Stokes' Theorem problem

gabbagabbahey
Homework Helper
Gold Member
Are we all integrating $$\int -4(sin\theta)^2 + (2cos\theta)^4$$ from 0 to 2pi?

$$\int_0^{2\pi}\sin^n\theta d\theta=\int_0^{2\pi}\cos^n\theta d\theta=0$$

is only true for odd $n$....Perhaps this is the source of your error?

ideasrule
Homework Helper
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.

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ideasrule
Homework Helper
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.