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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
                                                         .   .      .   .   .   .
Announcements




    ◮   Office hours Tuesday, Wednesday 2–4pm SC 323
    ◮   Problem Sessions: Mon, 8:30; Thur, 7:30; SC 103b




                                                .   .      .   .   .   .
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




                                           .   .   .   .   .   .
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




                                           .   .   .   .   .   .
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
                                                            .    .    .      .
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




                                                     .     .    .    .    .      .
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




                                                     .     .    .    .    .      .
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




                                                        .      .   .   .   .   .
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




                                                    .    .    .    .    .      .
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.




                                                              .   .   .   .   .   .
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.)




                                                     .     .        .   .   .   .
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.




                                                   .    .    .     .    .   .
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




                                                   .    .    .     .    .   .
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




                                                         .      .    .    .        .   .
Worksheet #2




  Problem
  Use the Midpoint Rule to evaluate the volume of the solid in Problem 1.




                                                    .    .    .    .    .   .
Worksheet #2




  Problem
  Use the Midpoint Rule to evaluate the volume of the solid in Problem 1.

  Answer
  144




                                                    .    .    .    .    .   .
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




                                           .   .   .   .   .   .
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




                                                   .    .    .    .      .   .
Worksheet #3




  Problem
  Calculate
        ∫ 3∫    1                           ∫   1∫ 3
                    (1 + 4xy) dx dy   and              (1 + 4xy) dy dx.
        1   0                               0     1




                                                  .       .    .    .     .   .
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.




                                                                  .       .       .     .     .   .
Worksheet #4




  Problem
  Evaluate the volume of the solid in Problem 1 by computing an iterated
  integral.




                                                   .    .    .    .        .   .
Worksheet #4




  Problem
  Evaluate the volume of the solid in Problem 1 by computing an iterated
  integral.

  Answer
  144




                                                   .    .    .    .        .   .
Meet the mathematician: Guido Fubini




   ◮   Italian, 1879–1943
   ◮   graduated Pisa 1900
   ◮   professor in Turin,
       1908–1938
   ◮   escaped to US and died
       five years later




                                       .   .   .   .   .   .
Worksheet #5



  Problem
  Calculate                     ∫∫
                                       xy2
                                            dA
                                     x2 + 1
                                R

  where R = [0, 1] × [−3, 3].




                                                 .   .   .   .   .   .
Worksheet #5



  Problem
  Calculate                     ∫∫
                                       xy2
                                            dA
                                     x2 + 1
                                R

  where R = [0, 1] × [−3, 3].

  Answer
  ln 512 = 9 ln 2




                                                 .   .   .   .   .   .
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




                                                         .    .   .   .   .   .
Worksheet #6



  Problem
  Find the average value of f(x, y) = x2 y over the rectangle
  R = [−1, 1] × [0, 5].




                                                      .    .    .   .   .   .
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




                                                          .   .   .   .   .   .

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Lesson18 Double Integrals Over Rectangles Slides

  • 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. Announcements ◮ Office hours Tuesday, Wednesday 2–4pm SC 323 ◮ Problem Sessions: Mon, 8:30; Thur, 7:30; SC 103b . . . . . .
  • 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. Worksheet #2 Problem Use the Midpoint Rule to evaluate the volume of the solid in Problem 1. . . . . . .
  • 16. Worksheet #2 Problem Use the Midpoint Rule to evaluate the volume of the solid in Problem 1. Answer 144 . . . . . .
  • 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. 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. Worksheet #3 Problem Calculate ∫ 3∫ 1 ∫ 1∫ 3 (1 + 4xy) dx dy and (1 + 4xy) dy dx. 1 0 0 1 . . . . . .
  • 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. Worksheet #4 Problem Evaluate the volume of the solid in Problem 1 by computing an iterated integral. . . . . . .
  • 22. Worksheet #4 Problem Evaluate the volume of the solid in Problem 1 by computing an iterated integral. Answer 144 . . . . . .
  • 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. Worksheet #5 Problem Calculate ∫∫ xy2 dA x2 + 1 R where R = [0, 1] × [−3, 3]. . . . . . .
  • 25. Worksheet #5 Problem Calculate ∫∫ xy2 dA x2 + 1 R where R = [0, 1] × [−3, 3]. Answer ln 512 = 9 ln 2 . . . . . .
  • 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. Worksheet #6 Problem Find the average value of f(x, y) = x2 y over the rectangle R = [−1, 1] × [0, 5]. . . . . . .
  • 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 . . . . . .