14 Multiple Integration

14.1 Iterated Integrals and Area

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 fx(x,y)=2xy, we can treat y as staying constant and integrate to obtain f(x,y):

f(x,y) =fx(x,y)dx

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)=x2y+siny+y3+17, then fx(x,y)=2xy. To signify that C is actually a function of y, we write:


Using this process we can even evaluate definite integrals.

Example 14.1.1 Integrating functions of more than one variable

Evaluate the integral 12y2xydx.

SolutionWe find the indefinite integral as before, then apply the Fundamental Theorem of Calculus to evaluate the definite integral:

12y2xydx =x2y|12y

We can also integrate with respect to y. In general,




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.

Example 14.1.2 Integrating functions of more than one variable

Evaluate 1x(5x3y-3+6y2)dy.

SolutionWe consider x as staying constant and integrate with respect to y:

1x(5x3y-3+6y2)dy =(5x3y-2-2+6y33)|1x

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.

Example 14.1.3 Integrating an integral

Evaluate 12(1x(5x3y-3+6y2)dy)dx.

SolutionWe follow a standard “order of operations” and perform the operations inside parentheses first (which is the integral evaluated in Example 14.1.2.)

12(1x(5x3y-3+6y2)dy)dx =12([5x3y-2-2+6y33]|1x)dx

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.

Definition 14.1.1 Iterated Integration

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 g1(x), g2(x), h1(y) and h2(y) be functions of x and y, respectively. Then:

  1. (a)


  2. (b)


Again make note of the bounds of these iterated integrals.
With cdh1(y)h2(y)f(x,y)dxdy, x varies from h1(y) to h2(y), whereas y varies from c to d. That is, the bounds of x are curves, the curves x=h1(y) and x=h2(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.

Area of a plane region

Consider the plane region R bounded by axb and g1(x)yg2(x), shown in Figure 14.1.1. We learned in Section 6.1 that the area of R is given by

margin: Ry=g1(x)y=g2(x)abxy Figure 14.1.1: Calculating the area of a plane region R with an iterated integral. Λ

We can view the expression (g2(x)-g1(x)) as


meaning we can express the area of R as an iterated integral:

area of R=ab(g2(x)-g1(x))dx=ab(g1(x)g2(x)dy)dx=abg1(x)g2(x)dydx.

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 cyd and h1(y)xh2(y), as shown in Figure 14.1.2. Using a process similar to that above, we have

the area of R=cdh1(y)h2(y)dxdy.
margin: Rx=h1(y)x=h2(y)cdxy Figure 14.1.2: Calculating the area of a plane region R with an iterated integral. Λ

We state this formally in a theorem.

Theorem 14.1.1 Area of a plane region

(a) Let R be a plane region bounded by axb and g1(x)yg2(x), where g1 and g2 are continuous functions on [a,b]. The area A of R is A=abg1(x)g2(x)dydx. (b) Let R be a plane region bounded by cyd and h1(y)xh2(y), where h1 and h2 are continuous functions on [c,d]. The area A of R is A=cdh1(y)h2(y)dxdy.

The following examples should help us understand this theorem.

Example 14.1.4 Area of a rectangle

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.

margin: R-1123123xy Figure 14.1.3: Calculating the area of a rectangle with an iterated integral in Example 14.1.4. Λ

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


We could also integrate with respect to x first, giving:


Clearly there are simpler ways to find this area, but it is interesting to note that this method works.

Example 14.1.5 Area of a triangle

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: y=1y=2x-5y=xR4512312345xy Figure 14.1.4: Calculating the area of a triangle with iterated integrals in Example 14.1.5. 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=y+52, 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 =15yy+52dxdy

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 =131x1dydx +  352x-5x1dydx
=13(y)|1xdx +  35(y)|2x-5xdx
=13(x-1)dx +  35(-x+5)dx
=2 +  2

As expected, we get the same answer both ways. This equality will also be justified by Theorem 14.2.2 in the next section.

Example 14.1.6 Area of a plane region

Find the area of the region enclosed by y=2x and y=x2, as shown in Figure 14.1.5. margin: Ry=2xy=x2121234xy Figure 14.1.5: Calculating the area of a plane region with iterated integrals in Example 14.1.6. Λ

SolutionOnce again we’ll find the area of the region using both orders of integration.

Using dydx:


Using dxdy:


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 14.1.4), and so:


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.

Example 14.1.7 Changing the order of integration

Rewrite the iterated integral 060x/31dydx with the order of integration dxdy.

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.

margin: y=x/3R24612xy Figure 14.1.6: Sketching the region R described by the iterated integral in Example 14.1.7. Λ

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 023y61dxdy.

Example 14.1.8 Changing the order of integration

Change the order of integration of 04y2/4(y+4)/21dxdy.

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=y2/4 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.

margin: Rx=y2/4x=(y+4)/22424xy Figure 14.1.7: Drawing the region determined by the bounds of integration in Example 14.1.8. Λ

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 0x2, and y=2x-4 on 2x4. Thus we need two double integrals. The upper bound for each is y=2x. Thus we have


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.

Exercises 14.1


Terms and Concepts

  1. 1.

    When integrating fx(x,y) with respect to x, the constant of integration C is really which: C(x) or C(y)? What does this mean?

  2. 2.

    Integrating an integral is called                          .

  3. 3.

    When evaluating an iterated integral, we integrate from              to             , then from              to             .

  4. 4.
    One understanding of an iterated integral is that abg1(x)g2(x)dydx gives the              of a plane region.


In Exercises 5–10., evaluate the integral and subsequent iterated integral.

  1. 5.
    (a) 25(6x2+4xy-3y2)dy (b) -3-225(6x2+4xy-3y2)dydx
  2. 6.
    (a) 0π(2xcosy+sinx)dx (b) 0π/20π(2xcosy+sinx)dxdy
  3. 7.
    (a) 1x(x2y-y+2)dy (b) 021x(x2y-y+2)dydx
  4. 8.
    (a) yy2(x-y)dx (b) -11yy2(x-y)dxdy
  5. 9.
    (a) 0y(cosxsiny)dx (b) 0π0y(cosxsiny)dxdy
  6. 10.
    (a) 0x(11+x2)dy (b) 120x(11+x2)dydx

In Exercises 11–16., a graph of a planar region R is given. Give the iterated integrals, with both orders of integration dydx and dxdy, that give the area of R. Evaluate one of the iterated integrals to find the area.

  1. 11.
  2. 12.
  3. 13.
  4. 14.
  5. 15.
  6. 16.

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.

  1. 17.


  2. 18.


  3. 19.


  4. 20.


  5. 21.


  6. 22.


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