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)


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

but it is wrong.

  • 1
    $\begingroup$ 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


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.