Answer
$=\pi \left(-e^{-4}+1\right) $
Work Step by Step
$\int_{-2}^{2} \int_{\sqrt{4-y^{2}}}^{\sqrt{4+y^{2}}} e^{-\left(y^{2}+x^{2}\right)} d x d y$
changing to polar coordinates $R=\{(r . \theta): 0 \leqslant \theta \leqslant 2 \pi, 0 \leqslant r \leqslant 2\}$
so
$=\int_{0}^{2 \pi} \int_{0}^{2} e^{-r^{2}} r d r d \theta$
integrate with respect to $r$ $=\left.\int_{0}^{2 \pi}\left(-\frac{1}{2} e^{-r^{2}}\right)\right|_{0} ^{2} d \theta$
$=-\frac{1}{2} \int_{0}^{2 \pi}\left(e^{-4}-1\right) d \theta$
integrate
$=\frac{-e^{-4}+1}{2}[\theta]_{0}^{2 \pi}$
$=(2 \pi) \frac{-e^{4}+1}{2}$
$=\pi \left(-e^{-4}+1\right) $
