解：在极坐标系中，闭区域D可表示为
$0\le\rho\le a\qquad 0\le\theta\le 2\pi \\ \iint_{D}e^{-x^2-y^2}dxdy = \iint_{D}e^{-\rho^2}\rho d\rho d\theta =\int_0^{2\pi}[\int_0^ae^{-\rho^2}\rho d\rho]d\theta \\ = \int_0^{2\pi}[\int_0^ae^{-\rho^2}]_0^ad\theta= \frac{1}{2}(1-e^{-a^2})\int_0^{2\pi}d\theta=\pi(1-e^{-a^2})$0≤ρ≤a0≤θ≤2π∬D​e−x2−y2dxdy=∬D​e−ρ2ρdρdθ=∫02π​[∫0a​e−ρ2ρdρ]dθ=∫02π​[∫0a​e−ρ2]0a​dθ=21​(1−e−a2)∫02π​dθ=π(1−e−a2)