I'm trying to calculate this integral : $$\int_{D}\int{\sin(x-y)dxdy} $$ where $D=${$(x,y)|x^2+y^2\le1$} ,I've changed the parameters to: $$x(u,v)=r\cos(\theta+\frac{\pi}4),\ y(u,v)=r\sin(\theta+\frac{\pi}4)$$ and finally I got stuck at the integral: $$\int_{0}^{2\pi}\int_{0}^{1}|r|\sin(-\sqrt2r\sin(\theta))drd\theta$$ which I believe is very hard to solve.

Can anyone help?

  • $\begingroup$ hint: Set $r=x-y$,$R=x+y$ $\endgroup$ – tired Jan 24 '17 at 19:30
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    $\begingroup$ what are the limits on the original double integral? $\endgroup$ – Umberto P. Jan 24 '17 at 19:32
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    $\begingroup$ $\sin(x-y)=\sin x \cos y - \cos x \sin y$ $\endgroup$ – W.R.P.S Jan 24 '17 at 19:41
  • $\begingroup$ Umberto, unit circle $x^2+y^2\le1$ $\endgroup$ – CodeHoarder Jan 24 '17 at 19:45
  • $\begingroup$ Guys, i don't understand how these clues help in this problem. $\endgroup$ – CodeHoarder Jan 25 '17 at 5:58

The integral of an odd integrable function over a symmetric (with respect to the origin) domain is always zero.


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