What is the surface area of the solid created by revolving f ( x ) = 1 x 2 for x ∈ [ 1 , 2 ] around the x-axis?

Question

What is the surface area of the solid created by revolving   f  (  x  )  =      1          x              2             for   x  ∈  [  1  ,  2  ] around the x-axis?

Leland Ochoa · Accepted Answer

The first step is to check to make sure that f(x) is defined on the interval [1,2]. The only discontinuity of the the function is at x=0, so we do not have to concern ourselves with this.The surface area of a curve on [a,b] rotated around the x-axis is defined by  S  =  2  π      ∫          a              b        f  (  x  )      1    +    (          f      ′        (    x    )          )              2              d  x  . The derivative of f(x) is       f    ′    (  x  )  =  −      2          x              3            . This means that   (      f    ′    (  x  )      )          2        =  (  −      2          x              3                  )          2        =      4          x              6              S  =  2  π      ∫          1              2            1          x              2                  1    +          4              x                  6                      d  x  S  =  2  π      ∫          1              2            1          x              2                                          x                      6                          +        4                    x                  6                      d  x  S  =  2  π      ∫          1              2            1          x              5                        x              6              +    4    d  xThis integral has no elementary solution.But according to Wolfram Alpha, the surface area is approximately 4.46 square units.

What is the surface area of the solid created by revolving f ( x ) = 1

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