{{g}_{1}(x)}^{{g}_{2}(x)}}f(x,y)\mathrm{d}y\right)\mathrm{d}x.$
Again make note of the bounds of these iterated integrals.
With ${\int}_{c}^{d}}{\displaystyle {\int}_{{h}_{1}(y)}^{{h}_{2}(y)}}f(x,y)\mathrm{d}x\mathrm{d}y$, $x$ varies from ${h}_{1}(y)$ to ${h}_{2}(y)$, whereas $y$ varies from $c$ to $d$. That is, the bounds of $x$ are curves, the curves $x={h}_{1}(y)$ and $x={h}_{2}(y)$, whereas the bounds of $y$ are constants, $y=c$ and $y=d$. It is useful to remember that
after integrating with respect to a variable, that variable is no longer present.
We now begin to investigate why we are interested in iterated integrals and what they mean.
Consider the plane region $R$ bounded by $a\le x\le b$ and ${g}_{1}(x)\le y\le {g}_{2}(x)$, shown in Figure 14.1.1. We learned in Section 6.1 that the area of $R$ is given by
$${\int}_{a}^{b}\left({g}_{2}(x)-{g}_{1}(x)\right)\mathrm{d}x.$$ |
We can view the expression $\left({g}_{2}(x)-{g}_{1}(x)\right)$ as
$$\left({g}_{2}(x)-{g}_{1}(x)\right)={\int}_{{g}_{1}(x)}^{{g}_{2}(x)}1\mathrm{d}y={\int}_{{g}_{1}(x)}^{{g}_{2}(x)}\mathrm{d}y,$$ |
meaning we can express the area of $R$ as an iterated integral:
$$\text{area of}R={\int}_{a}^{b}\left({g}_{2}(x)-{g}_{1}(x)\right)\mathrm{d}x={\int}_{a}^{b}\left({\int}_{{g}_{1}(x)}^{{g}_{2}(x)}\mathrm{d}y\right)\mathrm{d}x={\int}_{a}^{b}{\int}_{{g}_{1}(x)}^{{g}_{2}(x)}\mathrm{d}y\mathrm{d}x.$$ |
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 $c\le y\le d$ and ${h}_{1}(y)\le x\le {h}_{2}(y)$, as shown in Figure 14.1.2. Using a process similar to that above, we have
$$\text{the area of}R={\int}_{c}^{d}{\int}_{{h}_{1}(y)}^{{h}_{2}(y)}\mathrm{d}x\mathrm{d}y.$$ |
We state this formally in a theorem.
(a)
Let $R$ be a plane region bounded by $a\le x\le b$ and ${g}_{1}(x)\le y\le {g}_{2}(x)$, where ${g}_{1}$ and ${g}_{2}$ are continuous functions on $[a,b]$. The area $A$ of $R$ is
$$A={\int}_{a}^{b}{\int}_{{g}_{1}(x)}^{{g}_{2}(x)}\mathrm{d}y\mathrm{d}x.$$
(b)
Let $R$ be a plane region bounded by $c\le y\le d$ and ${h}_{1}(y)\le x\le {h}_{2}(y)$, where ${h}_{1}$ and ${h}_{2}$ are continuous functions on $[c,d]$. The area $A$ of $R$ is
$$A={\int}_{c}^{d}{\int}_{{h}_{1}(y)}^{{h}_{2}(y)}\mathrm{d}x\mathrm{d}y.$$
The following examples should help us understand this theorem.
Find the area $A$ of the rectangle with corners $(-1,1)$ and $(3,3)$, as shown in Figure 14.1.3.
SolutionMultiple integration is obviously overkill in this situation, but we proceed to establish its use.
The region $R$ is bounded by $x=-1$, $x=3$, $y=1$ and $y=3$. Choosing to integrate with respect to $y$ first, we have
$$A={\int}_{-1}^{3}{\int}_{1}^{3}1\mathrm{d}y\mathrm{d}x={\int}_{-1}^{3}\left({y|}_{1}^{3}\right)\mathrm{d}x={\int}_{-1}^{3}2\mathrm{d}x={2x|}_{-1}^{3}=8.$$ |
We could also integrate with respect to $x$ first, giving:
$$A={\int}_{1}^{3}{\int}_{-1}^{3}1\mathrm{d}x\mathrm{d}y={\int}_{1}^{3}\left({x|}_{-1}^{3}\right)\mathrm{d}y={\int}_{1}^{3}4\mathrm{d}y={4y|}_{1}^{3}=8.$$ |
Clearly there are simpler ways to find this area, but it is interesting to note that this method works.
Find the area $A$ of the triangle with vertices at $(1,1)$, $(3,1)$ and $(5,5)$, as shown in Figure 14.1.4. ^{†}^{†}margin: Λ
SolutionThe triangle is bounded by the lines as shown in the figure. Choosing to integrate with respect to $x$ first gives that $x$ is bounded by $x=y$ to $x=\frac{y+5}{2}$, while $y$ is bounded by $y=1$ to $y=5$. (Recall that since $x$-values increase from left to right, the leftmost curve, $x=y$, is the lower bound and the rightmost curve, $x=(y+5)/2$, is the upper bound.) The area is
$A$ | $={\displaystyle {\int}_{1}^{5}}{\displaystyle {\int}_{y}^{\frac{y+5}{2}}}\mathrm{d}x\mathrm{d}y$ | ||
$={\displaystyle {\int}_{1}^{5}}\left({x|}_{y}^{\frac{y+5}{2}}\right)\mathrm{d}y$ | |||
$={\displaystyle {\int}_{1}^{5}}\left(-{\displaystyle \frac{1}{2}}y+{\displaystyle \frac{5}{2}}\right)\mathrm{d}y$ | |||
$={\left(-{\displaystyle \frac{1}{4}}{y}^{2}+{\displaystyle \frac{5}{2}}y\right)|}_{1}^{5}$ | |||
$=4.$ |
We can also find the area by integrating with respect to $y$ first. In this situation, though, we have two functions that act as the lower bound for the region $R$, $y=1$ and $y=2x-5$. This requires us to use two iterated integrals. Note how the $x$-bounds are different for each integral:
$A$ | $={\displaystyle {\int}_{1}^{3}}{\displaystyle {\int}_{1}^{x}}1\mathrm{d}y\mathrm{d}x$ | $+\mathit{\hspace{1em}\hspace{1em}}{\displaystyle {\int}_{3}^{5}}{\displaystyle {\int}_{2x-5}^{x}}1\mathrm{d}y\mathrm{d}x$ | ||
$={{\displaystyle {\int}_{1}^{3}}\left(y\right)|}_{1}^{x}\mathrm{d}x$ | $+\mathit{\hspace{1em}\hspace{1em}}{{\displaystyle {\int}_{3}^{5}}\left(y\right)|}_{2x-5}^{x}\mathrm{d}x$ | |||
$={\displaystyle {\int}_{1}^{3}}\left(x-1\right)\mathrm{d}x$ | $+\mathit{\hspace{1em}\hspace{1em}}{\displaystyle {\int}_{3}^{5}}\left(-x+5\right)\mathrm{d}x$ | |||
$=2$ | $+\mathit{\hspace{1em}\hspace{1em}}2$ | |||
$=4.$ |
As expected, we get the same answer both ways. This equality will also be justified by Theorem 14.2.2 in the next section.
Find the area of the region enclosed by $y=2x$ and $y={x}^{2}$, as shown in Figure 14.1.5. ^{†}^{†}margin: Λ
SolutionOnce again we’ll find the area of the region using both orders of integration.
Using $\mathrm{d}y\mathrm{d}x$:
$${\int}_{0}^{2}{\int}_{{x}^{2}}^{2x}1\mathrm{d}y\mathrm{d}x={\int}_{0}^{2}(2x-{x}^{2})\mathrm{d}x={\left({x}^{2}-\frac{1}{3}{x}^{3}\right)|}_{0}^{2}=\frac{4}{3}.$$ |
Using $\mathrm{d}x\mathrm{d}y$:
$${\int}_{0}^{4}{\int}_{y/2}^{\sqrt{y}}1\mathrm{d}x\mathrm{d}y={\int}_{0}^{4}(\sqrt{y}-y/2)\mathrm{d}y={\left(\frac{2}{3}{y}^{3/2}-\frac{1}{4}{y}^{2}\right)|}_{0}^{4}=\frac{4}{3}.$$ |
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 14.1.4), and so:
$${\int}_{a}^{b}{\int}_{c}^{d}1\mathrm{d}y\mathrm{d}x={\int}_{c}^{d}{\int}_{a}^{b}1\mathrm{d}x\mathrm{d}y.$$ |
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.
Rewrite the iterated integral ${\int}_{0}^{6}}{\displaystyle {\int}_{0}^{x/3}}1\mathrm{d}y\mathrm{d}x$ with the order of integration $\mathrm{d}x\mathrm{d}y$.
SolutionWe need to use the bounds of integration to determine the region we are integrating over.
The bounds tell us that $y$ is bounded by $0$ and $x/3$; $x$ is bounded by 0 and 6. We plot these four curves: $y=0$, $y=x/3$, $x=0$ and $x=6$ to find the region described by the bounds. Figure 14.1.6 shows these curves, indicating that $R$ is a triangle.
To change the order of integration, we need to consider the curves that bound the $x$-values. We see that the lower bound is $x=3y$ and the upper bound is $x=6$. The bounds on $y$ are $0$ to $2$. Thus we can rewrite the integral as ${\int}_{0}^{2}}{\displaystyle {\int}_{3y}^{6}}1\mathrm{d}x\mathrm{d}y$.
Change the order of integration of ${\int}_{0}^{4}}{\displaystyle {\int}_{{y}^{2}/4}^{(y+4)/2}}1\mathrm{d}x\mathrm{d}y$.
SolutionWe sketch the region described by the bounds to help us change the integration order. We see $x$ is bounded below and above (i.e., to the left and right) by $x={y}^{2}/4$ and $x=(y+4)/2$ respectively, and $y$ is bounded between 0 and 4. Graphing the previous curves, we find the region $R$ to be that shown in Figure 14.1.7.
To change the order of integration, we need to establish curves that bound $y$. The figure makes it clear that there are two lower bounds for $y$: $y=0$ on $0\le x\le 2$, and $y=2x-4$ on $2\le x\le 4$. Thus we need two double integrals. The upper bound for each is $y=2\sqrt{x}$. Thus we have
$${\int}_{0}^{4}{\int}_{{y}^{2}/4}^{(y+4)/2}1\mathrm{d}x\mathrm{d}y={\int}_{0}^{2}{\int}_{0}^{2\sqrt{x}}1\mathrm{d}y\mathrm{d}x+{\int}_{2}^{4}{\int}_{2x-4}^{2\sqrt{x}}1\mathrm{d}y\mathrm{d}x.$$ |
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.
When integrating ${f}_{x}(x,y)$ with respect to $x$, the constant of integration $C$ is really which: $C\left(x\right)$ or $C\left(y\right)$? What does this mean?
Integrating an integral is called .
When evaluating an iterated integral, we integrate from to , then from to .
In Exercises 5–10., evaluate the integral and subsequent iterated integral.
In Exercises 11–16., a graph of a planar region $R$ is given. Give the iterated integrals, with both orders of integration $\mathrm{d}y\mathrm{d}x$ and $\mathrm{d}x\mathrm{d}y$, that give the area of $R$. Evaluate one of the iterated integrals to find the area.
In Exercises 17–22., iterated integrals are given that compute the area of a region $R$ in the $x$-$y$ plane. Sketch the region $R$, and give the iterated integral(s) that give the area of $R$ with the opposite order of integration.
${\int}_{-2}^{2}}{\displaystyle {\int}_{0}^{4-{x}^{2}}}\mathrm{d}y\mathrm{d}x$
${\int}_{0}^{1}}{\displaystyle {\int}_{5-5x}^{5-5{x}^{2}}}\mathrm{d}y\mathrm{d}x$
${\int}_{-2}^{2}}{\displaystyle {\int}_{0}^{2\sqrt{4-{y}^{2}}}}\mathrm{d}x\mathrm{d}y$
${\int}_{-3}^{3}}{\displaystyle {\int}_{-\sqrt{9-{x}^{2}}}^{\sqrt{9-{x}^{2}}}}\mathrm{d}y\mathrm{d}x$
${\int}_{0}^{1}}{\displaystyle {\int}_{-\sqrt{y}}^{\sqrt{y}}}\mathrm{d}x\mathrm{d}y+{\displaystyle {\int}_{1}^{4}}{\displaystyle {\int}_{y-2}^{\sqrt{y}}}\mathrm{d}x\mathrm{d}y$
${\int}_{-1}^{1}}{\displaystyle {\int}_{\left(x-1\right)/2}^{\left(1-x\right)/2}}\mathrm{d}y\mathrm{d}x$