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I have no idea how to calculate this kind of double integral.

$\int_D \cos(y) ~dA$ where $D=\{ 0 \leq x \leq 2\pi,~ |y|\leq x \}$.

Any help with this?


(added from now deleted answer)

Okay,

I tried this $\int_{0}^{ 2\pi\ } \int_{0}^{\ x} \cos y \, dydx = 0$

but it is wrong.

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    $\begingroup$ Please, show what you tried. People in this place like to see your efforts before they can help you. Also, use $\tt LaTeX$-MathJax. $\endgroup$ – Felix Marin Feb 16 '14 at 10:43
  • $\begingroup$ Welcome and i seriously recommend you to read the faq to understand the norms of this site. Normally you write your attempted solution in the question. $\endgroup$ – Lost1 Feb 16 '14 at 12:44
  • $\begingroup$ The approach taken (at Don Antonio's suggestion) gives twice the integral shown (added from now-deleted Answer) by symmetry (cosine being even, as Don Antonio said). But twice zero is... $\endgroup$ – hardmath Feb 16 '14 at 16:18
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Hints:

Since $\;|y|\le x\iff -x\le y\le x\;$ , as $\;\;x\ge 0\;$, this gives the following integration region in $\;\Bbb R^2\;$:

$$\int\int\limits_D\cos y\,dA=\int\limits_0^{2\pi}\int\limits_{-x}^x\cos y\;dydx=\int\limits_0^{2\pi}\left(\sin x-\sin(-x)\right)dx=\left.-2\cos x\right|_0^{2\pi}=0$$

which doesn't surprise since $\;\cos x\;$ , the original integrand, is being integrated on a symmetric interval around zero.

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  • $\begingroup$ Me neither, @hardmath...yet I'm almost positive that was the original formulation and the OP perhaps changed within the first 5 minutes of having posted his question, so that it didn't get registered as "editing", and in the meanwhile I began trying to solve the question and didn't notice the change...or there wasn't any change and I just misread: go figure! $\endgroup$ – DonAntonio Feb 16 '14 at 11:20
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    $\begingroup$ Anyway, thanks @hardmath. The question's now edited. $\endgroup$ – DonAntonio Feb 16 '14 at 11:24
  • $\begingroup$ Thanks DonAntonio ! :) Now i got the right answer and it really was 0. $\endgroup$ – user129180 Feb 16 '14 at 12:12

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