Solution
= Solution
{c}
The <annulus> is $1\leq r\leq2$ in <polar coordinates>. Their <Jacobian determinant> supplies $dx\,dy=r\,dr\,d\theta$, so
$$
\begin{aligned}
\iint_D e^{-x^2-y^2}\,dx\,dy
&=\int_0^{2\pi}\int_1^2 e^{-r^2}r\,dr\,d\theta\\
&=\pi\left[-e^{-r^2}\right]_1^2
=\boxed{\pi(e^{-1}-e^{-4})}.
\end{aligned}
$$