3
$\begingroup$

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? $\iint_{E} {e}^{-(r^2)}r drd\theta$

  • 3
    I would suggest doing things in polar coordinates....2012-11-29

2 Answers 2

3

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.

  • 2
    I don't think $\sin t\leq \cos t$ on $[3\pi/4,\pi]$, so you probably want to just integrate for $t\in [0,\pi/4]$.2012-11-29
1

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.

  • 0
    Oh yes, I forgot the integral for $r$. Now I edited it. Thanks.2012-11-29