PDF

Chapter 15

Multiple Integrals

15.1

Double Integrals over Rectangles, page 988

Review of the Definite Integral, page 988

Suppose that f (x) is defined for a ≤ x ≤ b.

(1) Divide the interval [a, b] into n subintervals [xi−1, xi] of equal width ∆x = b−a_n .

(2) In each subinterval [xi−1, xi], choose a sample point x ∗

i ∈ [xi−1, xi].

(3) Define the Riemann sum (黎曼和) =

n

P

i=1

f(x∗ i)∆x.

(4) Define the definite integral (定積分) of f (x) from a to b by Z b a f(x) dxdef.= lim n→∞ n X i=1 f(x∗ i)∆x. x y a b

Figure 1: Definition of a definite integral.



Volumes and Double Integrals, page 988

Similarly, we consider a function f (x, y) defined on a closed rectangle R _{= [a, b] × [c, d] = {(x, y) ∈ R}2|a ≤ x ≤ b, c ≤ y ≤ d},

and we first suppose that f (x, y) ≥ 0. The graph of f is a surface with equation z = f (x, y). Let S be the solid that lies above R and under the graph of f , that is

S _{= {(x, y, z) ∈ R}3

|0 ≤ z ≤ f (x, y), (x, y) ∈ R}, The goal is to find the volume of S.

(1) Divide the rectangle R into subrectangles:

– Divide [a, b] into m subinterval [xi−1, xi] with width ∆x = b−a_m .

– Divide [c, d] into n subinterval [yi−1, yi] with width ∆y = d−c_n .

– We form the subrectangles:

Rij = [xi−1, xi] × [yi−1, yi] = {(x, y)|xi−1≤ x ≤ xi, yj−1 ≤ y ≤ yj}.

Each Rij with area ∆A = ∆x∆y.

(2) Choose a sample point (x∗ ij, y

∗

ij) (樣本點) in each Rij.

(3) We get an approximation to the total volume of S by double Riemann sum (二重黎曼和): V ≈ m X i=1 n X j=1 f(x∗ ij, y ∗ ij)∆A.

(4) Define the volume (體積) of the solid S by

V = lim m,n→∞ m X i=1 n X j=1 f(x∗ ij, y ∗ ij)∆A.

Definition 1 (page 990). The double integral (重積分, 二重積分) of f (x, y) over the rectangle R is Z Z R f(x, y) dAdef.= lim m,n→∞ m X i=1 n X j=1 f(x∗ ij, y ∗ ij)∆A

if this limit exists.







Iterated Integrals, page 993

Goal: Compute the double integrals by iterated integrals.

Recall that it is usually difficult to evaluate single integrals directly from the defi-nition of an integral. The evaluation of double integrals from the defidefi-nition is even more difficult. In this section, we will see how to express a double integral as an iterated integral, which can be evaluated by calculating two single integrals.

Suppose that f (x, y) is integrable on the rectangle R = [a, b] × [c, d]. §15.1-2

Definition 2 (page 993). Define the partial integration of f (x, y) with respect to y, denoted by Rd

c f(x, y) dy to mean that x is fixed and f (x, y) is integrated with

respect to y from y = c to y = d. After partial integration, Rd

c f(x, y) dy depends on x, so we denote it by A(x).

Definition 3 (page 993). If we integrate A(x) = Rd

c f(x, y) dy with respect to x

from x = a to x = b, we get the iterated integral (先對 y 後對 x 的二次積分): Z b a A(x) dx = Z b a Z d c f(x, y) dy  dx = Z b a Z d c f(x, y) dy dx



Similarly, the iterated integral (先對 x 後對y 的二次積分)

Z d c Z b a f(x, y) dx dy = Z d c Z b a f(x, y) dx  dy = Z d c B(y) dy.

means that we first integrate with respect to x (fixed y) from x = a to x = b and then integrate the resulting function B(y) =Rb

a f(x, y) dx from y = c to y = d.

Fubini’s Theorem (page 994). If f (x, y) is continuous on the rectangles R = {(x, y)|a ≤ x ≤ b, c ≤ y ≤ d}, then Z Z R f(x, y) dA = Z b a Z d c f(x, y) dy  dx = Z d c Z b a f(x, y) dx  dy.

In general, this is true if we assume that f(x, y) is bounded on R, f (x, y) is discon-tinuous only on a finite number of smooth curves, and the iterated integrals exist. Example 4 (page 995). Evaluate RR

Rysin(xy) dA, where R = [1, 2] × [0, π].

Solution. If we first integrate with respect to x, we get Z Z

R

ysin(xy) dA =

Solution 2. If we reverse the order of integration, we get Z Z

R

ysin(xy) dA =

We use and get

So



Example 5 (page 996). Find the volume of the solid S that is bounded by the elliptic paraboloid x2

+ 2y2

+ z = 16, the plane x = 2 and y = 2, and the three coordinate planes.

Solution.

In the special case, where f (x, y) = g(x)h(y) is the product of a function of x only and a function of y only, by Fubini’s Theorem, we get

Example 6 (page 996). If R = [0,π 2] × [0, π 2], then Z Z R sin x cos y dA = §15.1-4