So I know the length L of the curve y = R 2 − x 2 from x = 0 to x = a where | a | &lt; R is given by: L = ∫ 0 a R R 2 − x 2 d xNow I must set up the arc length integral and simplify it so that it is in the form listed above. L = ∫ 0 a 1 + ( d y d x ) 2 d xand d y d x = − x R 2 − x 2 ( d y d x ) 2 = x 2 R 2 − x 2 so L = ∫ 0 a 1 + x 2 R 2 − x 2 d xI am unsure where to go from here to simplify into the first integral, any help would be greatly appreciated. Thanks

Question

So I know the length L of the curve   y  =            R              2              −          x              2             from   x  =  0 to   x  =  a where       |    a      |    &amp;lt;  R is given by:  L  =      ∫    0    a        R                  R                  2                    −              x                  2                      d  xNow I must set up the arc length integral and simplify it so that it is in the form listed above.  L  =      ∫    0    a        1    +                  (                              d            y                                d            x                          )                    2              d  xand            d      y              d      x        =  −      x                  R                  2                    −              x                  2                                (                        d          y                          d          x                    )              2        =            x              2                            R                  2                    −              x                  2                    so  L  =      ∫    0    a        1    +                  x                  2                                      R                      2                          −                  x                      2                                d  xI am unsure where to go from here to simplify into the first integral, any help would be greatly appreciated. Thanks

beentjie8e · Accepted Answer

Assume   R  &amp;gt;  a and   a  ,  R  ∈            R      +      :  L  =      ∫          0              a            1    +                  (                  ∂                      ∂            x                                    [                                    R              2                        −                          x              2                                ]                )            2           d  x  =      ∫          0              a            1    +                  (        −                  x                                    R              2                        −                          x              2                                      )            2           d  x  =      ∫          0              a            1    +                  x        2                              R          2                −                  x          2                         d  x  =      ∫          0              a                                    R          2                −                  x          2                                      R          2                −                  x          2                      +                  x        2                              R          2                −                  x          2                         d  x  =      ∫          0              a                                    R          2                −                  x          2                +                  x          2                                      R          2                −                  x          2                         d  x  =      ∫          0              a                          R        2                              R          2                −                  x          2                         d  x  =      ∫          0              a                                    R          2                          R          2                                                  R            2                                R            2                          −                              x            2                                R            2                                   d  x  =      ∫          0              a            1          1      −                        x          2                          R          2                         d  x  =      ∫          0              a                                    R          2                −                  x          2                +                  x          2                                      R          2                −                  x          2                         d  x  =      ∫          0              a                          R        2                              R          2                −                  x          2                         d  x  =      ∫          0              a                                    R          2                          R          2                                                  R            2                                R            2                          −                              x            2                                R            2                                   d  x  =      ∫          0              a            1          1      −                        x          2                          R          2                         d  x  =Substitute   u  =      x    R   and   d  u  =      1    R       d  x.This gives a new lower bound   u  =      0    R    =  0 and upper bound   u  =      a    R  :  R      ∫          0                      a        R                  1          1      −              u        2                 d  u  =  R            [      arcsin      ⁡      (      u      )      ]              0                      a        R              =  R      (    arcsin    ⁡          (              a        R            )        −    arcsin    ⁡    (    0    )    )    =  R  arcsin  ⁡      (          a      R        )

motsetjela · Accepted Answer

Ah never mind I solved it, assuming R and x are positive,      1    +                  x                  2                                      R                      2                          −                  x                      2                                =  R            1                        R                      2                          −                  x                      2                                =      R                  R                  2                    −              x                  2

So I know the length L of the curve y=sqrt(R^2−x^2) from x=0 to x=a where |a|<R is given by: L=int^a_0 (R)/(sqrt(R^2−x^2))dx

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