单选题 （共 10 题 ），每题只有一个选项正确
设函数 $f(x, y)$ 连续，则二次积分 $\int_{\frac{\pi}{2}}^\pi \mathrm{d} x \int_{\sin x}^1 f(x, y) \mathrm{d} y$ 等于
$\text{A.}$ $\int_0^1 \mathrm{~d} y \int_{\pi+\arcsin y}^\pi f(x, y) \mathrm{d} x$ ．
$\text{B.}$ $\int_0^1 \mathrm{~d} y \int_{\pi-\arcsin y}^\pi f(x, y) \mathrm{d} x$ ．
$\text{C.}$ $\int_0^1 \mathrm{~d} y \int_{\frac{\pi}{2}}^{\pi+\arcsin y} f(x, y) \mathrm{d} x$ ．
$\text{D.}$ $\int_0^1 \mathrm{~d} y \int_{\frac{\pi}{2}}^{\pi-\arcsin y} f(x, y) \mathrm{d} x$ ．
设函数 $f(x, y)$ 连续，则 $\int_1^2 \mathrm{~d} x \int_x^2 f(x, y) \mathrm{d} y+\int_1^2 \mathrm{~d} y \int_y^{4-y} f(x, y) \mathrm{d} x=$
$\text{A.}$ $\int_1^2 \mathrm{~d} x \int_1^{4-x} f(x, y) \mathrm{d} y$ ．
$\text{B.}$ $\int_1^2 \mathrm{~d} x \int_x^{4-x} f(x, y) \mathrm{d} y$ ．
$\text{C.}$ $\int_1^2 \mathrm{~d} y \int_1^{4-y} f(x, y) \mathrm{d} x$ ．
$\text{D.}$ $\int_1^2 \mathrm{~d} y \int_y^2 f(x, y) \mathrm{d} x$ ．
累次积分 $\int_0^{\frac{\pi}{2}} \mathrm{~d} \theta \int_0^{\cos \theta} f(r \cos \theta, r \sin \theta) r \mathrm{~d} r$ 可以写成
$\text{A.}$ $\int_0^1 \mathrm{~d} y \int_0^{\sqrt{y-y^2}} f(x, y) \mathrm{d} x$ ．
$\text{B.}$ $\int_0^1 \mathrm{~d} y \int_0^{\sqrt{1-y^{\prime}}} f(x, y) \mathrm{d} x$ ．
$\text{C.}$ $\int_0^1 \mathrm{~d} x \int_0^1 f(x, y) \mathrm{d} y$ ．
$\text{D.}$ $\int_0^1 \mathrm{~d} x \int_0^{\sqrt{x-x}} f(x, y) \mathrm{d} y$ ．
设 $f(x, y)$ 是连续函数，则 $\int_0^1 \mathrm{~d} y \int_{-\sqrt{1-y^2}}^{1-y} f(x, y) \mathrm{d} x=$
$\text{A.}$ $\int_0^1 \mathrm{~d} x \int_0^{x-1} f(x, y) \mathrm{d} y+\int_{-1}^0 \mathrm{~d} x \int_0^{\sqrt{1-x^2}} f(x, y) \mathrm{d} y$ ．
$\text{B.}$ $\int_0^1 \mathrm{~d} x \int_0^{1-x} f(x, y) \mathrm{d} y+\int_{-1}^0 \mathrm{~d} x \int_{-\sqrt{1-x}}^0 f(x, y) \mathrm{d} y$ ．
$\text{C.}$ $\int_0^{\frac{\pi}{2}} \mathrm{~d} \theta \int_0^{\frac{1}{\cos \theta+\sin \theta}} f(r \cos \theta, r \sin \theta) \mathrm{d} r+\int_{\frac{\pi}{2}}^\pi \mathrm{d} \theta \int_0^1 f(r \cos \theta, r \sin \theta) \mathrm{d} r$ ．
$\text{D.}$ $\int_0^{\frac{\pi}{2}} \mathrm{~d} \theta \int_0^{\frac{1}{\cos \theta+\sin \theta}} f(r \cos \theta, r \sin \theta) r \mathrm{~d} r+\int_{\frac{\pi}{2}}^{\mathrm{x}} \mathrm{d} \theta \int_0^1 f(r \cos \theta, r \sin \theta) r \mathrm{~d} r$.
设函数 $f(u)$ 连续，区域 $D=\left\{(x, y) \mid x^2+y^2 \leqslant 2 y\right\}$ ，则 $\iint_D f(x y) \mathrm{d} x \mathrm{~d} y$ 等于 ()
$\text{A.}$ $\int_{-1}^1 \mathrm{~d} x \int_{-\sqrt{1-x^x}}^{\sqrt{1-x^x}} f(x y) \mathrm{d} y$ ．
$\text{B.}$ $2 \int_0^2 \mathrm{~d} y \int_0^{\sqrt{2 y-y^2}} f(x y) \mathrm{d} x$ ．
$\text{C.}$ $\int_0^\pi \mathrm{d} \theta \int_0^{2 \sin \theta} f\left(r^2 \sin \theta \cos \theta\right) \mathrm{d} r$ ．
$\text{D.}$ $\int_0^\pi \mathrm{d} \theta \int_0^{2 \sin \theta} f\left(r^2 \sin \theta \cos \theta\right) \mathrm{rd} r$ ．
设 $D$ 是 $x O y$ 平面上以 $(1,1),(-1,1)$ 和 $(-1,-1)$ 为顶点的三角形区域，$D_1$是 $D$ 在第一象限的部分，则 $\iint_D(x y+\cos x \sin y) \mathrm{d} x \mathrm{~d} y$ 等于
$\text{A.}$ $2 \iint_{D_1} \cos x \sin y \mathrm{~d} x \mathrm{~d} y$ ．
$\text{B.}$ $2 \iint_{D_1} x y \mathrm{~d} x \mathrm{~d} y$ ．
$\text{C.}$ $4 \iint_{D_1}(x y+\cos x \sin y) \mathrm{d} x \mathrm{~d} y$ ．
$\text{D.}$ 0 ．
设 $f(x, y)$ 连续，且 $f(x, y)=x y+\iint_D f(u, v) \mathrm{d} u \mathrm{~d} v$ ，其中 $D$ 是由 $y=0, y=x^2, x=1$所围成的区域，则 $f(x, y)$ 等于
$\text{A.}$ $x y$ ．
$\text{B.}$ $2 x y$ ．
$\text{C.}$ $x y+\frac{1}{8}$ ．
$\text{D.}$ $x y+1$ ．
设有空间区域 $\Omega_1: x^2+y^2+z^2 \leqslant R^2, z \geqslant 0$ 及 $\Omega_2: x^2+y^2+z^2 \leqslant R^2, x \geqslant 0, y \geqslant 0, z \geqslant 0$ ，则
$\text{A.}$ $\iiint_{\Omega_1} x \mathrm{~d} v=4 \iiint_{\Omega_2} x \mathrm{~d} v$ ．
$\text{B.}$ $\iiint_{\Omega_1} y \mathrm{~d} v=4 \iiint_{\Omega_2} y \mathrm{~d} v$ ．
$\text{C.}$ $\iiint_{\Omega_1} z \mathrm{~d} v=4 \iiint_{\Omega_2} z \mathrm{~d} v$ ．
$\text{D.}$ $\iiint_{\Omega_1} x y z \mathrm{~d} v=4 \iiint_{\Omega_2} x y z \mathrm{~d} v$ ．
设函数 $Q(x, y)=\frac{x}{y^2}$ ．如果对上半平面 $(y>0)$ 内的任意有向光滑封闭曲线 $C$ 都有 $\oint_c P(x, y) \mathrm{d} x+Q(x, y) \mathrm{d} y=0$ ，那么函数 $P(x, y)$ 可取为
$\text{A.}$ $y-\frac{x^2}{y^3}$ ．
$\text{B.}$ $\frac{1}{y}-\frac{x^2}{y^3}$ ．
$\text{C.}$ $\frac{1}{x}-\frac{1}{y}$ ．
$\text{D.}$ $x-\frac{1}{y}$ ．
设 $S: x^2+y^2+z^2=a^2(z \geqslant 0), S_1$ 为 $S$ 在第一卦限中的部分，则有
$\text{A.}$ $\iint_S x \mathrm{~d} S=4 \iint_{S_1} x \mathrm{~d} S$ ．
$\text{B.}$ $\iint_S y \mathrm{~d} S=4 \iint_{S_1} x \mathrm{~d} S$ ．
$\text{C.}$ $\iint_S z \mathrm{~d} S=4 \iint_{S_1} x \mathrm{~d} S$ ．
$\text{D.}$ $\iint_S x y z \mathrm{~d} S=4 \iint_{S_1} x y z \mathrm{~d} S$ ．