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 $(\quad)$
$\text{A.}$ $\int_{0}^{\frac{\sqrt{2}}{2}} \mathrm{~d} x \int_{x}^{\sqrt{1-x^{2}}} f(x, y) \mathrm{d} y$.
$\text{B.}$ $\int_{0}^{\frac{\sqrt{2}}{2}} \mathrm{~d} x \int_{0}^{\sqrt{1-x^{2}}} f(x, y) \mathrm{d} y$.
$\text{C.}$ $\int_{0}^{\frac{\sqrt{2}}{2}} \mathrm{~d} y \int_{y}^{\sqrt{1-y^{2}}} f(x, y) \mathrm{d} x$.
$\text{D.}$ $\int_{0}^{\frac{\sqrt{2}}{2}} \mathrm{~d} y \int_{0}^{\sqrt{1-y^{2}}} f(x, y) \mathrm{d} x$.