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?
 A: Yes, using polar coordinates the boundaries are:
$$0 \leq r \leq R \\
  0 \leq \theta \leq 2 \pi$$
Since $D(R)$ is the disk of radius $R$ with center at $(0,0)$:
$D(R)=\{(x,y): x^2+y^2 \leq R^2\}$
So we have the following:
$$x = r\cos \theta, y = r \sin\theta \\
  \renewcommand{\intd}{\,\mathrm{d}}
  \intd x \intd y = r \intd r \intd \theta$$
$$
\begin{align}
  \iint_{D(R)} e^{-(x^2+y^2)} \intd x \intd y
  &= \int_0^{2 \pi} \int_0^R e^{-r^2} r \intd r \intd \theta \\
  &= \int_0^{2 \pi} \int_0^R \frac{\mathrm{d}}{\intd r} \left( -\frac{1}{2} e^{-r^2} \right) \intd r \intd \theta \\
  &= \int_0^{2 \pi} \left( -\frac{1}{2} e^{-r^2} \right )_0^R \intd \theta \\
  &= \int_0^{2 \pi} -\frac{1}{2} \left( e^{-R^2} - 1 \right) \intd \theta \\
  &= -\frac{1}{2} \left( e^{-R^2} - 1 \right) 2 \pi \\
  &= -\pi \left( e^{-R^2} - 1 \right) \\
  &= \pi \left( 1 - e^{-R^2} \right)
\end{align}$$
A: Hint: Let $x = r\cos \theta$ and $y = r\sin \theta$, then
$$I = \int_{0}^{2\pi} \int_{0}^R r e^{-r^2} \,\mathrm{d}r \,\mathrm{d}\theta$$
A: You have a defined region to integrate about, which is $D(R)=\{(x,y) \in \mathbb{R}^{2}| x^{2}+y^{2} \leq R^{2}\}$ as the disc with radius $R$.
At first, you should choose an adequate parametrization for making the problem easier. Here you might use the transformation $(r, \theta) \mapsto (x, y)=(r \cos \theta, r \sin \theta)$ with $r \in [0, R]$, $\theta \in [0, 2\pi)$.
