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Stokes' Theorem problem

  • Thread starter mathman44
  • Start date
  • #26
gabbagabbahey
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Are we all integrating [tex]
\int -4(sin\theta)^2 + (2cos\theta)^4
[/tex] from 0 to 2pi?
[tex]\int_0^{2\pi}\sin^n\theta d\theta=\int_0^{2\pi}\cos^n\theta d\theta=0[/tex]

is only true for odd [itex]n[/itex]....Perhaps this is the source of your error?
 
  • #27
ideasrule
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No, the source of my error was just sloppiness. (Specifically, I thought that cos^2 (x)=(cos(2x)-1)/2. It took an amazingly long time to fix that.) Now I get 8*pi, in agreement with everybody else.
 
Last edited:
  • #28
ideasrule
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By an odd coincidence, you actually get the same result either way.
I don't think it's a coincidence. z doesn't change across the flat surface, so it's a constant, and whether you set a constant before or after integration shouldn't matter.
 

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