设 $f(x, y)$ 为连续函数, 则 $\int_{0}^{\frac{\pi}{4}} \mathrm{d}\theta \int_{0}^{1} f(r \cos \theta, r \sin \theta) r \mathrm{d}r$ 等于( ) $\int_{0}^{\frac{\sqrt{2}}{2}} \mathrm{d}x \int_{x}^{\sqrt{1 - x^2}} f(x, y) \, \mathrm{d}y.$ (B) $\int_{0}^{\frac{\sqrt{2}}{2}}\mathrm{d}x\int_{0}^{\sqrt{1 - x^2}}f(x,y)\mathrm{d}y.$ (C) $\int_0^{\frac{\sqrt{2}}{2}}\mathrm{d}y\int_{y}^{\sqrt{1 - y^2}}f(x,y)\mathrm{d}x.$ $\int_0^{\frac{\sqrt{2}}{2}}\mathrm{d}y\int_0^{\sqrt{1 - y^2}}f(x,y)\mathrm{d}x.$