【答案】A

【解析】由题易知，此二重积分积分区域为 

$D = \left\{(x,y)\vert 4 - x^2 \leq y \leq 4, -2 \leq x \leq 2\right\}$，对应图像为上图所示。

记$D_1 = \left\{(x,y)\vert 4 - x^2 \leq y \leq 4, -2 \leq x \leq 0\right\}$，$D_2 = \left\{(x,y)\vert 4 - x^2 \leq y \leq 4, 0 \leq x \leq 2\right\}$，且 $I = \int_{-2}^{2}dx\int_{4 - x^2}^{4}f(x,y)dy$，则$I = \iint\limits_{D_1} f(x,y)d\sigma + \iint\limits_{D_2} f(x,y)d\sigma$，交换积分次序得 

$I = \int_{0}^{4}dy\int_{-2}^{-\sqrt{4 - y}} f(x,y)dx + \int_{0}^{4}dy\int_{\sqrt{4 - y}}^{2} f(x,y)dx$ 

$= \int_{0}^{4}\left[\int_{-2}^{-\sqrt{4 - y}} f(x,y)dx + \int_{\sqrt{4 - y}}^{2} f(x,y)dx\right]dy$ 

故 A 正确。