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File: Multiple Integrals Pdf 170523 | Lecture32
jim lambers mat460 560 fall semeseter 2009 10 lecture 32 notes these notes correspond to section 4 8 in the text multiple integrals double integrals as many problems in scientic ...

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                                                            Jim Lambers
                                                            MAT460/560
                                                      Fall Semeseter 2009-10
                                                          Lecture 32 Notes
                These notes correspond to Section 4.8 in the text.
                Multiple Integrals
                Double Integrals
                As many problems in scientific computing involve two-dimensional domains, it is essential to be
                able to compute integrals over such domains. Such integrals can be evaluated using the following
                strategies:
                    ∙ If a two-dimensional domain Ω can be decomposed into rectangles, then the integral of a
                       function f(x,y) over Ω can be computed by evaluating integrals of the form
                                                         I(f) = ZabZcdf(x,y)dydx.
                       Then, to evaluate I(f), one can use a Cartesian product rule, whose nodes and weights are
                       obtained by combining one-dimensional quadrature rules that are applied to each dimension.
                       For example, if functions of x are integrated along the line between x = a and x = b using
                       nodes x and weights w , for i = 1,...,n, and if functions of y are integrated along the line
                               i                i
                       between y = c and y = d using nodes y and weights z , for i = 1,...,m, then the resulting
                                                                 i               i
                       Cartesian product rule
                                                                    n   m
                                                       Q (f)=XXf(x,y)wz
                                                         n,m                  i  j   i j
                                                                   i=1 j=1
                       has nodes (x ,y ) and corresponding weights w z for i = 1,...,n and j = 1,...,m.
                                    i  j                                i j
                    ∙ If the domain Ω can be described as the region between two curves y (x) and y (x) for
                                                                                                    1           2
                       x∈[a,b], then we can write                  Z Z
                                                           I(f) =        f(x,y)dA
                                                                       Ω
                       as an iterated integral                 Z Z
                                                                  b  y2(x)
                                                       I(f) =             f(x,y)dydx
                                                                 a  y (x)
                                                                     1
                                                                    1
                       which can be evaluated by applying a one-dimensional quadrature rule to compute the outer
                       integral                                       Z
                                                                         b
                                                              I(f) =      g(x)dx
                                                                        a
                       where g(x) is evaluated by using a one-dimensional quadrature rule to compute the inner
                       integral                                    Z
                                                                     y2(x)
                                                           g(x) =         f(x,y)dy.
                                                                    y (x)
                                                                     1
                    ∙ For various simple regions such as triangles, there exist cubature rules that are not combi-
                       nations of one-dimensional quadrature rules. Cubature rules are more direct generalizations
                       of quadrature rules, in that they evaluate the integrand at selected nodes and use weights
                       determined by the geometry of the domain and the placement of the nodes.
                 It should be noted that all of these strategies apply to certain special cases. The first algorithm
                 capable of integrating over a general two-dimensional domain was developed by Lambers and Rice.
                 This algorithm combines the second and third strategies described above, decomposing the domain
                 into subdomains that are either triangles or regions between two curves.
                 Example We will use the Composite Trapezoidal Rule with m = n = 2 to evaluate the double
                 integral                                Z     Z
                                                            1/2   1/2
                                                                     ey−xdydx.
                                                           0     0
                 The Composite Trapezoidal Rule with n = 2 subintervals is
                                      Z b            ℎ            a+b                     b −a
                                       a f(x)dx ≈ 2 f(a)+2f            2     +f(b) ,     ℎ= n .
                 If a = 0 and b = 1/2, then ℎ = (1/2−0)/2 = 1/4 and this simplifies to
                                              Z 1/2f(x)dx ≈ 1[f(0)+2f(1/4)+f(1/2)].
                                               0               8
                 Wefirst use this rule to evaluate the “single” integral
                                                              Z01/2g(x)dx
                 where                                             Z
                                                                     1
                                                           g(x) =      ey−xdy.
                                                                    0
                                                                    2
               This yields
                       Z 1/2Z 1/2 y−x            Z 1/2
                                  e   dydx =          g(x)dx
                        0     0                   0
                                             ≈ 1[g(0)+2g(1/4)+g(1/2)]
                                                 8 "Z                Z               Z             #
                                                 1     1/2             1/2             1/2
                                                           y−0             y−1/4           y−1/2
                                             ≈ 8 0 e dy+2 0 e                   dy + 0    e      dy .
               Now, to evaluate each of these integrals, we use the Composite Trapezoidal Rule in the y-direction
               with m = 2. If we let k denote the step size in the y-direction, we have k = (1/2−0)/2 = 1/4, and
               therefore we have
                        Z 1/2Z 1/2 y−x            1 "Z 1/2 y−0       Z 1/2 y−1/4     Z 1/2 y−1/2   #
                         0    0   e    dydx ≈ 8       0   e   dy +2 0     e     dy + 0     e     dy
                                                  1 1 h 0−0     1/4−0   1/2−0i
                                              ≈ 8 8 e       +2e       +e        +
                                                   1 h 0−1/4    1/4−1/4   1/2−1/4i
                                                 28 e       +2e        +e          +
                                                                                 
                                                  1 h 0−1/2    1/4−1/2    1/2−1/2i
                                                  8 e      +2e        +e
                                                  1 h 0     1/4    1/2i
                                              ≈ 64 e +2e       +e      +
                                                  1 h −1/4     0    1/4i
                                                  32 e     +2e +e       +
                                                  1 h −1/2     −1/4    0i
                                                  64 e     +2e      +e
                                              ≈ 3e0+ 1e−1/4+ 1 e−1/2+ 1 e1/4+ 1 e1/2
                                                  32     16        64         16       64
                                              ≈ 0.25791494889765.
               The exact value, to 15 digits, is 0.255251930412762. The error is 2.66 × 10−3, which is to be
               expected due to the use of few subintervals, and the fact that the Composite Trapezoidal Rule is
               only second-order-accurate. □
               ExampleWewillusetheComposite Simpson’s Rule with n = 2 and m = 4 to evaluate the double
               integral                             Z Z
                                                      1  2x
                                                             2    3
                                                            x +y dydx.
                                                     0  x
               In this case, the domain of integration described by the limits is not a rectangle, but a triangle
               defined by the lines y = x, y = 2x, and x = 1. The Composite Simpson’s Rule with n = 2
                                                              3
                 subintervals is      Z                                          
                                        b            ℎ               a+b                      b −a
                                       a f(x)dx ≈ 3 f(a)+4f            2     +f(b) ,     ℎ= n .
                 If a = 0 and b = 1, then ℎ = (1 − 0)/2 = 1/2, and this simplifies to
                                               Z 1/2f(x)dx ≈ 1[f(0)+4f(1/2)+f(1)].
                                                0               6
                 Wefirst use this rule to evaluate the “single” integral
                                                               Z 1g(x)dx
                                                                0
                 where                                           Z
                                                                   2x
                                                                       2    3
                                                         g(x) =       x +y dy.
                                                                  x
                 This yields
                       Z 1Z 2xx2 +y3dydx = Z 1g(x)dx
                         0   x                        0
                                                ≈ 1[g(0)+4g(1/2)+g(1)]
                                                     6 "Z                  Z                   Z              #
                                                     1    0                   1   1 2               2
                                                             2    3                       3            2    3
                                                ≈ 6 0 0 +y dy+4 1/2 2                  +y dy+ 1 1 +y dy .
                 The first integral will be zero, since the limits of integration are equal. To evaluate the second
                 and third integrals, we use the Composite Simpson’s Rule in the y-direction with m = 4. If we let
                 k denote the step size in the y-direction, we have k = (2x − x)/4 = x/4, and therefore we have
                 k = 1/8 for the second integral and k = 1/4 for the third. This yields
                   Z 1Z 2x 2      3             1 " Z 1 12        3      Z 2 2      3   #
                    0   x   x +y dydx ≈ 6 4 1/2 2                +y dy+ 1 1 +y dy
                                                  (     "         !                 !                 !
                                                1     1     1      1 3          1     5 3          1      3 3
                                            ≈ 6 424         4 + 2        +4 4+ 8             +2 4+ 4            +
                                                               !              #        "                          !
                                                                                                         
                                                    1      7 3        1             1                       5 3
                                                4     +          +      +13      +       1+13 +4 1+                 +
                                                    4      8          4             12                        4
                                                          !                !               #)
                                                          3 3                7 3             
                                                2 1+             +4 1+              + 1+23
                                                          2                  4
                                            ≈ 1.03125.
                                                                    4
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...Jim lambers mat fall semeseter lecture notes these correspond to section in the text multiple integrals double as many problems scientic computing involve two dimensional domains it is essential be able compute over such can evaluated using following strategies if a domain decomposed into rectangles then integral of function f x y computed by evaluating form i zabzcdf dydx evaluate one use cartesian product rule whose nodes and weights are obtained combining quadrature rules that applied each dimension for example functions integrated along line between b w n c d z m resulting q xxf wz j has corresponding described region curves we write da an iterated which applying outer g dx where inner dy various simple regions triangles there exist cubature not combi nations more direct generalizations they integrand at selected determined geometry placement should noted all apply certain special cases rst algorithm capable integrating general was developed rice this combines second third above de...

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