单选题 （共 4 题 ），每题只有一个选项正确
设 $f$ 是连续函数, 积分区域 $D: x^2+y^2 \leq 1$ 且 $y \geq 0$, 则 $\iint_D f\left(\sqrt{x^2+y^2}\right) \mathrm{d} x \mathrm{~d} y$ 可化为
$\text{A.}$ $\pi \int_0^1 r f(r) \mathrm{d} r$
$\text{B.}$ $2 \pi \int_0^1 r f(r) \mathrm{d} r$
$\text{C.}$ $2 \pi \int_0^1 f(r) \mathrm{d} r$
$\text{D.}$ $\pi \int_0^1 f(r) d r$
累次积分 $\int_0^{\frac{\pi}{4}} d \theta \int_0^{2 \cos \theta} f(\rho \cos \theta, \rho \sin \theta) \rho d \rho$ 等于
$\text{A.}$ $\int_0^1 d y \int_y^{1-\sqrt{1-y^2}} f(x, y) d x$
$\text{B.}$ $\int_0^2 d x \int_0^{\sqrt{2 x-x^2}} f(x, y) d y$
$\text{C.}$ $\int_0^2 d \rho \int_0^{\frac{\pi}{4}} f(\rho \cos \theta, \rho \sin \theta) d \theta$
$\text{D.}$ $\int_0^{\sqrt{2}} d \rho \int_0^{\frac{\pi}{4}} f(\rho \cos \theta, \rho \sin \theta) \rho d \theta+\int_{\sqrt{2}}^2 d \rho \int_0^{\arccos \frac{\rho}{2}} f(\rho \cos \theta, \rho \sin \theta) \rho d \theta$
设 $D$ 是以 $A(1,1), B(-1,1), C(-1,-1)$ 为三顶点的三角形, 则 $I=$ $\iint_D\left[\sin (x y) \sqrt{x^2+3 y^2+1}+3 x+3 y\right] \mathrm{d} x \mathrm{~d} y=$
$\text{A.}$ 4
$\text{B.}$ 3
$\text{C.}$ 2
$\text{D.}$ 0
设 $I_1=\iint_D \sin \left|\frac{x-y}{2}\right| \mathrm{d} x \mathrm{~d} y, I_2=\iint_D \sin \left(\frac{x-y}{2}\right)^2 \mathrm{~d} x \mathrm{~d} y, I_3=\iint_D \sin \left(\frac{x-y}{2}\right)^3 \mathrm{~d} x \mathrm{~d} y$, 其中 $D=$ $\left\{(x, y) \mid(x-1)^2+(y-1)^2 \leqslant 2\right\}$, 则
$\text{A.}$ $I_1 < I_2 < I_3$
$\text{B.}$ $I_2 < I_3 < I_1$
$\text{C.}$ $I_3 < I_1 < I_2$
$\text{D.}$ $I_3 < I_2 < I_1$
填空题 （共 2 题 ），请把答案直接填写在答题纸上
交换积分次序后 $\int_0^1 \mathrm{~d} y \int_{-\sqrt{y}}^{\sqrt{y}} f(x, y) \mathrm{d} x=$
$\int_0^\pi d \theta \int_0^{\frac{1}{\cos \theta}} \rho^2 d \rho+\int_1^{\sqrt{2}} d x \int_0^{\sqrt{2-x^2}} \sqrt{x^2+y^2} d y=$