Stewart J. Calculus

Подождите немного. Документ загружается.

982

||||

CHAPTER 15 PARTIAL DERIVATIVES

35. If , where , , and

, use the Chain Rule to ﬁnd .

36. If , where and , use the

Chain Rule to ﬁnd and when and .

37. Suppose , where , ,

, , , ,

, , , and .

Find and when and .

38. Use a tree diagram to write out the Chain Rule for the case

where , , and

are all differentiable functions.

39. If , where is differentiable, show that

40. The length of a side of a triangle is increasing at a rate of

3in兾s, the length of another side is decreasing at a rate of

2in兾s, and the contained angle is increasing at a rate of

radian兾s. How fast is the area of the triangle changing

when in, in, and ?

41. If , where , , and has continuous

second partial derivatives, show that

42. If , ﬁnd and .

43. Find the gradient of the function .

44. (a) When is the directional derivative of a maximum?

(b) When is it a minimum?

(d) When is it half of its maximum value?

45– 46 Find the directional derivative of at the given point in

the indicated direction.

45. ,,

in the direction toward the point

46. ,,

in the direction of

47. Find the maximum rate of change of

at the point . In which direction does it occur?

48. Find the direction in which increases most

rapidly at the point . What is the maximum rate of

increase?

49. The contour map shows wind speed in knots during Hurri-

cane Andrew on August 24, 1992. Use it to estimate the

共0, 1, 2兲

f 共x, y, z兲苷 ze

共2, 1兲

f 共x, y兲苷 x

y ⫹

v 苷 2i ⫹ j ⫺ 2k

共1, 2, 3兲

f 共x, y, z兲苷 x

y ⫹ x

1 ⫹ z

共4, 1兲

共1, 5兲

f 共x, y兲苷 2

⫺ y

f 共x, y, z兲苷 z

⭸z

⭸y

⭸z

⭸x

⫹ x

苷 e

xyz

⭸

⭸x

⫺ y

⭸

⭸y

苷 ⫺4uv

⭸

⭸u ⭸v

⫹ 2v

⭸z

⭸v

fv 苷 y兾xu 苷 xyz 苷 f 共u, v兲

␪

苷

␲

兾6y 苷 50x 苷 40

0.05

␪

⭸z

⭸x

⫹ x

⭸z

⭸y

苷 x

fz 苷 y ⫹ f 共x

⫺ y

兲

v 苷 v共 p, q, r, s兲

u 苷 u共 p, q, r, s兲

w 苷 f 共t, u, v兲, t 苷 t共 p, q, r, s兲

t 苷 2s 苷 1⭸z兾⭸t⭸z兾⭸s

共3, 6兲苷 8f

共3, 6兲苷 7h

共1, 2兲苷 10h

共1, 2兲苷 ⫺5

h共1, 2兲苷 6t

共1, 2兲苷 4t

共1, 2兲苷 ⫺1t共1, 2兲苷 3

y 苷 h共s, t兲x 苷 t共s, t兲z 苷 f 共x, y兲

t 苷 1s 苷 0⭸

v兾⭸t⭸v兾⭸s

y 苷 stx 苷 s ⫹ 2t

v 苷 x

sin y ⫹ ye

du兾dpz 苷 p sin p

y 苷 pe

x 苷 p ⫹ 3p

u 苷 x

⫹ z

18. The speed of sound traveling through ocean water is a func-

tion of temperature, salinity, and pressure. It has been

modeled by the function

where is the speed of sound (in meters per second), is the

temperature (in degrees Celsius), is the salinity (the concen-

tration of salts in parts per thousand, which means the num-

ber of grams of dissolved solids per 1000 g of water), and

is the depth below the ocean surface (in meters). Compute

, , and when , parts

per thousand, and m. Explain the physical signiﬁ-

cance of these partial derivatives.

19–22 Find all second partial derivatives of .

19. 20.

21. 22.

23. If , show that .

24. If , show that

25–29 Find equations of (a) the tangent plane and (b) the normal

line to the given surface at the speciﬁed point.

25. ,

26. ,

27. ,

28. ,

29. ,

;

30. Use a computer to graph the surface and its

tangent plane and normal line at on the same screen.

Choose the domain and viewpoint so that you get a good

view of all three objects.

31. Find the points on the hyperboloid where

the tangent plane is parallel to the plane .

32. Find if .

33. Find the linear approximation of the function

at the point (2, 3, 4) and use it

to estimate the number .

34. The two legs of a right triangle are measured as 5 m and

12 m with a possible error in measurement of at most

cm in each. Use differentials to estimate the maximum

error in the calculated value of (a) the area of the triangle and

(b) the length of the hypotenuse.

0.2

共1.98兲

共3.01兲

⫹ 共3.97兲

f 共x, y, z兲苷 x

⫹ z

u 苷 ln共1 ⫹ se

兲du

2x ⫹ 2y ⫹ z 苷 5

⫹ 4y

⫺ z

苷 4

共1, 1, 2兲

z 苷 x

⫹ y

共2, ⫺1, 0兲sin共xyz兲苷 x ⫹ 2y ⫹ 3z

共1, 1, 1兲xy ⫹ yz ⫹ zx 苷 3

共2, ⫺1, 1兲x

⫹ 2y

⫺ 3z

苷 3

共0, 0, 1兲z 苷 e

cos y

共1, ⫺2, 1兲z 苷 3x

⫺ y

⫹ 2x

⭸z

⭸x

⭸

⭸x ⭸t

苷

⭸z

⭸t

⭸

⭸x

z 苷 sin共x ⫹ sin t兲

⭸z

⭸x

⫹ y

⭸z

⭸y

苷 xy ⫹ zz 苷 xy ⫹ xe

y兾x

v 苷 r cos共s ⫹ 2t兲f 共x, y, z兲苷 x

z 苷 xe

⫺2y

f 共x, y兲苷 4x

⫺ xy

D 苷 100

S 苷 35T 苷 10⬚C⭸C兾⭸D⭸C兾⭸S⭸C兾⭸T

⫹ 共1.34 ⫺ 0.01T兲共S ⫺ 35兲 ⫹ 0.016D

C 苷 1449.2 ⫹ 4.6T ⫺ 0.055T

⫹ 0.00029T

CHAPTER 15 REVIEW

||||

983

;

58. Use a graphing calculator or computer (or Newton’s method

or a computer algebra system) to ﬁnd the critical points of

correct to three

decimal places. Then classify the critical points and ﬁnd

the highest point on the graph.

59–62 Use Lagrange multipliers to ﬁnd the maximum and mini-

mum values of subject to the given constraint(s).

59. ;

60. ;

61. ;

62. ;

63. Find the points on the surface that are closest to

the origin.

64. A package in the shape of a rectangular box can be mailed by

the US Postal Service if the sum of its length and girth (the

perimeter of a cross-section perpendicular to the length) is at

most 108 in. Find the dimensions of the package with largest

volume that can be mailed.

65. A pentagon is formed by placing an isosceles triangle on a

rectangle, as shown in the ﬁgure. If the pentagon has ﬁxed

perimeter , ﬁnd the lengths of the sides of the pentagon that

maximize the area of the pentagon.

66. A particle of mass moves on the surface . Let

and be the - and -coordinates of the

particle at time .

(a) Find the velocity vector and the kinetic energy

of the particle.

(b) Determine the acceleration vector .

the velocity vector, the kinetic energy, and the

acceleration vector.

y共t兲苷 t sin tx共t兲苷 t cos tz 苷 x

⫹ y

K 苷

ⱍ

yxy 苷 y共t兲x 苷 x共t兲

z 苷 f 共x, y兲m

苷 2

x ⫺ y ⫹ 2z 苷 2x ⫹ y ⫹ z 苷 1

f 共x, y, z兲苷 x

⫹ 2y

⫹ 3z

⫹ y

⫹ z

苷 3f 共x, y, z兲苷 xyz

⫹

苷 1f 共x, y兲苷

⫹

⫹ y

苷 1f 共x, y兲苷 x

f 共x, y兲苷 12 ⫹ 10y ⫺ 2x

⫺ 8xy ⫺ y

value of the directional derivative of the wind speed at

Homestead, Florida, in the direction of the eye of the

hurricane.

50. Find parametric equations of the tangent line at the point

to the curve of intersection of the surface

and the plane .

51–54 Find the local maximum and minimum values and saddle

points of the function. If you have three-dimensional graphing

software, graph the function with a domain and viewpoint that

reveal all the important aspects of the function.

51.

52.

53.

54.

55–56 Find the absolute maximum and minimum values of on

the set .

55. ; is the closed triangular

region in the -plane with vertices , , and

56. ; is the disk

;

57. Use a graph andⲐor level curves to estimate the local

maximum and minimum values and saddle points of

. Then use calculus to ﬁnd

these values precisely.

f 共x, y兲苷 x

⫺ 3x ⫹ y

⫺ 2y

⫹ y

艋 4Df 共x, y兲苷 e

⫺x

⫺y

共x

⫹ 2y

兲

共6, 0兲共0, 6兲共0, 0兲xy

Df 共x, y兲苷 4xy

⫺ x

⫺ xy

f 共x, y兲苷共x

⫹ y兲e

y兾2

f 共x, y兲苷 3xy ⫺ x

y ⫺ xy

f 共x, y兲苷 x

⫺ 6xy ⫹ 8y

f 共x, y兲苷 x

⫺ xy ⫹ y

⫹ 9x ⫺ 6y ⫹ 10

z 苷 4z 苷 2x

⫺ y

共⫺2, 2, 4兲

Key West

(Distance in miles)

10 20 30 40

Homestead

NOAA / AOML / Hurricane Research Division

984

1. A rectangle with length and width is cut into four smaller rectangles by two lines parallel

to the sides. Find the maximum and minimum values of the sum of the squares of the areas of

the smaller rectangles.

2. Marine biologists have determined that when a shark detects the presence of blood in the water,

it will swim in the direction in which the concentration of the blood increases most rapidly.

Based on certain tests, the concentration of blood (in parts per million) at a point on

the surface of seawater is approximated by

where and are measured in meters in a rectangular coordinate system with the blood source

at the origin.

(a) Identify the level curves of the concentration function and sketch several members of this

family together with a path that a shark will follow to the source.

(b) Suppose a shark is at the point when it ﬁrst detects the presence of blood in the

water. Find an equation of the shark’s path by setting up and solving a differential equation.

3. A long piece of galvanized sheet metal with width is to be bent into a symmetric form with

three straight sides to make a rain gutter. A cross-section is shown in the ﬁgure.

(a) Determine the dimensions that allow the maximum possible ﬂow; that is, ﬁnd the dimen-

sions that give the maximum possible cross-sectional area.

(b) Would it be better to bend the metal into a gutter with a semicircular cross-section?

4. For what values of the number is the function

continuous on ?

5. Suppose is a differentiable function of one variable. Show that all tangent planes to the

surface intersect in a common point.

6. (a) Newton’s method for approximating a root of an equation (see Section 4.8)

can be adapted to approximating a solution of a system of equations and

. The surfaces and intersect in a curve that intersects thez 苷 t共x, y兲z 苷 f 共x, y兲t共x, y兲苷 0

f 共x, y兲苷 0

f 共x兲苷 0

z 苷 xf共y兾x兲

⺢

f 共x, y, z兲苷

再

共x ⫹ y ⫹ z兲

⫹ y

⫹ z

共x, y, z兲苷 0

¨¨

w-2x

共x

, y

兲

C共x, y兲苷 e

⫺共x

⫹2y

兲兾10

P共x, y兲

PROBLEMS PLUS

985

-plane at the point , which is the solution of the system. If an initial approximation

is close to this point, then the tangent planes to the surfaces at intersect in a

straight line that intersects the -plane in a point , which should be closer to .

(Compare with Figure 2 in Section 4.8.) Show that

where , , and their partial derivatives are evaluated at . If we continue this proce-

dure, we obtain successive approximations .

(b) It was Thomas Simpson (1710–1761) who formulated Newton’s method as we know it

today and who extended it to functions of two variables as in part (a). (See the biography

of Simpson on page 538.) The example that he gave to illustrate the method was to solve

the system of equations

In other words, he found the points of intersection of the curves in the ﬁgure. Use the

method of part (a) to ﬁnd the coordinates of the points of intersection correct to six deci-

mal places.

7. If the ellipse is to enclose the circle , what values of and

minimize the area of the ellipse?

8. Among all planes that are tangent to the surface , ﬁnd the ones that are

farthest from the origin.

苷 1

bax

⫹ y

苷 2yx

兾a

⫹ y

兾b

苷 1

=1000

=100

⫹ y

苷 100x

⫹ y

苷 1000

共x

, y

兲

共x

, y

兲tf

苷 y

⫺

t ⫺ ft

⫺ f

andx

苷 x

⫺

⫺ f

共r, s兲共x

, y

兲xy

共x

, y

兲共x

, y

兲

共r, s兲xy

PROBLEMS PLUS

986

In this chapter we extend the idea of a deﬁnite integral to double and triple integrals

of functions of two or three variables. These ideas are then used to compute volumes,

masses, and centroids of more general regions than we were able to consider in

Chapters 6 and 8. We also use double integrals to calculate probabilities when two

random variables are involved.

We will see that polar coordinates are useful in computing double integrals over some

types of regions. In a similar way, we will introduce two new coordinate systems in

three-dimensional space––cylindrical coordinates and spherical coordinates––that

greatly simplify the computation of triple integrals over certain commonly occurring

solid regions.

A double integral of a positive function is a volume, which

is the limit of sums of volumes of rectangular columns.

MULTIPLE INTEGRALS

DOUBLE INTEGRALS OVER RECTANGLES

In much the same way that our attempt to solve the area problem led to the deﬁnition of a

deﬁnite integral, we now seek to ﬁnd the volume of a solid and in the process we arrive at

the deﬁnition of a double integral.

REVIEW OF THE DEFINITE INTEGRAL

First let’s recall the basic facts concerning deﬁnite integrals of functions of a single vari-

able. If is deﬁned for , we start by dividing the interval into n sub-

intervals of equal width and we choose sample points in these

subintervals. Then we form the Riemann sum

and take the limit of such sums as to obtain the deﬁnite integral of from to :

In the special case where , the Riemann sum can be interpreted as the sum of the

areas of the approximating rectangles in Figure 1, and represents the area under

the curve from to .

VOLUMES AND DOUBLE INTEGRALS

In a similar manner we consider a function of two variables deﬁned on a closed rectangle

and we ﬁrst suppose that . The graph of f is a surface with equation .

Let S be the solid that lies above R and under the graph of f, that is,

(See Figure 2.) Our goal is to ﬁnd the volume of S.

The ﬁrst step is to divide the rectangle into subrectangles. We accomplish this by

dividing the interval into m subintervals of equal width

and dividing into n subintervals of equal width . By draw-⌬y 苷共d ⫺ c兲兾n关y

j⫺1

, y

兴关c, d 兴

⌬x 苷共b ⫺ a兲兾m关x

i⫺1

, x

兴关a, b兴

S 苷

兵

共x, y, z兲僆⺢

ⱍ

0 艋 z 艋 f 共x, y兲, 共x, y兲僆 R

其

z 苷 f 共x, y兲f 共x, y兲艌 0

R 苷关a, b兴 ⫻ 关c, d兴苷

兵

共x, y兲僆⺢

ⱍ

a 艋 x 艋 b, c 艋 y 艋 d

其

FIGURE 1

⁄‹¤x

i-1

x¡

x™

x£

** *

Îx

f(x

)

n-1

bay 苷 f 共x兲

f 共x兲 dx

f 共x兲艌 0

f 共x兲 dx 苷 lim

n l ⬁

兺

i苷1

f 共x

兲 ⌬x

bafn l ⬁

兺

i苷1

f 共x

兲 ⌬x

⌬x 苷共b ⫺ a兲兾n关x

i⫺1

, x

兴

关a, b兴a 艋 x 艋 bf 共x兲

16.1

987

FIGURE 2

z=f(x,y)

ing lines parallel to the coordinate axes through the endpoints of these subintervals, as in

Figure 3, we form the subrectangles

each with area .

If we choose a sample point in each , then we can approximate the part of

S that lies above each by a thin rectangular box (or “column”) with base and height

as shown in Figure 4. (Compare with Figure 1.) The volume of this box is the

height of the box times the area of the base rectangle:

If we follow this procedure for all the rectangles and add the volumes of the corresponding

boxes, we get an approximation to the total volume of S:

(See Figure 5.) This double sum means that for each subrectangle we evaluate at the cho-

sen point and multiply by the area of the subrectangle, and then we add the results.

,

FIGURE 4 FIGURE 5

f(x

)

V ⬇

兺

i苷1

兺

j苷1

f 共x

, y

兲 ⌬A

f 共x

, y

兲 ⌬A

f 共x

, y

兲

共x

, y

兲

FIGURE 3

Dividing R into subrectangles

j-1

£™,y

£™)

›

⁄¤

,y

)

,y

)

Îx

Îy

i-1

⌬A 苷 ⌬x ⌬y

苷关x

i⫺1

, x

兴 ⫻ 关y

j⫺1

, y

兴苷

兵

共x, y兲

ⱍ

i⫺1

艋 x 艋 x

, y

j⫺1

艋 y 艋 y

其

988

||||

CHAPTER 16 MULTIPLE INTEGRALS

Our intuition tells us that the approximation given in (3) becomes better as and

become larger and so we would expect that

We use the expression in Equation 4 to deﬁne the volume of the solid that lies under the

graph of and above the rectangle . (It can be shown that this deﬁnition is consistent with

our formula for volume in Section 6.2.)

Limits of the type that appear in Equation 4 occur frequently, not just in ﬁnding vol-

umes but in a variety of other situations as well—as we will see in Section 16.5—even

when is not a positive function. So we make the following deﬁnition.

DEFINITION The double integral of over the rectangle is

if this limit exists.

The precise meaning of the limit in Deﬁnition 5 is that for every number there is

an integer such that

for all integers and greater than and for any choice of sample points in

A function is called integrable if the limit in Deﬁnition 5 exists. It is shown in courses

on advanced calculus that all continuous functions are integrable. In fact, the double inte-

gral of exists provided that is “not too discontinuous.” In particular, if is bounded

[that is, there is a constant such that for all in ], and is con-

tinuous there, except on a ﬁnite number of smooth curves, then is integrable over .

The sample point can be chosen to be any point in the subrectangle but if

we choose it to be the upper right-hand corner of [namely , see Figure 3], then

the expression for the double integral looks simpler:

By comparing Deﬁnitions 4 and 5, we see that a volume can be written as a double

integral:

If , then the volume of the solid that lies above the rectangle and

below the surface is

V 苷

f 共x, y兲 dA

z 苷 f 共x, y兲

RVf 共x, y兲艌 0

f 共x, y兲 dA 苷 lim

m, n l ⬁

兺

i苷1

兺

j苷1

f 共x

, y

兲 ⌬A

共x

, y

兲R

,共x

, y

兲

fR共x, y兲

ⱍ

f 共x, y兲

ⱍ

艋 MM

fff

.共x

, y

兲Nnm

冟

f 共x, y兲 dA ⫺

兺

i苷1

兺

j苷1

f 共x

, y

兲 ⌬A

冟

⬍

␧

␧⬎0

f 共x, y兲 dA 苷 lim

m, n l ⬁

兺

i苷1

兺

j苷1

f 共x

, y

兲 ⌬A

V 苷 lim

m, n l ⬁

兺

i苷1

兺

j苷1

f 共x

, y

兲 ⌬A

SECTION 16.1 DOUBLE INTEGRALS OVER RECTANGLES

||||

989

N The meaning of the double limit in Equation 4

is that we can make the double sum as close as

we like to the number [for any choice of

in ] by taking and sufﬁciently

large.

nmR

共x

, y

兲

N Notice the similarity between Deﬁnition 5

and the deﬁnition of a single integral in

Equation 2.

N Although we have deﬁned the double integral

by dividing into equal-sized subrectangles, we

could have used subrectangles of unequal

size. But then we would have to ensure that all

of their dimensions approach in the limiting

process.

The sum in Deﬁnition 5,

is called a double Riemann sum and is used as an approximation to the value of the

double integral. [Notice how similar it is to the Riemann sum in (1) for a function of a

single variable.] If happens to be a positive function, then the double Riemann sum

represents the sum of volumes of columns, as in Figure 5, and is an approximation to the

volume under the graph of and above the rectangle .

EXAMPLE 1

Estimate the volume of the solid that lies above the square

and below the elliptic paraboloid . Divide

into four equal squares and choose the sample point to be the upper right corner of

each square . Sketch the solid and the approximating rectangular boxes.

SOLUTION

The squares are shown in Figure 6. The paraboloid is the graph of

and the area of each square is 1. Approximating the volume

by the Riemann sum with , we have

This is the volume of the approximating rectangular boxes shown in Figure 7.

We get better approximations to the volume in Example 1 if we increase the number of

squares. Figure 8 shows how the columns start to look more like the actual solid and the

corresponding approximations become more accurate when we use 16, 64, and 256

squares. In the next section we will be able to show that the exact volume is 48.

EXAMPLE 2

If , evaluate the integral

1 ⫺ x

R 苷兵共x, y兲

ⱍ

⫺1 艋 x 艋 1, ⫺2 艋 y 艋 2其

FIGURE 8

The Riemann sum approximations to

the volume under z=16-≈-2¥

become more accurate as m and

n increase.

苷 13共1兲 ⫹ 7共1兲 ⫹ 10共1兲 ⫹ 4共1兲苷 34

苷 f 共1, 1兲 ⌬A ⫹ f 共1, 2兲 ⌬A ⫹ f 共2, 1兲 ⌬A ⫹ f 共2, 2兲 ⌬A

V ⬇

兺

i苷1

兺

j苷1

f 共x

, y

兲 ⌬A

m 苷 n 苷 2

f 共x, y兲苷 16 ⫺ x

⫺ 2y

Rz 苷 16 ⫺ x

⫺ 2y

R 苷关0, 2兴 ⫻ 关0, 2兴

兺

i苷1

兺

j苷1

f 共x

, y

兲 ⌬A

990

||||

CHAPTER 16 MULTIPLE INTEGRALS

(2,2)

R¡™ R™™

R¡¡ R™¡

(2,1)

(1,1)

(1,2)

FIGURE 6

URE

z=1

≈

Openmirrors.com

SOLUTION It would be very difﬁcult to evaluate this integral directly from Deﬁnition 5 but,

because , we can compute the integral by interpreting it as a volume. If

, then and , so the given double integral represents the

volume of the solid S that lies below the circular cylinder and above the

rectangle R. (See Figure 9.) The volume of S is the area of a semicircle with radius 1

times the length of the cylinder. Thus

THE MIDPOINT RULE

The methods that we used for approximating single integrals (the Midpoint Rule, the

Trapezoidal Rule, Simpson’s Rule) all have counterparts for double integrals. Here we

consider only the Midpoint Rule for double integrals. This means that we use a double Rie-

mann sum to approximate the double integral, where the sample point in is

chosen to be the center of . In other words, is the midpoint of and

is the midpoint of .

MIDPOINT RULE FOR DOUBLE INTEGRALS

where is the midpoint of and is the midpoint of .

EXAMPLE 3 Use the Midpoint Rule with to estimate the value of the

integral , where , .

SOLUTION In using the Midpoint Rule with , we evaluate at

the centers of the four subrectangles shown in Figure 10. So , , , and

. The area of each subrectangle is . Thus

Thus we have M

In the next section we will develop an efficient method for com-

puting double integrals and then we will see that the exact value of the double integral in

Example 3 is . (Remember that the interpretation of a double integral as a volume is

valid only when the integrand is a positive function. The integrand in Example 3 is not

a positive function, so its integral is not a volume. In Examples 2 and 3 in Section 16.2 we

will discuss how to interpret integrals of functions that are not always positive in terms of

volumes.) If we keep dividing each subrectangle in Figure 10 into four smaller ones with

⫺12

NOTE

共x ⫺ 3y

兲 dA ⬇ ⫺11.875

苷 ⫺

苷 ⫺11.875

苷

(

⫺

)

⫹

(

⫺

139

)

⫹

(

⫺

)

⫹

(

⫺

123

)

苷 f

(

)

⌬A ⫹ f

(

)

⌬A ⫹ f

(

)

⌬A ⫹ f

(

)

⌬A

苷 f 共x

, y

兲 ⌬A ⫹ f 共x

, y

兲 ⌬A ⫹ f 共x

, y

兲 ⌬A ⫹ f 共x

, y

兲 ⌬A

共x ⫺ 3y

兲

dA ⬇

兺

i苷1

兺

j苷1

f 共x

, y

兲 ⌬A

⌬A 苷

苷

f 共x, y兲苷 x ⫺ 3y

m 苷 n 苷 2

1 艋 y 艋 2其R 苷兵共x, y兲

ⱍ

0 艋 x 艋 2xx

共x ⫺ 3y

兲 dA

m 苷 n 苷 2

关y

j⫺1

, y

兴y

关x

i⫺1

, x

兴x

f 共x, y兲 dA ⬇

兺

i苷1

兺

j苷1

f 共x

, y

兲 ⌬A

关y

j⫺1

, y

兴

关x

i⫺1

, x

兴x

共x

, y

兲

共x

, y

兲

1 ⫺ x

dA 苷

␲

共1兲

⫻ 4 苷 2

␲

⫹ z

苷 1

z 艌 0x

⫹ z

苷 1z 苷

1 ⫺ x

艌 0

SECTION 16.1 DOUBLE INTEGRALS OVER RECTANGLES

||||

991

FIGURE 9

(0,0,1)

(1,0,0)

(0,2,0)

(2,2)

R¡™ R™™

R¡¡ R™¡

FIGURE 10