设函数 $f(x, y)$ 连续, 则累次积分 $\int_0^1 \mathrm{~d} x \int_{x-1}^{\sqrt{x-x^2}} f(x, y) \mathrm{d} y$ 等于
A. $\int_{-1}^1 {~d} y \int_0^{y+1} f(x, y) {d} x+\int_0^{\frac{1}{2}} {~d} y \int_0^{\frac{1}{2}-\sqrt{\frac{1}{4}-y^2}} {~d} x$
B. $\int_{-1}^1 {~d} y \int_0^{y+1} f(x, y) {d} x+\int_0^{\frac{1}{2}} {~d} y \int_0^{\frac{1}{2}+\sqrt{\frac{1}{4}-y^2}} {~d} x$
C. $\int_{-\frac{\pi}{2}}^0 {~d} \theta \int_0^{\frac{1}{\cos \theta-\sin \theta}} f(r \cos \theta, r \sin \theta) r {~d} r+\int_0^{\frac{\pi}{2}} {~d} \theta \int_0^{\cos \theta} f(r \cos \theta, r \sin \theta) r {~d} r$
D. $\int_{-\frac{\pi}{2}}^0 \mathrm{~d} \theta \int_0^{\frac{1}{\cos \theta+\sin \theta}} f(r \cos \theta, r \sin \theta) r \mathrm{~d} r+\int_0^{\frac{\pi}{2}} {~d} \theta \int_0^{\sin \theta} f(r \cos \theta, r \sin \theta) r {~d} r$