ITERATED INTEGRALS AND FUBINI’S THEOREM - Multiple Integrals - Advanced Calculus of Several Variables

Advanced Calculus of Several Variables (1973)

Part IV. Multiple Integrals

Chapter 4. ITERATED INTEGRALS AND FUBINI'S THEOREM

If f is a continuous function on the rectangle Image, then

Image

by Exercise 3.6. According to the main theorem of this section, the integral ∫Rf is equal to the common value in (1). Thus ∫Rf may be computed as the result of two “iterated” single-variable integrations (which are easy, using the fundamental theorem of calculus, provided that the necessary antiderivatives can be found). In a similar way, the integral over an interval in Imagen of a continuous function of n variables can be computed as the result of n iterated single-variable integrations (as we shall see).

In introductory calculus, the fact, that

Image

is usually presented as an application of a “volume by cross-sections” approach. Given Image, let

Image

so At is the “cross-section” in which the plane y = t intersects the ordinate set of f : RImage (see Fig. 4.19). One envisions the ordinate set of f as being swept

Image

Figure 4.19

out by At as t increases from c to d. Let A(t) denote the area of At, and V(t) the volume over [a, b] × [c, t] = Rt, that is,

Image

If one accepts the assertion that

Image

then (2) follows easily. Since

Image

(3) and the fundamental theorem of calculus give

Image

So

Image

because V(c) = 0 gives C = 0. Then

Image

as desired.

It is the fact, that V′(t) = A(t), for which a heuristic argument is sometimes given at the introductory level. To prove this, assuming that f is continuous on R, let

Image

Then φ and ψ are continuous functions of (x, h) and

Image

We temporarily fix t and h, and regard φ and ψ as functions of x, for x Image [a, b]. It is clear from the definitions of φ and ψ that the volume between At and At+h under z = f(x, y) contains the set

Image

and is contained in the set

Image

The latter two sets are both “cylinders” of height h, with volumes

Image

respectively. Therefore

Image

We now divide through by h and take limits as h → 0. Since Exercise 3.3 permits us to take the limit under the integral signs, substitution of (4) gives

Image

so V′(t) = A(t) as desired. This completes the proof of (2) if f is continuous on R.

However the hypothesis that f is continuous on R is too restrictive for most applications. For example, in order to reduce ∫Af to iterated integrals, we would choose an interval R containing the contented set A, and define

Image

Then ∫Af = ∫R g. But even if f is continuous on A, g will in general fail to be continuous at points of ∂A (see Fig. 4.20).

Image

Figure 4.20

The following version of what is known as “Fubini's Theorem” is sufficiently general for our purposes. Notice that the integrals in its statement exist by the remark that, if g : ImagekImage is integrable and Image is contented, then ∫C g exists, because the product of the integrable functions g and φC is integrable by Exercise 3.1.

Theorem 4.1 Let f : Imagem+n = Imagem × ImagenImage be an integrable function such that, for each Image, the function fx : ImagenImage, defined by fx(y) = f(x, y), is integrable. Given contented sets Image and Image, let F : ImagemImage be defined by

Image

Then F is integrable, and

Image

That is, in the usual notation,

Image

REMARK The roles of x and y in this theorem can, of course, be interchanged. That is,

Image

under the hypothesis that, for each Image, the function fy : ImagenImage, defined by fy(x) = f(x, y), is integrable.

PROOFWe remark first that it suffices to prove the theorem under the assumption that f(x, y) = 0 unless Image. For if f* = A × B, fx* = fx φB, and F* = ∫ fx*, then

Image

In essence, therefore, we may replace A, B, and A × B by Imagem, Imagen, and Imagem × n throughout the proof.

We shall employ step functions via Theorem 3.3. First note that, if φ is the characteristic function of an interval Image, then

Image

From this it follows that, if Image is a step function, then

Image

So the theorem holds for step functions.

Now, given ε > 0, let h and k be step functions such that Image and ∫ (h − k) < ε. Then for each x we have Image. Hence, if

Image

then H and K are step functions on Imagem such that Image and

Image

by the fact that we have proved the theorem for step functions. Since ε > 0 is arbitrary, Theorem 3.3 now implies that F is integrable, with Image. Since we now have ∫f and ∫ F both between ∫ h = ∫ H and ∫ k = ∫ K, and the latter integrals differ by < ε, it follows that

Image

This being true for all ε > 0, the proof is complete.

Image

The following two applications of Fubini's theorem are generalizations of methods often used in elementary calculus courses either to define or to calculate volumes. The first one generalizes to higher dimensions the method of “volumes by cross-sections.”

Theorem 4.2 (Cavalieri's Principle) Let A be a contented subset of Imagen+1, with Image, where Image and Image are intervals. Suppose

Image

is contented for each t Image [a, b], and write A(t) = v(At). Then

Image

PROOF(see Fig. 4.21).

Image

Image

Image

Figure 4.21

The most typical application of Cavalieri's principle is to the computation of volumes of revolution in Image3. Let f : [a, b] → Image be a positive continuous function. Denote by A the set in Image3 obtained by revolving about the x-axis the ordinate set of f (Fig. 4.22). It is clear that A(t) = π[f(t)]2, so Theorem 4.2 gives

Image

For example, to compute the volume of the 3-dimensional ball Br3 of radius r, take f(x) = (r2x2)1/2 on [−r, r]. Then we obtain

Image

Image

Figure 4.22

Theorem 4.3If Image is a contented set, and f1 and f2 are continuous functions on A such that Image, then

Image

is a contented set. If g : CImage is continuous, then

Image

PROOF The fact that C is contented follows easily from Corollary 2.3. Let B be a closed interval in Image containing Image, so that Image (see Fig. 4.23).

Image

Figure 4.23

If h gφc, then it is clear that the functions h : Imagen+1Image and hx : ImageImage are admissible, so Fubini's theorem gives

Image

because hx(y) = 1 if Image, and 0 otherwise.

Image

For example, if g is a continuous function on the unit ball Image, then

Image

where Image.

The case m = n = 1 of Fubini's theorem, with A = [a, b] and B = [c, d], yields the formula

Image

alluded to at the beginning of this section. Similarly, if f : QImage is continuous, and Q = [a1, b1] × · · · × [an, bn] is an interval in Imagen, then n − 1 applications of Fubini's theorem yield the formula

Image

This is the computational significance of Fubini's theorem—it reduces a multivariable integration over an interval in Imagen to a sequence of n successive single-variable integrations, in each of which the fundamental theorem of calculus can be employed.

If Q is not an interval, it may be possible, by appropriate substitutions, to “transform” ∫Q f to an integral ∫R g, where R is an interval (and then evaluate ∫R g by use of Fubini's theorem and the fundamental theorem of calculus). This is the role of the “change of variables formula” of the next section.

Exercises

4.1Let Image and Image be contented sets, and f : ImagemImage and g : ImagenImage integrable functions. Define h : Imagem + nImage by h(x, y) = f(x)g(y), and prove that

Image

Conclude as a corollary that v(A × B) = v(A)v(B).

4.2Use Fubini‘s theorem to give an easy proof that 2f/∂x ∂y = ∂2f/∂y ∂x if these second derivatives are both continuous. Hint: If D1 D2fD2 D1f > 0 at some point, then there is a rectangle R on which it is positive. However use Fubini’s theorem to calculate ∫R (D1 D2f − D2 D1f) = 0.

4.3Define f : I × IImage, I = {0, 1} by

Image

Then show that

(a) Image

(b) Image

(c) Image if y is irrational, but does not exist if y is rational.

4.4Let T be the solid torus in Image3 obtained by revolving the circle Image, in the yz-plane, about the z-axis. Use Cavalieri's principle to compute v(T) = 2π2ab2.

4.5Let S be the intersection of the cylinders Image and Image. Use Cavalieri's principle to compute Image.

4.6The area of the ellipse Image, with semiaxes a and b, is A = πab. Use this fact and Cavalieri's principle to show that the volume enclosed by the ellipsoid

Image

is Image. Hint: What is the area A(t) of the ellipse of intersection of the plane x = t and the ellipsoid? What are the semiaxes of this ellipse?

4.7Use the formula Image, for the volume of a 3-dimensional ball of radius r, and Cavalieri's principle, to show that the volume of the 4-dimensional unit ball Image is π2/2. Hint: What is the volume A(t) of the 3-dimensional ball in which the hyperplane x4 = t intersects B4?

4.8Let C be the 4-dimensional “solid cone” in Image4 that is bounded above by the 3-dimensional ball of radius a that is centered at (0, 0, 0, h) in the hyperplane x4 = h, and below by the conical “surface” x4 = (x12 + x22 + x22)1/2. Show that Image. Note that this is one-fourth times the height of the cone C times the volume of its base.