Page 235 - Calculus Demystified
P. 235

CHAPTER 8
                     222
                                   EXAMPLE 8.3                  Applications of the Integral
                                   A solid has base in the x-y plane consisting of a unit square with center at
                                   the origin and vertices on the axes. The vertical cross-section at position x
                                   isitself a square. Refer to Fig. 8.9. What isthe volume of thissolid?
















                                                               Fig. 8.9

                                   SOLUTION
                                     It is sufficient to calculate the volume of the right half of this solid, and to
                                                                                         √
                                   double the answer. Of course the extent of x is then 0 ≤ x ≤ 1/ 2.At position
                                                                                  √
                                   x, the height of the upper edge of the square base is 1/ 2 − x. So the base of
                                                             √
                                   the vertical square slice is 2(1/ 2−x) (Fig. 8.10). The area of the slice is then
                                                        
   √          2   √        2
                                                 A(x) = 2 1/ 2 − x     =    2 − 2x .






                                                                               _
                                                                          2(1/√2   x)




                                                                _
                                                           2(1/√2   x)
                                                              Fig. 8.10

                                   It follows that
                                                                 √
                                                                1/ 2
                                                      V = 2 ·       A(x) dx
                                                              0
                                                                √

                                                              1/ 2  √        2
                                                        = 2          2 − 2x  dx
                                                             0
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