Finding the volume of the solid generated by revolving the given curve.The objective is to find the volume of the solid generated by revolving the curve y = a 3 a 2 + x 2 about its asymptote.Observing the given function yields that y ≠ 0, hence y = 0 is the asymptote to the given curve. Thus, the volume of the solid formed by revolving the given curve about x − a x i s is gives as V = 2 π ∫ 0 ∞ ( f ( x ) 2 − 0 ) d x.Which gives: 2 π ∫ 0 ∞ a 6 ( a 2 + x 2 ) 2 d xNow, this integral is quite tedious and I don't know why the result tends to infinity. The integral takes the form 1 x 4 + 2 x 2 + 1 for a = 1, which is transformed into 1 x 2 x 2 + 2 + 1 x 2 = 2 [ 1 + 1 x 2 x 2 + 2 + 1 x 2 − ( 1 − 1 x 2 ) x 2 + 2 + 1 x 2 ] which can be integrated, but this integral is tending to infinity.

Question

Finding the volume of the solid generated by revolving the given curve.The objective is to find the volume of the solid generated by revolving the curve   y  =                    a        3                              a          2                +                  x          2                     about its asymptote.Observing the given function yields that   y  ≠  0, hence   y  =  0 is the asymptote to the given curve. Thus, the volume of the solid formed by revolving the given curve about   x  −  a  x  i  s is gives as   V  =  2  π      ∫    0          ∞        (  f  (  x      )    2    −  0  )  d  x.Which gives:   2  π      ∫    0    ∞                      a        6                    (                  a          2                +                  x          2                          )          2                      d  xNow, this integral is quite tedious and I don&#039;t know why the result tends to infinity. The integral takes the form             1                        x          4                +        2                  x          2                +        1             for   a  =  1, which is transformed into                     1                  x          2                                      x          2                +        2        +                  1                      x            2                                =  2  [                    1        +                  1                      x            2                                                x          2                +        2        +                  1                      x            2                                −                    (        1        −                  1                      x            2                          )                              x          2                +        2        +                  1                      x            2                                ] which can be integrated, but this integral is tending to infinity.

Kenley Rasmussen · Accepted Answer

Step 1Neither of your two integrals in the last step are divergent. The first one is      I    1    =      ∫    0    ∞              1      +              1                  x          2                                    x        2            +      2      +              1                  x          2                        d  x  =      ∫    0    ∞                      x        2            +      1                      x        4            +      2              x        2            +      1          d  x  =      ∫    0    ∞              d      x                      x        2            +      1        =            [      arctan      ⁡      x      ]        0    ∞    =      π    2    .Step 2So that&#039;s nice and convergent. The other one is a bit trickier, but we can find it with a quick jaunt into the complex plane:  ∫            1      −              1                  x          2                                    x        2            +      2      +              1                  x          2                        d  x  =  ∫                    x        2            −      1                      x        4            +      2              x        2            +      1          d  x  =      1    2        [    ∫          1              (        x        +        i                  )          2                      +    ∫          1              (        x        −        i                  )          2                      ]      =  −      1    2        [          1              x        +        i              +          1              x        −        i              ]    =  −      1    2              2      x                      x        2            +      1        =  −      x                  x        2            +      1      and so       ∫    0    ∞              1      −              1                  x          2                                    x        2            +      2      +              1                  x          2                        d  x  =            [      −              x                              x            2                    +          1                    ]        0    ∞    =  0..

Raiden Barr · Accepted Answer

Step 1This seems like a trig substitution problem to me, albeit a not-so-straightforward one.Let   x  =  a  tan  ⁡  θ. Then   d  x  =  a      sec    2    ⁡  θ    d  θ, and       a    2    +      x    2    =      a    2        sec    2    ⁡  θ. Since   θ  =      tan          −      1        ⁡      (          x      a        )  ,   θ  =  0 when   x  =  0, and   θ  →      π    2   as   x  →  ∞. Therefore                    2        π                  ∫          0          ∞                                      a            6                                (                          a              2                        +                          x              2                                      )              2                                              d        θ                            =        2        π                  ∫          0                      π                          /                        2                                                a            6                                              a              4                                      sec              4                        ⁡            θ                                  a                  sec          2                ⁡        θ                d        θ                                          =        2        π                  a          3                          ∫          0                      π                          /                        2                                    cos          2                ⁡        θ                d        θ                                          =        π                  a          3                          ∫          0                      π                          /                        2                          (        1        +        cos        ⁡        2        θ        )                d        θ                                          =        π                  a          3                                      [            θ            +                          1              2                        sin            ⁡            2            θ            ]                    0                      π                          /                        2                                                            =                                            π              2                                      a              3                                2                    Step 2I can&#039;t quite figure out how you intended to integrate  ∫            1      +              1                  x          2                                    x        2            +      2      +              1                  x          2                        d  x, but it could be that you were trying to represent a convergent improper integral as the difference of two divergent improper integrals. That leads to trouble, since   ∞  −  ∞ is an indeterminate limit form.

The objective is to find the volume of the solid generated by revolving the curve y=a^3/(a^2+x^2) about its asymptote.

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