How can we integrate a function for surface area of revolution? S = 2 π ∫ 0 8 ( − 1 24 x 2 + 2 3 x ) 1 + ( − 1 12 x + 2 3 ) 2 d xThis is the equation (with my values plugged in) for surface area of revolution which I found online and I understand the derivation, thus I wanted to use it for a math project I am currently working on. I attempted to use the quadratic function (outside the radical) and inside the radical is the first derivative squared. However, I am stuck in evaluating the integral. I tried u substitution where u is equal to − 1 / 12 x + 2 / 3 but was unable to finish it.Is trigonometry required and does the u substitution I attempted not work in this case? What is the Surface area (answer to the integral)?

Question

How can we integrate a function for surface area of revolution?  S  =  2  π      ∫          0              8        (  −      1    24        x    2    +      2    3    x  )      1    +    (    −          1      12        x    +          2      3              )      2        d  xThis is the equation (with my values plugged in) for surface area of revolution which I found online and I understand the derivation, thus I wanted to use it for a math project I am currently working on. I attempted to use the quadratic function (outside the radical) and inside the radical is the first derivative squared. However, I am stuck in evaluating the integral. I tried u substitution where u is equal to   −  1      /    12  x  +  2      /    3 but was unable to finish it.Is trigonometry required and does the u substitution I attempted not work in this case? What is the Surface area (answer to the integral)?

Houston Ochoa · Accepted Answer

Step 1Note that                    S                            =        2        π                  ∫          0          8                          (                      2            3                    x          −                      1            12                                x            2                    )                          1          +                                    (                              2                3                            −                              1                12                            x              )                        2                                  d        x                                                         =                              2            π                    12                          ∫          0          8                          (          8          x          −                      x            2                    )                          1          +                                    (                              2                3                            −                              1                12                            x              )                        2                                  d        x                                                         =                              2            π                    12                          ∫          0          8                          (          8          x          −                      x            2                    )                                      1                          12              2                                            (                          x              2                        −            16            x            +            208            )                                  d        x                                                         =                              2            π                                12            2                                    ∫          0          8                          (          8          x          −                      x            2                    )                                      x            2                    −          16          x          +          208                        d        x                                                         =                  π          72                          ∫          0          8                x        (        8        −        x        )                              12            2                    +          (          8          −          x                      )            2                                  d        x            and with   y  =  8  −  x, noting that   d  y  =  −  d  x, but we also need to swap the integration boundaries, so we get                    S                            =        −                  π          72                          ∫          8          0                (        8        −        y        )        y                              12            2                    +                      y            2                                  d        y                                                         =                  π          72                          ∫          0          8                          (          8          y          −                      y            2                    )                                      12            2                    +                      y            2                                  d        y            Step 2I would continue by integration by parts, via                    S                            =                  π          72                          ∫          0          8                f        (        y        )                g        (        y        )                d        y                            f        (        y        )                            =        8        y        −                  y          2                                    g        (        y        )                            =                              12            2                    +                      y            2                              The exact result is                    S                            =        36        π        log        ⁡                  (                      2            3                    +                      1            +                                          (                                  2                  3                                )                            2                                )                +                              π                        424                                      13                                27                −        64        π                                                         =        36        π        log        ⁡                  (                      2            3                    +                                    13                        3                    )                +                              π                        424                                      13                                27                −        64        π                                                         =        36        π        asinh        ⁡                  (                      2            3                    )                +                              π                        424                                      13                                27                −        64        π                                                         ≈        47.519

How can we integrate a function for surface area of revolution? S=2pi int_0^8(-1/(24)x^2+2/3x)sqrt{1+(-1/(12)x+2/3)^2}dx

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