设 $f(x, y)$ 是连续函数，则 $\int_0^1 \mathrm{~d} y \int_{-\sqrt{1-y^2}}^{1-y} f(x, y) \mathrm{d} x=$
A. $\int_0^1 \mathrm{~d} x \int_0^{x-1} f(x, y) \mathrm{d} y+\int_{-1}^0 \mathrm{~d} x \int_0^{\sqrt{1-x^2}} f(x, y) \mathrm{d} y$
B. $\int_0^1 \mathrm{~d} x \int_0^{1-x} f(x, y) \mathrm{d} y+\int_{-1}^0 \mathrm{~d} x \int_{-\sqrt{1-x^2}}^0 f(x, y) \mathrm{d} y$
C. $\int_0^{\frac{\pi}{2}} \mathrm{~d} \theta \int_0^{\frac{1}{\cos \theta+\sin \theta}} f(r \cos \theta, r \sin \theta) \mathrm{d} r$+ $ \int_{\frac{\pi}{2}}^\pi \mathrm{d} \theta \int_0^1 f(r \cos \theta, r \sin \theta) \mathrm{d} r$
D. $\int_0^{\frac{\pi}{2}} \mathrm{~d} \theta \int_0^{\frac{1}{\cos \theta+\sin \theta}} f(r \cos \theta, r \sin \theta) \mathrm{d} r $+ $\int_{\frac{\pi}{2}}^\pi \mathrm{d} \theta \int_0^1 f(r \cos \theta, r \sin \theta) r \mathrm{~d} r$