Arclength in polar coordinates

  • Thread starter nate808
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  • #1
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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?

Thanks
-nate808
 

Answers and Replies

  • #2
102
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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:
  • #3
vsage
prace said:
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).

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!
 
  • #4
102
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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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