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Solve $$\iint_{D} {e}^{-(x^2+y^2)}dxdy$$ when $ D=\left \{ \left ( x,y \right ):x^2+y^2\leq R^2, 0\leq y\leq x \right \}$.

I have problems with how i should decide the intervals of the integrals? And what they are? $$\iint_{E} {e}^{-(r^2)}r drd\theta$$

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I would suggest doing things in polar coordinates.... –  froggie Nov 29 '12 at 13:43

2 Answers 2

Froggie's hint is right on the money:

$$x=r\cos t\,\,,\,y=r\sin t\,\,,\,r^2\leq R^2\,\,,\,\,0\leq r\sin t\leq r\cos t\Longrightarrow 0\leq\sin t\leq\cos t$$

So $\,t\,$ must be chosen s.t.

$$0\leq t\leq \pi\,\,\,(\text{because}\,\,\sin t\geq 0)\,\,,\,\,\text{and also}\,\,t\in\left[0,\frac{\pi}{4}\right]\cup\left[\frac{3\pi}{4},\pi\right]\,\,(\text{so that}\,\,\sin t\leq\cos t)$$

Now try to take it from here.

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okay thank you! –  stuck Nov 29 '12 at 14:03
    
+1 Nice approach. I hope the OP can do the rest. –  Babak S. Nov 29 '12 at 14:12
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I don't think $\sin t\leq \cos t$ on $[3\pi/4,\pi]$, so you probably want to just integrate for $t\in [0,\pi/4]$. –  froggie Nov 29 '12 at 15:17

Following the suggestion of froggie, we can calculate the integral using polar coordinates. In terms of polar coordinates $(r,\theta)$, we have $$D=\{(r,\theta):0\leq r\leq R, 0\leq \theta\leq\frac{\pi}{4}\}.$$ To see this, note that $x^2+y^2\leq R^2$ gives $r^2\leq R^2$. On the other hand, $0\leq y\leq x$ gives $0\leq \sin\theta\leq \cos\theta$, which implies that $0\leq\tan\theta\leq 1$.

Note also that the area element $dxdy=rdrd\theta$. Therefore, the integral can be written as $$\iint_{D} {e}^{-(x^2+y^2)}dxdy=\int_0^{\frac{\pi}{4}}\int_0^Re^{-r^2}rdrd\theta.$$ I think you can solve it from here.

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I think you also want to integrate in the $r$ variable from $0$ to $R$ in your last integral. –  froggie Nov 29 '12 at 15:30
    
but what interval have r? [0,R]? How do i solve that? –  stuck Nov 29 '12 at 15:54
    
@stuck: I think Paul just miswrote his last integral. Should be $$\int_0^{\pi/4}\int_0^R e^{-r^2}r\,dr\,d\theta.$$ –  froggie Nov 29 '12 at 15:59
    
Oh yes, I forgot the integral for $r$. Now I edited it. Thanks. –  Paul Nov 29 '12 at 21:20

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