Solve $\iint_{D} \exp(-(x^2+y^2))dxdy$ when $ D= \{ ( x,y ):x^2+y^2\leq R^2, 0\leq y\leq x \}$ 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?
$$\iint_{E} {e}^{-(r^2)}r drd\theta$$
 A: 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.
A: 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. 
