Recall that we defined the double integral of a continuous function over a rectangle as
where the different variables and notation are as described in Section 11.1. Thus is 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 11.2.1, 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.
As with partial derivatives, we may treat one of the variables in 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.
Choose a fixed value of in the interior of . Let
What is the geometric meaning of the value of relative to the surface defined by . (Hint: Think about the trace determined by the fixed value of , and consider how is related to the image at left in Figure 11.2.2.)
Figure11.2.2.Left: A cross section with fixed . Right: A cross section with fixed and .
Since is continuous on , we can define the function at every value of in . Now think about subdividing the -interval into subintervals, and choosing a value in each of those subintervals. What will be the meaning of the sum ?
The ideas that we explored in Preview Activity 11.2.1 work more generally and lead to the idea of an iterated integral. Let be a continuous function on a rectangular domain , and let
The function determines the value of the cross sectional area (by area we mean “signed” area) in the direction for the fixed value of of the solid bounded between the surface defined by and the -plane.
The value of this cross sectional area is determined by the input in . Since is a function of , it follows that we can integrate with respect to . In doing so, we use a partition of and make an approximation to the integral given by
where is any number in the subinterval . Each term in the sum represents an approximation of a fixed cross sectional slice of the surface in the direction with a fixed width of as illustrated in Figure 11.2.3. We add the signed volumes of these slices as shown in the frames in Figure 11.2.3 to obtain an approximation of the total signed volume.
As we let the number of subintervals in the direction approach infinity, we can see that the Riemann sum approaches a limit and that limit is the sum of signed volumes bounded by the function on . Therefore, since is itself determined by an integral, we have
Hence, we can compute the double integral of over by first integrating with respect to on , then integrating the resulting function of with respect to on . The nested integral
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 . The same argument shows that we can also find the double integral as an iterated integral integrating with respect to first, or
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.
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.
The temperature at any point on a metal plate in the plane is given by , where and are measured in inches and in degrees Celsius. Consider the portion of the plate that lies on the rectangular region .
Write an iterated integral whose value represents the volume under the surface over the rectangle .
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?