What is the surface area of the solid created by revolving f ( x ) = 2 x 2 − 4 x + 8 , x ∈ [ 1 , 2 ] around the x axis?

Question

What is the surface area of the solid created by revolving  f  (  x  )  =  2      x          2        −  4  x  +  8  ,  x  ∈  [  1  ,  2  ] around the x axis?

Charlize Manning · Accepted Answer

The surface area S created by revolving the function f(x) around the x-axis on the interval   x  ∈  [  a  ,  b  ] can be found through:  S  =  2  π      ∫          a              b        f  (  x  )      1    +    [          f      ′        (    x    )          ]              2              d  xUsing   f  (  x  )  =  2      x          2        −  4  x  +  8 and f&#039;(x)=4x-4 on the interval   x  ∈  [  1  ,  2  ] gives:  S  =  2  π      ∫          1              2        (  2      x          2        −  4  x  +  8  )      1    +    (    4    x    −    4          )              2              d  xActually integrating this is very complex and far beyond the scope of this problem, so put this into a calculator-be mindful of parentheses.The answer you receive (don&#039;t forget to multiply by   2  π ) should be  S  ≈  100.896

What is the surface area of the solid created by revolving f ( x ) = 2 x <m

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