Evaluation of the integral of $e^{-(x^2+y^2)}$ over a disk 
Show that
  $$\renewcommand{\intd}{\,\mathrm{d}}
    \iint_{D(R)} e^{-(x^2+y^2)} \intd x \intd y = \pi \left(1 - e^{-R^2}\right)$$
  where $D(R)$ is the disc of radius $R$ with center $(0,0).$

I have never been asked to calculate a double integral without a defined region, so I don't even know where to start. I don't know the boundaries. 
This is my guess:
$$0 < r < R\\
0 < \theta < 2\pi $$
Is this correct?

Hint: Let $x = r\cos \theta$ and $y = r\sin \theta$, then
$$I = \int_{0}^{2\pi} \int_{0}^R r e^{-r^2} \,\mathrm{d}r \,\mathrm{d}\theta$$
