Page 223 - Schaum's Outline of Theory and Problems of Advanced Calculus
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214                            MULTIPLE INTEGRALS                          [CHAP. 9



                                 We can also consider the double integral as representing the mass of the region r assuming a
                                            2
                                                2
                              density varying as x þ y .
                          (c)  Method 1:  The double integral can be expressed as the iterated integral
                                                                                         2
                                          x
                                                              x
                                      ð 2  ð 2           ð  2  ( ð 2    )     ð 2     3 x
                                              2  2               2   2            2  y
                                                                                          dx
                                            ðx þ y Þ dy dx ¼    ðx þ y Þ dy dx ¼  x y þ
                                       x¼1 y¼1            x¼1  y¼1            x¼1     3    y¼1
                                                                         !
                                                          2      x      1     1006
                                                         ð        6
                                                              4      2
                                                       ¼     x þ     x    dx ¼
                                                          x¼1    3      3      105
                                                                                       2
                                 The integration with respect to y (keeping x constant) from y ¼ 1to y ¼ x corresponds formally
                              to summing in a vertical column (see Fig. 9-6). The subsequent integration with respect to x from x ¼ 1
                              to x ¼ 2corresponds to addition of contributions from all such vertical columns between x ¼ 1 and
                              x ¼ 2.
                              Method 2:  The double integral can also be expressed as the iterated integral
                                                           (             )
                                     ð 4  ð  2           ð 4  ð 2             ð  4  3     2
                                              2  2                2   2          x
                                            ðx þ y Þ dx dy ¼     ðx þ y Þ dx dy ¼  þ xy  2    dy
                                          p
                                                               p
                                      y¼1  x¼ y ffiffi       y¼1  x¼ y ffiffi          y¼1 3      x¼ y ffiffi
                                                                                          p
                                                                           !
                                                             8   2  y    5=2    1006
                                                         ð 4         3=2
                                                         y¼1 3       3           105
                                                       ¼      þ 2y       y   dy ¼
                                 In this case the vertical column of region r in Fig. 9-6 above is replaced by a horizontal column as
                                                                                                y to x ¼ 2
                                                                                               p ffiffiffi
                              in Fig. 9-7 above. Then the integration with respect to x (keeping y constant) from x ¼
                              corresponds to summing in this horizontal column.  Subsequent integration with respect to y from
                              y ¼ 1to y ¼ 4corresponds to addition of contributions for all such horizontal columns between y ¼ 1
                              and y ¼ 4.
                                                                                      2
                                                                                         1 2
                      9.2. Find the volume of the region bound by the elliptic paraboloid z ¼ 4   x   y and the plane
                                                                                         4
                          z ¼ 0.
                              Because of the symmetry of the elliptic paraboloid, the result can be obtained by multiplying the first
                          octant volume by 4.
                                              2
                                                  2
                              Letting z ¼ 0yields 4x þ y ¼ 16. The limits of integration are determined from this equation. The
                          required volume is
                                                                                   p ffiffiffiffiffiffiffiffi
                                             p ffiffiffiffiffiffiffiffi                          ! 2 4 x 2
                                           2
                                          ð ð 2 4 x 2    1          ð 2        1 y 3
                                                      2
                                                                           2
                                         4        4   x   y 2  dy dx ¼ 4  4y   x y     dx
                                           0  0          4          0          4 3
                                                                                   0
                                                                 ¼ 16
                              Hint: Use trigonometric substitutions to complete the integrations.
                                                                                      2       p ffiffiffiffiffiffiffiffiffiffiffiffiffi 2
                      9.3. The geometric model of a material body is a plane region R bound by y ¼ x and y ¼  2   x on
                          the interval 0 @ x @ 1, and with a density function   ¼ xy (a) Draw the graph of the region.
                          (b) Find the mass of the body. (c) Find the coordinates of the center of mass. (See Fig. 9-8.)
                          (a)

                                                                Fig. 9-8
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