单选题 （共 5 题 ），每题只有一个选项正确
设 $D_k$ 是圆域 $D=\left\{(x, y) \mid x^2+y^2 \leq 1\right\}$ 的第 $k$ 象限的部分，记 $I_k=\iint_{D_k}(y-x) \mathrm{d} x \mathrm{~d} y(k=1,2,3,4)$ ，则
$\text{A.}$ $I_1>0$
$\text{B.}$ $I_2>0$
$\text{C.}$ $I_3>0$
$\text{D.}$ $I_4>0$
设函数 $f(x, y)$ 连续, 则 $\int_{-2}^2 d x \int_{4-x^2}^4 f(x, y) d y= $
$\text{A.}$ $\int_0^4\left[\int_{-2}^{-\sqrt{4-y}} f(x, y) d x+\int_{\sqrt{4-y}}^2 f(x, y) d x\right] d y$
$\text{B.}$ $\int_0^4\left[\int_{-2}^{\sqrt{4-y}} f(x, y) d x+\int_{\sqrt{4-y}}^2 f(x, y) d x\right] d y$
$\text{C.}$ $\int_0^4\left[\int_{-2}^{-\sqrt{4-y}} f(x, y) d x+\int_2^{\sqrt{4-y}} f(x, y) d x\right] d y$
$\text{D.}$ $2 \int_0^4 d y \int_{\sqrt{4-y}}^2 f(x, y) d x$
设函数 $f(x)=\iint_{u^2+v^2 \leqslant x^2} \arctan \left(1+\sqrt{u^2+v^2}\right) d u d v(x>0)$, 则 $\lim _{x \rightarrow 0^{+}} \frac{f(x)}{ e ^{-2 x}-1+2 x}=$
$\text{A.}$ $-\frac{\pi^2}{8}$.
$\text{B.}$ $-\frac{\pi^2}{4}$.
$\text{C.}$ $\frac{\pi^2}{4}$.
$\text{D.}$ $\frac{\pi^2}{8}$.
累次积分 $\int_0^{\frac{\pi}{2}} d \theta \int_0^{\cos \theta} f(\rho \cos \theta, \rho \sin \theta) \rho d \rho$ 可写成
$\text{A.}$ $\int_0^1 d y \int_0^{\sqrt{y-y^2}} f(x, y) d x$
$\text{B.}$ $\int_0^1 d y \int_0^{\sqrt{y}} f(x, y) d x$
$\text{C.}$ $\int_0^1 d x \int_0^{\sqrt{x-x^2}} f(x, y) d y$
$\text{D.}$ $\int_0^1 d x \int_0^{\sqrt{x}} f(x, y) d y$
$\int_0^1 \int_{y-1}^{1-y} f(x, y) d x d y=$
$\text{A.}$ $\int_{-1}^0 \int_0^{x+1} f(x, y) d x d y+\int_0^1 \int_0^{1-x} f(x, y) d x d y$
$\text{B.}$ $\int_0^1 d x \int_0^x f(x, y) d y$
$\text{C.}$ $\int_{-1}^1 \int_0^1 f(x, y) d x d y$
$\text{D.}$ $\int_0^1 \int_0^{2-x} f(x, y) d x d y$ ．
填空题 （共 1 题 ），请把答案直接填写在答题纸上
设 $a>0 ， f(x)=g(x)=\left\{\begin{array}{cc}a, & 0 \leq x \leq 1, \\ 0, & \text { 其他， }\end{array} \quad D\right.$ 表示全平面，则 $I=\iint_D f(x) g(y-x) \mathrm{d} x \mathrm{~d} y=$