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