设函数 $f(t)$ 连续, 则二次积分 $\int_0^{\frac{\pi}{2}} \mathrm{~d} \theta \int_{2 \cos \theta}^2 f\left(r^2\right) r \mathrm{~d} r=$
A. $\int_0^2 \mathrm{~d} x \int_{\sqrt{2 x-x^2}}^{\sqrt{4-x^2}} \sqrt{x^2+y^2} f\left(x^2+y^2\right) \mathrm{d} y$
B. $\int_0^2 \mathrm{~d} x \int_{\sqrt{2 x-x^2}}^{\sqrt{4-x^2}} f\left(x^2+y^2\right) \mathrm{d} y$
C. $\int_0^2 \mathrm{~d} y \int_{1+\sqrt{1-y^2}}^{\sqrt{4-y^2}} \sqrt{x^2+y^2} f\left(x^2+y^2\right) \mathrm{d} x$
D. $\int_0^2 \mathrm{~d} y \int_{1+\sqrt{1-y^2}}^{\sqrt{4-y^2}} f\left(x^2+y^2\right) \mathrm{d} x$