Page 217 - How To Solve Word Problems In Calculus
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                                   The sphere is generated by rotating the upper half of the circle
                                    2
                                         2
                                   x + y = 9 about the x axis. We solve this problem using the shell
                                   method, using shells of radius y and length 2x.
                                                                 b

                                                        V = 2π    hr dr
                                                                a
                                                                 3

                                                          = 2π    2xy dy
                                                                2
                                   Since the variable of integration is y, we must replace x in terms of y.

                                         2    2                                             2
                                   Since x + y = 9, it follows that in the first quadrant x =  9 − y .
                                   Hence
                                                              3


                                                                        2
                                                     V = 2π    2y 9 − y dy
                                                             2
                                                                                        2
                                   We evaluate this integral by making the substitution u = 9 − y .
                                   It follows that du =−2ydy. We change the limits of integration
                                   by observing that u = 5 when y = 2 and u = 0 when y = 3.
                                                                  0

                                                                   √
                                                       V =−2π        udu
                                                                 5
                                                                5

                                                         = 2π     u 1/2  du
                                                                0
                                                                      5
                                                               2
                                                         = 2π    u 3/2
                                                               3
                                                                     0
                                                            4   3/2
                                                         =   π(5   − 0)
                                                            3
                                                               √
                                                            20π 5
                                                         =
                                                              3












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