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Arclength in polar coordinates

  1. Oct 29, 2006 #1
    I am working on a problem regarding arclength-which asks to find the arclength for r=2-2sinx (x=theta) I worked out the integral to the integral of the square root of 8-8sinx but i didnt know how to integrate from there--any help?

  2. jcsd
  3. Oct 29, 2006 #2
    I am not 100% sure, but can't you factor out the 8 and then take that outside the integral sign which would leave you with the integral of 1-sinx dx?, or 8(integral of (1-sinx)dx).
    Last edited: Oct 29, 2006
  4. Oct 30, 2006 #3
    I don't think this will produce anything productive. Judging by the answer my 89 spewed out, it's a difficult substitution and/or trig identity problem. Wish I could help!
  5. Oct 30, 2006 #4
    Well, I integrated 8-8sinx dx and got 8cosx+8x+C. I failed to see that it was the square root of that. Sorry about that and thanks for catching that vsage.
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