设函数$f(x,y)$ 连续，则$\int_{-2}^{2}dx\int_{4-x^{2}}^{4}f(x,y)dy=$

题目

设函数

f(x,y)

连续，则

\int_{-2}^{2}dx\int_{4-x^{2}}^{4}f(x,y)dy=

选项

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

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

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

2\int_{0}^{4}dy[\int_{\sqrt{4-y}}^{2}f(x,y)dx

正确答案