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Coordinated Multivariable Calculus

Section 3.2 Iterated Integrals

Recall that in Section 3.1 we defined the double integral of a continuous function f=f(x,y) over a rectangle R=[a,b]×[c,d] as a limit of double Riemann sums, but while this definition tells us exactly what a double integral is, it is not very helpful for determining the value of a double integral. Fortunately, there is a way to view a double integral as an iterated integral, which will make computations feasible in many cases.
The viewpoint of an iterated integral is closely connected to an important idea from single-variable calculus. When we studied solids of revolution, such as the one shown in Figure 3.2.1, we saw that in some circumstances we could slice the solid perpendicular to an axis and have each slice be approximately a circular disk. From there, we were able to find the volume of each disk, and then use an integral to add the volumes of the slices. In what follows, we are able to use single integrals to generalize this approach to handle even more general geometric shapes.
Figure 3.2.1. A solid of revolution.

Preview Activity 3.2.1.

Let f(x,y)=25x2y2 on the rectangular domain R=[3,3]×[4,4].
As with partial derivatives, we may treat one of the variables in f as constant and think of the resulting function as a function of a single variable. Now we investigate what happens if we integrate instead of differentiate.
  1. Choose a fixed value of x in the interior of [3,3]. Let
    A(x)=44f(x,y)dy.
    What is the geometric meaning of the value of A(x) relative to the surface defined by f. (Hint: Think about the trace determined by the fixed value of x, and consider how A(x) is related to the image at left in Figure 3.2.2.)
    Figure 3.2.2. Left: A cross section with fixed x. Right: A cross section with fixed x and Δx.
  2. For a fixed value of x, say xi, what is the geometric meaning of A(xi) Δx? (Hint: Consider how A(xi)Δx is related to the image at right in Figure 3.2.2.)
  3. Since f is continuous on R, we can define the function A=A(x) at every value of x in [3,3]. Now think about subdividing the x-interval [3,3] into m subintervals, and choosing a value xi in each of those subintervals. What will be the meaning of the sum i=1mA(xi) Δx?
  4. Explain why 33A(x)dx will determine the exact value of the volume under the surface z=f(x,y) over the rectangle R.

Subsection 3.2.1 Iterated Integrals

The ideas that we explored in Preview Activity 3.2.1 work more generally and lead to the idea of an iterated integral. Let f be a continuous function on a rectangular domain R=[a,b]×[c,d], and let
A(x)=cdf(x,y)dy.
The function A=A(x) determines the value of the cross sectional area (by area we mean “signed” area) in the y direction for the fixed value of x of the solid bounded between the surface defined by f and the xy-plane.
Figure 3.2.3. Summing volumes of cross section slices.
The value of this cross sectional area is determined by the input x in A. Since A is a function of x, it follows that we can integrate A with respect to x. In doing so, we use a partition of [a,b] and make an approximation to the integral given by
abA(x)dxi=1mA(xi)Δx,
where xi is any number in the subinterval [xi1,xi]. Each term A(xi)Δx in the sum represents an approximation of a fixed cross sectional slice of the surface in the y direction with a fixed width of Δx as illustrated in Figure 3.2.3. We add the signed volumes of these slices as shown in the frames in Figure 3.2.3 to obtain an approximation of the total signed volume.
As we let the number of subintervals in the x direction approach infinity, we can see that the Riemann sum i=1mA(xi)Δx approaches a limit and that limit is the sum of signed volumes bounded by the function f on R. Therefore, since A(x) is itself determined by an integral, we have
Rf(x,y)dA=limmi=1mA(xi)Δx=abA(x)dx=ab(cdf(x,y)dy)dx.
Hence, we can compute the double integral of f over R by first integrating f with respect to y on [c,d], then integrating the resulting function of x with respect to x on [a,b]. The nested integral
ab(cdf(x,y)dy)dx=abcdf(x,y)dydx
is called an iterated integral, and we see that each double integral may be represented by two single integrals.
We made a choice to integrate first with respect to y. The same argument shows that we can also find the double integral as an iterated integral integrating with respect to x first, or
Rf(x,y)dA=cd(abf(x,y)dx)dy=cdabf(x,y)dxdy.
The fact that integrating in either order results in the same value is known as Fubini’s Theorem.

Fubini’s Theorem.

If f=f(x,y) is a continuous function on a rectangle R=[a,b]×[c,d], then
Rf(x,y)dA=cdabf(x,y)dxdy=abcdf(x,y)dydx.
Fubini’s theorem enables us to evaluate iterated integrals without resorting to the limit definition. Instead, working with one integral at a time, we can use the Fundamental Theorem of Calculus from single-variable calculus to find the exact value of each integral, starting with the inner integral.

Activity 3.2.2.

Let f(x,y)=25x2y2 on the rectangular domain R=[3,3]×[4,4].
  1. Viewing x as a fixed constant, use the Fundamental Theorem of Calculus to evaluate the integral
    A(x)=44f(x,y)dy.
    Note that you will be integrating with respect to y, and holding x constant. Your result should be a function of x only.
  2. Next, use your result from (a) along with the Fundamental Theorem of Calculus to determine the value of 33A(x)dx.
  3. What is the value of Rf(x,y)dA? What are two different ways we may interpret the meaning of this value?

Example 3.2.4.

Integrate the function xexy over the rectangle [0,1]×[0,1] by first writing it as an iterated integral. Does one order of integration offer any advantages over the other?
Solution.
Notice that if we set this up as a dxdy integral, we’d have to integrate by parts:
y=01x=01xexydxdy
So let’s try the other order instead.
[0,1]×[0,1]xexydA=x=01[y=01xexydy]dx=x=01[xxexy|y=01]dx=x=01ex1dx=e2

Example 3.2.5.

Let f(x,y)=x+y2 on the rectangle R=[0,2]×[1,3].
  1. Evaluate Rf(x,y)dA using an iterated integral. Choose an order for integration by deciding whether you want to integrate first with respect to x or y.
  2. Evaluate Rf(x,y)dA using the iterated integral whose order of integration is the opposite of the order you chose in (a).
Solution.
  1. We’ll choose to integrate with respect to y first:
    [0,2]×[1,3]x+y2dA=x=02y=13x+y2dydx=x=02[xy+y33|y=13]dx=x=02(3x+273x13)dx=x=022x263dx=222632=403
  2. By Fubini’s Theorem, you should get the same answer by integrating in the opposite order.

Subsection 3.2.2 Summary

  • We can evaluate the double integral Rf(x,y)dA over a rectangle R=[a,b]×[c,d] as an iterated integral in one of two ways:
    • -.
      ab(cdf(x,y)dy)dx, or
    • -.
      cd(abf(x,y)dx)dy.
    This process works because each inner integral represents a cross-sectional (signed) area and the outer integral then sums all of the cross-sectional (signed) areas. Fubini’s Theorem guarantees that the resulting value is the same, regardless of the order in which we integrate.

Exercises 3.2.3 Exercises

1.

Evaluate the iterated integral 02034x2y3dxdy

2.

Evaluate the iterated integral 3423(x+y)2dydx

3.

Find 410511(x+lny)dydx

4.

Find 4859xyex+ydydx

5.

Calculate the double integral R(10x+4y+40)dA where R is the region: 0x2,0y5.

6.

Calculate the double integral Rxcos(x+y)dA where R is the region: 0xπ3,0yπ2

7.

Consider the solid that lies above the square (in the xy-plane) R=[0,2]×[0,2],
and below the elliptic paraboloid z=25x23y2.
(A) Estimate the volume by dividing R into 4 equal squares and choosing the sample points to lie in the lower left hand corners.
(B) Estimate the volume by dividing R into 4 equal squares and choosing the sample points to lie in the upper right hand corners..
(C) What is the average of the two answers from (A) and (B)?
(D) Using iterated integrals, compute the exact value of the volume.

8.

If 32f(x)dx=2 and 13g(x)dx=4, what is the value of Df(x)g(y)dA where D is the rectangle: 3x2,  1y3?

9.

Find the average value of f(x,y)=3x6y5 over the rectangle R with vertices (2,0),(2,2),(2,0),(2,2).
Average value =

10.

Find the average value of f(x,y)=5eyx+ey over the rectangle R=[0,4]×[0,5].
Average value =

11.

Evaluate each of the following double or iterated integrals exactly.
  1. 13(25xydy)dx
  2. 0π/4(0π/3sin(x)cos(y)dx)dy
  3. 01(01e2x3ydy)dx
  4. R2x+5ydA, where R=[0,2]×[0,3].

12.

The temperature at any point on a metal plate in the xy plane is given by T(x,y)=1004x2y2, where x and y are measured in inches and T in degrees Celsius. Consider the portion of the plate that lies on the rectangular region R=[1,5]×[3,6].
  1. Write an iterated integral whose value represents the volume under the surface T over the rectangle R.
  2. Evaluate the iterated integral you determined in (a).
  3. Find the area of the rectangle, R.
  4. Determine the exact average temperature, TAVG(R), over the region R.

13.

Consider the box with a sloped top that is given by the following description: the base is the rectangle R=[1,4]×[2,5], while the top is given by the plane z=p(x,y)=30x2y.
  1. Write an iterated integral whose value represents the volume under p over the rectangle R.
  2. Evaluate the iterated integral you determined in (a).
  3. What is the exact average value of p over R?
  4. If you wanted to build a rectangular box (with an identical base) that has the same volume as the box with the sloped top described here, how tall would the rectangular box have to be?