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

Question

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

abadunayj6f · Accepted Answer

The surface area for a revolution around the x axis is given by:      S    =    2    π                  ∫                  a                          b                            f              (        x        )                            1        +                              (                                                            d                  y                                                                              d                  x                                                      )                    2                            d      x      (which is basically a projection of the circumference along the function f(x) whose arc length you could have found.)In this case,                     (                                            d              y                                                          d              x                                      )            2       is given by:                    (                                            d              y                                                          d              x                                      )            2        =                  (        -                  e                      2            -            x                          )            2      And then we have:      S    =    2    π                  ∫        1        2                    e              2        -        x                            1        +                              (            -                          e                              2                -                x                                      )                    2                            d      x      First, let       u    =    -          e              2        -        x            . Therefore,       d    u    =          e              2        -        x                    d      x       and:      S    =    2    π    ∫                  1        +                  u          2                      d    u  where we omit the integral bounds for now. Then we can see it looks like the form                               a          2                +                  x          2                    , then let       u    =          tan      θ       to get       d    u    =                  sec        2            θ        d    θ  . Therefore:      S    =    2    π    ∫                  1        +                              tan            2                    θ                                    sec        2            θ        d    θ        =    2    π    ∫                  sec        3            θ        d    θ  And you should have written down the following integral in class:      =    2    π                                        [                          1              2                                      (                              sec                θ                                            tan                θ                            +              ln                              |                                  sec                  θ                                +                                  tan                  θ                                |                            )                        ]                          ⏞                    ∫                              sec            3                    θ                d        θ                  =    π          sec      θ              tan      θ        +    π    ln          |              sec        θ            +              tan        θ            |      Next, back-substitute.      ⇒    π    u                  1        +                  u          2                      +    π    ln          |                        1          +                      u            2                              +      u      |            =    π    u                  1        +                              (            -                          e                              2                -                x                                      )                    2                      +    π    ln          |                        1          +                                    (              -                              e                                  2                  -                  x                                            )                        2                              +              (        -                  e                      2            -            x                          )            |            =    -    π          e              2        -        x                            1        +                  e                      4            -            2            x                                +    π    ln          |                        1          +                      e                          4              -              2              x                                          -              e                  2          -          x                    |      As it turns out,                     1        +                  e                      4            -            2            x                                -          e              2        -        x              ≥    0  , so we can remove the absolute values.      =    -    π          e              2        -        x                            1        +                  e                      4            -            2            x                                +    π          ln              (                              1            +                          e                              4                -                2                x                                                    -                  e                      2            -            x                          )            Thus, we evaluate from 1 to 2:      ⇒          [      -      π              e                  2          -          2                                      1          +                      e                          4              -              2              ⋅              2                                          +      π              ln                  (                                    1              +                              e                                  4                  -                  2                  ⋅                  2                                                              -                      e                          2              -              2                                )                    ]        -          [      -      π              e                  2          -          1                                      1          +                      e                          4              -              2              ⋅              1                                          +      π              ln                  (                                    1              +                              e                                  4                  -                  2                  ⋅                  1                                                              -                      e                          2              -              1                                )                    ]            =          [      -      π                        2                    +      π              ln                  (                                    2                                -          1          )                    ]        -          [      -      π      e                        1          +                      e            2                              +      π              ln                  (                                    1              +                              e                2                                              -          e          )                    ]            =    -    π                  2              +    π          ln              (                              2                          -        1        )              +    π    e                  1        +                  e          2                      -    π          ln              (                              1            +                          e              2                                      -        e        )                                22.943

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

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