设 $D=\left\{(x, y) \mid x^2+y^2 \leq 2 x, x^2+y^2 \leq 2 y\right\}$ ，函数 $f(x, y)$ 在 $D$ 上连续，则 $\iint_D f(x, y) \mathrm{d} x \mathrm{~d} y=(\quad)$
A. $\int_0^{\frac{\pi}{4}} \mathrm{~d} \theta \int_0^{2 \cos \theta} f( r \cos \theta, r \sin \theta) r \mathrm{~d} r +\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_0^{2 \sin \theta} f(r \cos \theta, r \sin \theta) r \mathrm{~d} r $
B. $ \int_0^{\frac{\pi}{4}} \mathrm{~d} \theta \int_0^{2 \sin \theta} f(r \cos \theta, r \sin \theta) r \mathrm{~d} r \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_0^{2 \cos \theta} f(r \cos \theta, r \sin \theta) r \mathrm{~d} r $
C. $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_0^{2 \cos \theta} f(r \cos \theta, r \sin \theta) r \mathrm{~d} r$
D. $2 \int_0^1 \mathrm{~d} x \int_{1-\sqrt{1-x^2}}^x f(x, y) \mathrm{d} y$