Upcoming SlideShare
×

# Lesson18 Double Integrals Over Rectangles Slides

8,624 views

Published on

We develop double integrals for measuring volume and iterated integrals for calculating double integrals.

Published in: Technology, Education
0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total views
8,624
On SlideShare
0
From Embeds
0
Number of Embeds
41
Actions
Shares
0
118
0
Likes
0
Embeds 0
No embeds

No notes for slide

### Lesson18 Double Integrals Over Rectangles Slides

1. 1. Section 12.1–12.2 Double Integrals over Rectangles Iterated Integrals Math 21a March 17, 2008 Announcements ◮ Ofﬁce hours Tuesday, Wednesday 2–4pm SC 323 ◮ Problem Sessions: Mon, 8:30; Thur, 7:30; SC 103b . . Image: Flickr user Cobalt123 . . . . . .
2. 2. Announcements ◮ Ofﬁce hours Tuesday, Wednesday 2–4pm SC 323 ◮ Problem Sessions: Mon, 8:30; Thur, 7:30; SC 103b . . . . . .
3. 3. Outline Last Time Double Integrals over Rectangles Recall the deﬁnite integral Deﬁnite integrals in two dimensions Iterated Integrals Partial Integration Fubini’s Theorem Average value . . . . . .
4. 4. Outline Last Time Double Integrals over Rectangles Recall the deﬁnite integral Deﬁnite integrals in two dimensions Iterated Integrals Partial Integration Fubini’s Theorem Average value . . . . . .
5. 5. Cavalieri’s method Let f be a positive function deﬁned on the interval [a, b]. We want to ﬁnd the area between x = a, x = b, y = 0, and y = f(x). For each positive integer n, divide up the interval into n pieces. Then b−a ∆x = . For each i between 1 and n, let xi be the nth step n between a and b. So x0 = a b−a x 1 = x 0 + ∆x = a + n b−a x 2 = x 1 + ∆x = a + 2 · n ······ b−a xi = a + i · n x x x .0 .1 .2 . i . n −1 . n xx x ······ . . . . . . . . . a b−a b . xn = a + n · =b . . n . . . .
6. 6. Forming Riemann sums We have many choices of how to approximate the area: Ln = f(x0 )∆x + f(x1 )∆x + · · · + f(xn−1 )∆x Rn = f(x1 )∆x + f(x2 )∆x + · · · + f(xn )∆x ( ) ( ) ( ) x0 + x 1 x1 + x2 x n −1 + x n Mn = f ∆x + f ∆x + · · · + f ∆x 2 2 2 . . . . . .
7. 7. Forming Riemann sums We have many choices of how to approximate the area: Ln = f(x0 )∆x + f(x1 )∆x + · · · + f(xn−1 )∆x Rn = f(x1 )∆x + f(x2 )∆x + · · · + f(xn )∆x ( ) ( ) ( ) x0 + x 1 x1 + x2 x n −1 + x n Mn = f ∆x + f ∆x + · · · + f ∆x 2 2 2 In general, choose x∗ to be a point in the ith interval [xi−1 , xi ]. Form i the Riemann sum Sn = f(x∗ )∆x + f(x∗ )∆x + · · · + f(x∗ )∆x 1 2 n ∑ n = f(x∗ )∆x i i=1 . . . . . .
8. 8. Deﬁnition The deﬁnite integral of f from a to b is the limit ∫ b ∑ n f(x) dx = lim f(x∗ )∆x i a n→∞ i=1 (The big deal is that for continuous functions this limit is the same no matter how you choose the x∗ ).i . . . . . .
9. 9. The problem Let R = [a, b] × [c, d] be a rectangle in the plane, f a positive function deﬁned on R, and S = { (x, y, z) | a ≤ x ≤ b, c ≤ y ≤ d, 0 ≤ z ≤ f(x, y) } Our goal is to ﬁnd the volume of S . . . . . .
10. 10. The strategy: Divide and conquer For each m and n, divide the interval [a, b] into m subintervals of equal width, and the interval [c, d] into n subintervals. For each i and j, form the subrectangles Rij = [xi−1 , xi ] × [yj−1 , yj ] Choose a sample point (x∗ , y∗ ) in each subrectangle and form the ij ij Riemann sum ∑∑m n Smn = f(x∗ , y∗ ) ∆A ij ij i=1 j=1 where ∆A = ∆x ∆y. . . . . . .
11. 11. Deﬁnition The double integral of f over the rectangle R is ∫∫ ∑∑ m n f(x, y) dA = lim f(x∗ , y∗ ) ∆A ij ij m,n→∞ R i=1 j=1 (Again, for continuous f this limit is the same regardless of method for choosing the sample points.) . . . . . .
12. 12. Worksheet #1 Problem Estimate the volume of the solid that lies below the surface z = xy and above the rectangle [0, 6] × [0, 4] in the xy-plane using a Riemann sum with m = 3 and n = 2. Take the sample point to be the upper right corner of each rectangle. . . . . . .
13. 13. Worksheet #1 Problem Estimate the volume of the solid that lies below the surface z = xy and above the rectangle [0, 6] × [0, 4] in the xy-plane using a Riemann sum with m = 3 and n = 2. Take the sample point to be the upper right corner of each rectangle. Answer 288 . . . . . .
14. 14. Theorem (Midpoint Rule) ∫∫ ∑∑ m n f(x, y) dA ≈ f(¯i , ¯j ) ∆A x y R i=1 j=1 where ¯i is the midpoint of [xi−1 , xi ] and ¯j is the midpoint of [yj−1 , yj ]. x y . . . . . .
15. 15. Worksheet #2 Problem Use the Midpoint Rule to evaluate the volume of the solid in Problem 1. . . . . . .
16. 16. Worksheet #2 Problem Use the Midpoint Rule to evaluate the volume of the solid in Problem 1. Answer 144 . . . . . .
17. 17. Outline Last Time Double Integrals over Rectangles Recall the deﬁnite integral Deﬁnite integrals in two dimensions Iterated Integrals Partial Integration Fubini’s Theorem Average value . . . . . .
18. 18. Partial Integration Let f be a function on a rectangle R = [a, b] × [c, d]. Then for each ﬁxed x we have a number ∫ d A(x) = f(x, y) dy c The is a function of x, and can be integrated itself. So we have an iterated integral ∫ b ∫ b [∫ d ] A(x) dx = f(x, y) dy dx a a c . . . . . .
19. 19. Worksheet #3 Problem Calculate ∫ 3∫ 1 ∫ 1∫ 3 (1 + 4xy) dx dy and (1 + 4xy) dy dx. 1 0 0 1 . . . . . .
20. 20. Fubini’s Theorem Double integrals look hard. Iterated integrals look easy/easier. The good news is: Theorem (Fubini’s Theorem) If f is continuous on R = [a, b] × [c, d], then ∫∫ ∫ b∫ d ∫ d∫ b f(x, y) dA = f(x, y) dy dx = f(x, y) dx dy a c c a R This is also true if f is bounded on R, f is discontinuous only on a ﬁnite number of smooth curves, and the iterated integrals exist. . . . . . .
21. 21. Worksheet #4 Problem Evaluate the volume of the solid in Problem 1 by computing an iterated integral. . . . . . .
22. 22. Worksheet #4 Problem Evaluate the volume of the solid in Problem 1 by computing an iterated integral. Answer 144 . . . . . .
23. 23. Meet the mathematician: Guido Fubini ◮ Italian, 1879–1943 ◮ graduated Pisa 1900 ◮ professor in Turin, 1908–1938 ◮ escaped to US and died ﬁve years later . . . . . .
24. 24. Worksheet #5 Problem Calculate ∫∫ xy2 dA x2 + 1 R where R = [0, 1] × [−3, 3]. . . . . . .
25. 25. Worksheet #5 Problem Calculate ∫∫ xy2 dA x2 + 1 R where R = [0, 1] × [−3, 3]. Answer ln 512 = 9 ln 2 . . . . . .
26. 26. Average value ◮ One variable: If f is a function deﬁned on [a, b], then ∫ b 1 fave = f(x) dx b−a a ◮ Two variables: If f is a function deﬁned on a rectangle R, then ∫∫ 1 fave = f(x, y) dA Area(R) R . . . . . .
27. 27. Worksheet #6 Problem Find the average value of f(x, y) = x2 y over the rectangle R = [−1, 1] × [0, 5]. . . . . . .
28. 28. Worksheet #6 Problem Find the average value of f(x, y) = x2 y over the rectangle R = [−1, 1] × [0, 5]. Answer ∫ 5∫ 1 1 5 x2 y dx dy = 10 0 −1 6 . . . . . .