Download presentation

Presentation is loading. Please wait.

1
MULTIPLE INTEGRALS 16

2
2 MULTIPLE INTEGRALS 16.9 Change of Variables in Multiple Integrals In this section, we will learn about: The change of variables in double and triple integrals.

3
3 CHANGE OF VARIABLES IN SINGLE INTEGRALS In one-dimensional calculus. we often use a change of variable (a substitution) to simplify an integral.

4
4 By reversing the roles of x and u, we can write the Substitution Rule (Equation 6 in Section 5.5) as: where x = g(u) and a = g(c), b = g(d). SINGLE INTEGRALS Formula 1

5
5 Another way of writing Formula 1 is as follows: SINGLE INTEGRALS Formula 2

6
6 A change of variables can also be useful in double integrals. We have already seen one example of this: conversion to polar coordinates. CHANGE OF VARIABLES IN DOUBLE INTEGRALS

7
7 The new variables r and θ are related to the old variables x and y by: x = r cos θ y = r sin θ DOUBLE INTEGRALS

8
8 The change of variables formula (Formula 2 in Section 16.4) can be written as: where S is the region in the rθ-plane that corresponds to the region R in the xy-plane. DOUBLE INTEGRALS

9
9 More generally, we consider a change of variables that is given by a transformation T from the uv-plane to the xy-plane: T(u, v) = (x, y) where x and y are related to u and v by: x = g(u, v)y = h(u, v) We sometimes write these as: x = x(u, v), y = y(u, v) TRANSFORMATION Equations 3

10
10 We usually assume that T is a C 1 transformation. This means that g and h have continuous first-order partial derivatives. C 1 TRANSFORMATION

11
11 A transformation T is really just a function whose domain and range are both subsets of. TRANSFORMATION

12
12 If T(u 1, v 1 ) = (x 1, y 1 ), then the point (x 1, y 1 ) is called the image of the point (u 1, v 1 ). If no two points have the same image, T is called one-to-one. IMAGE & ONE-TO-ONE TRANSFORMATION

13
13 The figure shows the effect of a transformation T on a region S in the uv-plane. T transforms S into a region R in the xy-plane called the image of S, consisting of the images of all points in S. CHANGE OF VARIABLES Fig. 16.9.1, p. 1049

14
14 If T is a one-to-one transformation, it has an inverse transformation T –1 from the xy–plane to the uv-plane. INVERSE TRANSFORMATION Fig. 16.9.1, p. 1049

15
15 Then, it may be possible to solve Equations 3 for u and v in terms of x and y : u = G(x, y) v = H(x, y) INVERSE TRANSFORMATION

16
16 A transformation is defined by: x = u 2 – v 2 y = 2uv Find the image of the square S = {(u, v) | 0 ≤ u ≤ 1, 0 ≤ v ≤ 1} TRANSFORMATION Example 1

17
17 The transformation maps the boundary of S into the boundary of the image. So, we begin by finding the images of the sides of S. TRANSFORMATION Example 1

18
18 The first side, S 1, is given by: v = 0 (0 ≤ u ≤ 1) TRANSFORMATION Example 1 Fig. 16.9.2, p. 1050

19
19 From the given equations, we have: x = u 2, y = 0, and so 0 ≤ x ≤ 1. Thus, S 1 is mapped into the line segment from (0, 0) to (1, 0) in the xy-plane. TRANSFORMATION Example 1

20
20 The second side, S 2, is: u = 1 (0 ≤ v ≤ 1) Putting u = 1 in the given equations, we get: x = 1 – v 2 y = 2v TRANSFORMATION E. g. 1—Equation 4 Fig. 16.9.2, p. 1050

21
21 Eliminating v, we obtain: which is part of a parabola. TRANSFORMATION E. g. 1—Equation 4

22
22 Similarly, S 3 is given by: v = 1 (0 ≤ u ≤ 1) Its image is the parabolic arc TRANSFORMATION E. g. 1—Equation 5 Fig. 16.9.2, p. 1050

23
23 Finally, S 4 is given by: u = 0(0 ≤ v ≤ 1) Its image is: x = –v 2, y = 0 that is, –1 ≤ x ≤ 0 TRANSFORMATION Example 1 Fig. 16.9.2, p. 1050

24
24 Notice that as, we move around the square in the counterclockwise direction, we also move around the parabolic region in the counterclockwise direction. TRANSFORMATION Example 1 Fig. 16.9.2, p. 1050

25
25 The image of S is the region R bounded by: The x-axis. The parabolas given by Equations 4 and 5. TRANSFORMATION Example 1 Fig. 16.9.2, p. 1050

26
26 Now, let’s see how a change of variables affects a double integral. DOUBLE INTEGRALS

27
27 We start with a small rectangle S in the uv-plane whose: Lower left corner is the point (u 0, v 0 ). Dimensions are ∆u and ∆v. DOUBLE INTEGRALS Fig. 16.9.3, p. 1050

28
28 The image of S is a region R in the xy-plane, one of whose boundary points is: (x 0, y 0 ) = T(u 0, v 0 ) DOUBLE INTEGRALS Fig. 16.9.3, p. 1050

29
29 The vector r(u, v) = g(u, v) i + h(u, v) j is the position vector of the image of the point (u, v). DOUBLE INTEGRALS Fig. 16.9.3, p. 1050

30
30 The equation of the lower side of S is: v = v 0 Its image curve is given by the vector function r(u, v 0 ). DOUBLE INTEGRALS Fig. 16.9.3, p. 1050

31
31 The tangent vector at (x 0, y 0 ) to this image curve is: DOUBLE INTEGRALS Fig. 16.9.3, p. 1050

32
32 Similarly, the tangent vector at (x 0, y 0 ) to the image curve of the left side of S (u = u 0 ) is: DOUBLE INTEGRALS Fig. 16.9.3, p. 1050

33
33 We can approximate the image region R = T(S) by a parallelogram determined by the secant vectors DOUBLE INTEGRALS Fig. 16.9.4, p. 1051

34
34 However, So, Similarly, DOUBLE INTEGRALS

35
35 This means that we can approximate R by a parallelogram determined by the vectors ∆u r u and ∆v r v DOUBLE INTEGRALS Fig. 16.9.5, p. 1051

36
36 Thus, we can approximate the area of R by the area of this parallelogram, which, from Section 13.4, is: |(∆u r u ) x (∆v r v )| = |r u x r v | ∆u ∆v DOUBLE INTEGRALS Equation 6

37
37 Computing the cross product, we obtain: DOUBLE INTEGRALS

38
38 The determinant that arises in this calculation is called the Jacobian of the transformation. It is given a special notation. JACOBIAN

39
39 The Jacobian of the transformation T given by x = g(u, v) and y = h(u, v) is: JACOBIAN OF T Definition 7

40
40 With this notation, we can use Equation 6 to give an approximation to the area ∆A of R: where the Jacobian is evaluated at (u 0, v 0 ). JACOBIAN OF T Approximation 8

41
41 The Jacobian is named after the German mathematician Carl Gustav Jacob Jacobi (1804–1851). The French mathematician Cauchy first used these special determinants involving partial derivatives. Jacobi, though, developed them into a method for evaluating multiple integrals. JACOBIAN

42
42 Next, we divide a region S in the uv-plane into rectangles S ij and call their images in the xy-plane R ij. DOUBLE INTEGRALS Fig. 16.9.6, p. 1052

43
43 Applying Approximation 8 to each R ij, we approximate the double integral of f over R as follows. DOUBLE INTEGRALS Fig. 16.9.6, p. 1052

44
44 where the Jacobian is evaluated at (u i, v j ). DOUBLE INTEGRALS

45
45 Notice that this double sum is a Riemann sum for the integral DOUBLE INTEGRALS

46
46 The foregoing argument suggests that the following theorem is true. A full proof is given in books on advanced calculus. DOUBLE INTEGRALS

47
47 Suppose: T is a C 1 transformation whose Jacobian is nonzero and that maps a region S in the uv-plane onto a region R in the xy-plane. f is continuous on R and that R and S are type I or type II plane regions. T is one-to-one, except perhaps on the boundary of S. CHG. OF VRBLS. (DOUBLE INTEG.) Theorem 9

48
48 Then, CHG. OF VRBLS. (DOUBLE INTEG.) Theorem 9

49
49 Theorem 9 says that we change from an integral in x and y to an integral in u and v by expressing x and y in terms of u and v and writing: CHG. OF VRBLS. (DOUBLE INTEG.)

50
50 Notice the similarity between Theorem 9 and the one-dimensional formula in Equation 2. Instead of the derivative dx/du, we have the absolute value of the Jacobian, that is, |∂(x, y)/∂(u, v)| CHG. OF VRBLS. (DOUBLE INTEG.)

51
51 As a first illustration of Theorem 9, we show that the formula for integration in polar coordinates is just a special case. CHG. OF VRBLS. (DOUBLE INTEG.)

52
52 Here, the transformation T from the rθ-plane to the xy-plane is given by: x = g(r, θ) = r cos θ y = h(r, θ) = r sin θ CHG. OF VRBLS. (DOUBLE INTEG.)

53
53 The geometry of the transformation is shown here. T maps an ordinary rectangle in the rθ-plane to a polar rectangle in the xy-plane. CHG. OF VRBLS. (DOUBLE INTEG.) Fig. 16.9.7, p. 1053

54
54 The Jacobian of T is: CHG. OF VRBLS. (DOUBLE INTEG.)

55
55 So, Theorem 9 gives: This is the same as Formula 2 in Section 16.4 CHG. OF VRBLS. (DOUBLE INTEG.)

56
56 Use the change of variables x = u 2 – v 2, y = 2uv to evaluate the integral where R is the region bounded by: The x-axis. The parabolas y 2 = 4 – 4x and y 2 = 4 + 4x, y ≥ 0. CHG. OF VRBLS. (DOUBLE INTEG.) Example 2

57
57 The region R is pictured here. CHG. OF VRBLS. (DOUBLE INTEG.) Example 2 Fig. 16.9.2, p. 1050

58
58 In Example 1, we discovered that T(S) = R where S is the square [0, 1] x [0, 1]. Indeed, the reason for making the change of variables to evaluate the integral is that S is a much simpler region than R. CHG. OF VRBLS. (DOUBLE INTEG.) Example 2

59
59 First, we need to compute the Jacobian: CHG. OF VRBLS. (DOUBLE INTEG.) Example 2

60
60 So, by Theorem 9, CHG. OF VRBLS. (DOUBLE INTEG.) Example 2

61
61 Example 2 was not very difficult to solve as we were given a suitable change of variables. If we are not supplied with a transformation, the first step is to think of an appropriate change of variables. CHG. OF VRBLS. (DOUBLE INTEG.) Note

62
62 If f(x, y) is difficult to integrate, The form of f(x, y) may suggest a transformation. If the region of integration R is awkward, The transformation should be chosen so that the corresponding region S in the uv-plane has a convenient description. CHG. OF VRBLS. (DOUBLE INTEG.) Note

63
63 Evaluate the integral where R is the trapezoidal region with vertices (1, 0), (2, 0), (0, –2), (0,–1) CHG. OF VRBLS. (DOUBLE INTEG.) Example 3

64
64 It isn’t easy to integrate e (x+y)/(x–y). So, we make a change of variables suggested by the form of this function: u = x + y v = x – y These equations define a transformation T –1 from the xy-plane to the uv-plane. CHG. OF VRBLS. (DOUBLE INTEG.) E. g. 3—Eqns. 10

65
65 Theorem 9 talks about a transformation T from the uv-plane to the xy-plane. It is obtained by solving Equations 10 for x and y: x = ½(u + v) y = ½(u – v) CHG. OF VRBLS. (DOUBLE INTEG.) E. g. 3—Equation 11

66
66 The Jacobian of T is: CHG. OF VRBLS. (DOUBLE INTEG.) Example 3

67
67 To find the region S in the uv-plane corresponding to R, we note that: The sides of R lie on the lines y = 0 x – y = 2 x = 0 x – y = 1 From either Equations 10 or Equations 11, the image lines in the uv-plane are: u = v v = 2 u = –v v = 1 CHG. OF VRBLS. (DOUBLE INTEG.) Example 3

68
68 Thus, the region S is the trapezoidal region with vertices (1, 1), (2, 2), (–2, 2), (–1,1) CHG. OF VRBLS. (DOUBLE INTEG.) Example 3 Fig. 16.9.8, p. 1054

69
69 S = {(u, v) | 1 ≤ v ≤ 2, –v ≤ u ≤ v} CHG. OF VRBLS. (DOUBLE INTEG.) Example 3 Fig. 16.9.8, p. 1054

70
70 So, Theorem 9 gives: CHG. OF VRBLS. (DOUBLE INTEG.) Example 3

71
71 There is a similar change of variables formula for triple integrals. Let T be a transformation that maps a region S in uvw-space onto a region R in xyz-space by means of the equations x = g(u, v, w) y = h(u, v, w) z = k(u, v, w) TRIPLE INTEGRALS

72
72 The Jacobian of T is this 3 x 3 determinant: TRIPLE INTEGRALS Equation 12

73
73 Under hypotheses similar to those in Theorem 9, we have this formula for triple integrals: TRIPLE INTEGRALS Formula 13

74
74 Use Formula 13 to derive the formula for triple integration in spherical coordinates. The change of variables is given by: x = ρ sin Φ cos θ y = ρ sin Φ sin θ z = ρ cos Φ TRIPLE INTEGRALS Example 4

75
75 We compute the Jacobian as follows: TRIPLE INTEGRALS Example 4

76
76 TRIPLE INTEGRALS Example 4

77
77 Since 0 ≤ Φ ≤ π, we have sin Φ ≥ 0. Therefore, TRIPLE INTEGRALS Example 4

78
78 Thus, Formula 13 gives: This is equivalent to Formula 3 in Section15.8 TRIPLE INTEGRALS Example 4

Similar presentations

© 2020 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google