Write and evaluate the definite integral that represents the area of the surface generated by revolving the curve y = 64 − x 2 , − 1 ≤ x ≤ 1

Question

Write and evaluate the definite integral that represents the area of the surface generated by revolving the curve  y  =      64    −          x      2        ,        −  1  ≤  x  ≤  1

Johnathon Mcmillan · Accepted Answer

6) For function   y  =  f  (  x  )  ,           a  ≤  x  ≤  bsurface area obtained obtained by rotating the curve is given by,  s  =  2  π      ∫    a    b    f  (  x  )      1    +                            d          y                          d          x                    2        d  x  y  =  f  (  x  )  =      64    −          x      2        ⇒      f    ′    (  x  )  =            −      2      x              2              64        −                  x          2                      =            −      x              64      −              x        2                ⇒  [      f    ′    (  x  )      ]    2    =            x      2              (      64      −              x        2            )          ⇒  1  +  (      f    ′    (  x  )      )    2    =  1  +            x      2              64      −              x        2              =      64          64      −              x        2                ⇒  f  (  x  )  ⋅      1    +    (                  d        y                    d        x                  )    2    =      64    −          x      2        ⋅      8          64      −              x        2              =  8Answer:  2  π      ∫          −      1        1    8  d  x  =  32  π

Write and evaluate the definite integral that represents the area of the surface generated by revolving the curve y=sqrt(64-x^2), -1<=x<=1

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