Double integral in polar coordinates

Change of variables.
$$x = r\cos\theta, \quad y = r\sin\theta, \quad dx\,dy = r\,dr\,d\theta.$$
The domain becomes $0 \le r \le 1$, $0 \le \theta \le 2\pi$.

Computation.
$$\iint_D e^{-(x^2+y^2)} \, dx \, dy = \int_0^{2\pi} \int_0^1 e^{-r^2} r \, dr \, d\theta.$$
The $r$-integral:
$$\int_0^1 r e^{-r^2} \, dr = \left[ -\frac12 e^{-r^2} \right]_0^1 = \frac12(1-e^{-1}).$$
Hence
$$\iint_D e^{-(x^2+y^2)} \, dx \, dy = 2\pi \cdot \frac12(1-e^{-1}) = \pi(1-e^{-1}).$$

Remark. This computation is the basis for proving $\int_{-\infty}^{+\infty} e^{-x^2}dx = \sqrt{\pi}$.