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                                   Chapter 12: Who Needs Freud? Using the Integral to Solve Your Problems


                Arc Length and Surfaces of Revolution



                          Integration determines the length of a curve by cutting it up into an infinite number of
                          infinitesimal segments, each of which is sort of the hypotenuse of a tiny right triangle.
                          Then your pedestrian Pythagorean Theorem does the rest. The same basic idea applies
                          to surfaces of revolution. Here are two handy formulas for solving these problems:
                             Arc length: The length along a function, f xh, from a to b is given by
                                                                  ^
                                               b
                                   Arc Length= #  1 + ` f l ^ hj 2  dx
                                                        x
                                              a
                             Surface of revolution: The surface area generated by revolving the portion of a
                              function, f xh, between x = a and x = b about the x-axis is given by
                                       ^
                                                   b               2
                                                 π
                                              =
                                                                x
                                                    f x
                                   Surface Area 2 # ^ h  1 + ` f l ^ hj  dx
                                                  a
                Q.   What’s the arc length along f x =  x  / 2 3  from  Q.  Find the surface area generated by revolving
                                              ^ h
                     x =  8 to x =  27?                                   1  3
                                                                    f x =   x  ^ 0 #  x #  2h about the x-axis.
                                                                     ^ h
                                                                          3
                A.   The arc length is about 19.65.            A.              π
                                                                    The area is  `
                                                                               9  17 17 -  1j.
                     1. Find f l ^ xh.
                                                                    1. Find the function’s derivative.
                                          2  -  / 1 3
                               / 2 3
                                      x =
                       f x =  x    f l ^ h  x                                1  3             2
                        ^ h
                                                                                        x =
                                                                       ^ h
                                          3                           f x =   x       f l ^ h  x
                                                                             3
                     2. Plug into the arc length formula.
                                                                    2. Plug into the surface area formula.
                                      27
                                  27 = #    4  -  / 2 3                                2  1        2
                                                                                    π
                                                                                                  2
                                                                                 =
                       Arc Length to8   1 +  9  x  dx                 Surface Area 2 #   x  3  1 + _ x i  dx
                                     8                                                  3
                                                                                     0
                     3. Integrate.                                                  π  2
                                                                                               4
                                                                                 =  2   x #  3  1 +  x dx
                       These arc length problems tend to pro-                       3
                                                                                      0
                       duce tricky integrals; I’m not going to
                       show all the work here.                        You can do this integral with
                                                                      u-substitution.
                           27
                        1
                       = #   9 +  4 x  -  / 2 3  dx
                        3                                              u 1 +  x  4  when  x =  0 , u 1
                                                                                             =
                                                                         =
                          8
                                                                            3
                                                                                             =
                                                                         =
                           27                                         du 4 x dx  when  x =  2 , u 17
                        1
                                   / 2 3
                       =    x #  -  / 1 3  9 x +  4  dx
                        3                                                     2
                          8                                             2 π 1
                                                                                       4
                                                                      =   $    4 #  x  3  1 +  x dx
                       You finish this with a u-substitution,           3  4  0
                                  / 2 3
                       where u 9=  x +  4.                                17
                                                                        π
                                                                      = #    / 1 2
                                  85                                       u du
                                1
                              = #   1  u du                             6  1
                                       / 1 2
                                3   6                                          17
                                 40                                     π 2  / 3 2
                                       85                             = ;   u E
                                1 2   / 3 2                             6 3
                              =   ;  u E                                       1
                               18 3                                     π
                                       40
                                                                      = ` 17 17 -  1j
                                85 85 -  80 10                          9
                              =
                                     27
                              .  19 .65
                       An eminently sensible answer, because
                       from x = 8 to x = 27, x  / 2 3  is very similar to
                       a straight line of length 27 – 8, which
                       equals 19.
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