一、单选题 （共 30 题，每小题 5 分，共 50 分，每题只有一个选项正确）
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 $f(x, y)=1+\frac{x y}{\sqrt{1+y^3}}$, 则积分 $I=\int_0^1 \mathrm{~d} x \int_{x^2}^1 f(x, y) \mathrm{d} y=$
$\text{A.}$ $\frac{1}{3}(\sqrt{2}+1)$.
$\text{B.}$ $\frac{1}{6}(\sqrt{2}-1)$.
$\text{C.}$ $\frac{1}{6}(\sqrt{2}+1)$.
$\text{D.}$ $\frac{1}{3}(\sqrt{2}-1)$.
设二重积分 $I_1=\iint_D \frac{x+y-1}{4} \mathrm{~d} x \mathrm{~d} y, I_2=\iint_D\left(\frac{x+y-1}{4}\right)^2 \mathrm{~d} x \mathrm{~d} y, I_3=\iint_D\left(\frac{x+y-1}{4}\right)^3 \mathrm{~d} x \mathrm{~d} y$, 其 中 $D=\left\{(x, y) \mid(x-2)^2+(y-1)^2 \leqslant 2\right\}$, 则 $I_1, I_2, I_3$ 的大小关系为
$\text{A.}$ $I_1 < I_2 < I_3$.
$\text{B.}$ $I_3 < I_2 < I_1$.
$\text{C.}$ $I_3 < I_1 < I_2$.
$\text{D.}$ $I_2 < I_3 < I_1$.
设曲面 $\Sigma$ 是上半球面 $x^2+y^2+z^2=R^2(z \geq 0)$, 曲面 $\Sigma_1$ 是 $\Sigma$ 在第一卦限中的部分, 则有
$\text{A.}$ $\iint_{\Sigma} x \mathrm{~d} S=4 \iint_{\Sigma_1} x \mathrm{~d} S$
$\text{B.}$ $\iint_{\Sigma} y \mathrm{~d} S=4 \iint_{\Sigma_1} y \mathrm{~d} S$
$\text{C.}$ $\iint_{\Sigma} z \mathrm{~d} S=4 \iint_{\Sigma_1} z \mathrm{~d} S$
$\text{D.}$ $\iint_{\Sigma} x y z \mathrm{~d} S=4 \iint_{\Sigma_1} x y z \mathrm{~d} S$
设 $\Sigma$ 为球面 $x^2+y^2+z^2=R^2$ 的下半球面的下侧, 将曲面 积分 $\iint_{\Sigma} x^2 y^2 z \mathrm{~d} x \mathrm{~d} y$ 化为二重积分为
$\text{A.}$ $-\iint_{D_{x y}} x^2 y^2\left(-\sqrt{R^2-x^2-y^2}\right) \mathrm{d} x \mathrm{~d} y, \quad D_{x y}: x^2+y^2 \leq R^2$
$\text{B.}$ $-\iint_{D_{x y}} x^2 y^2 \sqrt{R^2-x^2-y^2} \mathrm{~d} x \mathrm{~d} y$, $D_{x y}: x^2+y^2 \leq R^2$
$\text{C.}$ $\iint_{D_{x y}} x^2 y^2\left(R^2-x^2-y^2\right) \mathrm{d} x \mathrm{~d} y$, $D_{x y}: x^2+y^2 \leq R^2$
$\text{D.}$ $-\iint_{D_{x y}} x^2 y^2\left(R^2-x^2-y^2\right) \mathrm{d} x \mathrm{~d} y$, $D_{x y}: x^2+y^2 \leq R^2$
设 $f(x, y)=x^2+2 y+y^2+x-y+1$, 则下面结论正确的是
$\text{A.}$ 点 $\left(-\frac{1}{2},-\frac{1}{2}\right)$ 是 $f(x, y)$ 的驻点且为极大值点
$\text{B.}$ 点 $\left(-\frac{1}{2},-\frac{1}{2}\right)$ 是极小值点
$\text{C.}$ 点 $(0,0)$ 是 $f(x, y)$ 的驻点但不是极值点
$\text{D.}$ 点 $(0,0)$ 是$f(x, y)$ 极大值点
设 $D=\left\{(x, y) \mid x^2+y^2 \leq R^2\right\}$, 则 $\iint_D \sqrt{x^2+y^2} \mathrm{~d} \sigma=$.
$\text{A.}$ $\pi R^3$
$\text{B.}$ $\frac{2 \pi R^3}{3}$
$\text{C.}$ $\pi R^2$
$\text{D.}$ $2 \pi R^2$
设函数 $f(x)=\iint_{u^2+v^2 \leqslant x^2} \arctan \left(1+\sqrt{u^2+v^2}\right) \mathrm{d} u \mathrm{~d} v(x>0)$, 则 $\lim _{x \rightarrow 0^{+}} \frac{f(x)}{\mathrm{e}^{-2 x}-1+2 x}=$
$\text{A.}$ $-\frac{\pi^2}{8}$.
$\text{B.}$ $-\frac{\pi^2}{4}$.
$\text{C.}$ $\frac{\pi^2}{4}$.
$\text{D.}$ $\frac{\pi^2}{8}$.
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
$\text{A.}$ $I_1 < I_2 < I_3$
$\text{B.}$ $I_2 < I_3 < I_1$
$\text{C.}$ $I_3 < I_1 < I_2$
$\text{D.}$ $I_3 < I_2 < I_1$
设平面区域 $D$ 是由 $y=x, x=1$ 及 $x$ 轴所围成，二重积分 $\iint_D \frac{1}{\sqrt{x^2+y^2}} d \sigma$ 转换成平面极坐标系下的二次积分，可表示为?
$\text{A.}$ $\int_0^{\frac{\pi}{2}} d \theta \int_0^{\frac{1}{\cos \theta}} 1 d r$
$\text{B.}$ $\int_0^{\frac{\pi}{4}} d \theta \int_0^{\frac{1}{\cos \theta}} 1 d r$
$\text{C.}$ $\int_0^{\frac{\pi}{4}} d \theta \int_0^{\frac{1}{\sin\theta}} 1 d r$
$\text{D.}$ $\int_0^{\frac{\pi}{4}} d \theta \int_0^{\frac{1}{\sin\theta}} 1 d r$
函数 $f(x, y)$ 连续，交换二重积分 $\int_0^1 d y \int_y^{\sqrt{y}} f(x, y) d x$ 次序，该二重积分可表示为?
$\text{A.}$ $\int_0^1 d x \int_{x^3}^x f(x, y) d y$
$\text{B.}$ $\int_0^1 d x \int_{x^4}^x f(x, y) d y$
$\text{C.}$ $\int_0^1 d x \int_{x^2}^x f(x, y) d y$
$\text{D.}$ $\int_0^1 d x \int_{x^5}^x f(x, y) d y$
已知平面区域 $D_1=\left\{(x, y) \left\lvert\, 0 \leqslant y \leqslant x \leqslant \frac{\pi}{2}\right.\right\}, D_2=\left\{(x, y) \left\lvert\, 0 \leqslant x \leqslant y \leqslant \frac{\pi}{2}\right.\right\}$, $D_3=\left\{(x, y) \left\lvert\, \frac{\pi}{2} \leqslant x \leqslant y \leqslant \pi\right.\right\}$, 记 $I_1=\iint_{D_1} \mathrm{e}^{-x^2} \sin y \mathrm{~d} x \mathrm{~d} y, I_2=\iint_{D_2} \mathrm{e}^{-x^2} \sin y \mathrm{~d} x \mathrm{~d} y$, $I_3=\iint_{D_3} \mathrm{e}^{-x^2} \sin y \mathrm{~d} x \mathrm{~d} y$, 则
$\text{A.}$ $I_3 < I_1 < I_2$.
$\text{B.}$ $I_3 < I_2 < I_1$.
$\text{C.}$ $I_1 < I_3 < I_2$.
$\text{D.}$ $I_1 < I_2 < I_3$.
设区域 $D=\left\{(x, y) \mid x^2+y^2 \leq 4, x \geq 0, y \geq 0\right\} ， f(x)$为 $D$ 上的正值连续函数， $a, b$ 为常数，则
$$
\iint_D \frac{a \sqrt{f(x)}+b \sqrt{f(y)}}{\sqrt{f(x)}+\sqrt{f(y)}} \mathrm{d} \sigma=
$$
$\text{A.}$ $a b \pi$
$\text{B.}$ $\frac{a b}{2} \pi$
$\text{C.}$ $(a+b) \pi$
$\text{D.}$ $\frac{a+b}{2} \pi$
设 $f(x, y)$ 为连续函数，则
$$
\int_0^{\frac{\pi}{4}} \mathrm{~d} \theta \int_0^1 f(r \cos \theta, r \sin \theta) r \mathrm{~d} r \text { 等于 }
$$
$\text{A.}$ $\int_0^{\frac{\sqrt{2}}{2}} \mathrm{~d} x \int_x^{\sqrt{1-x^2}} f(x, y) \mathrm{d} y$
$\text{B.}$ $\int_0^{\frac{\sqrt{2}}{2}} \mathrm{~d} x \int_0^{\sqrt{1-x^2}} f(x, y) \mathrm{d} y$
$\text{C.}$ $\int_0^{\frac{\sqrt{2}}{2}} \mathrm{~d} y \int_y^{\sqrt{1-y^2}} f(x, y) \mathrm{d} x$
$\text{D.}$ $\int_0^{\frac{\sqrt{2}}{2}} \mathrm{~d} y \int_0^{\sqrt{1-y^2}} f(x, y) \mathrm{d} x$
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 $\int_1^2 \mathrm{~d} x \int_x^2 f(x, y) \mathrm{d} y+\int_1^2 \mathrm{~d} y \int_y^{4-y} f(x, y) \mathrm{d} x=$
$\text{A.}$ $\int_1^2 \mathrm{~d} x \int_1^{4-x} f(x, y) \mathrm{d} y$
$\text{B.}$ $\int_1^2 \mathrm{~d} x \int_x^{4-x} f(x, y) \mathrm{d} y$
$\text{C.}$ $\int_1^2 \mathrm{~d} y \int_1^{4-y} f(x, y) \mathrm{d} x$
$\text{D.}$ $\int_1^2 \mathrm{~d} y \int_y^2 f(x, y) \mathrm{d} x$
$\lim _{x \rightarrow \infty} \sum_{i=1}^n \sum_{j=1}^n \frac{n}{(n+i)\left(n^2+j^2\right)}=$
$\text{A.}$ $\int_0^1 \mathrm{~d} x \int_0^x \frac{1}{(1+x)\left(1+y^2\right)} \mathrm{d} y$
$\text{B.}$ $\int_0^1 \mathrm{~d} x \int_0^x \frac{1}{(1+x)(1+y)} \mathrm{d} y$
$\text{C.}$ $\int_0^1 \mathrm{~d} x \int_0^1 \frac{1}{(1+x)(1+y)} \mathrm{d} y$
$\text{D.}$ $\int_0^1 d x \int_0^1 \frac{1}{(1+x)\left(1+y^2\right)} \mathrm{d} y$
$\lim _{x \rightarrow \infty} \sum_{i=1}^n \sum_{j=1}^n \frac{n}{(n+i)\left(n^2+j^2\right)}=$
$\text{A.}$ $\int_0^1 \mathrm{~d} x \int_0^x \frac{1}{(1+x)\left(1+y^2\right)} \mathrm{d} y$
$\text{B.}$ $\int_0^1 \mathrm{~d} x \int_0^x \frac{1}{(1+x)(1+y)} \mathrm{d} y$
$\text{C.}$ $\int_0^1 \mathrm{~d} x \int_0^1 \frac{1}{(1+x)(1+y)} d y$
$\text{D.}$ $\int_0^1 d x \int_0^1 \frac{1}{(1+x)\left(1+y^2\right)} \mathrm{d} y$
设函数 $f(t)$ 连续, 则二次积分 $\int_0^{\frac{\pi}{2}} \mathrm{~d} \theta \int_{2 \cos \theta}^2 f\left(r^2\right) r \mathrm{~d} r=$
$\text{A.}$ $\int_0^2 \mathrm{~d} x \int_{\sqrt{2 x-x^2}}^{\sqrt{4-x^2}} \sqrt{x^2+y^2} f\left(x^2+y^2\right) \mathrm{d} y$
$\text{B.}$ $\int_0^2 \mathrm{~d} x \int_{\sqrt{2 x-x^2}}^{\sqrt{4-x^2}} f\left(x^2+y^2\right) \mathrm{d} y$
$\text{C.}$ $\int_0^2 \mathrm{~d} y \int_{1+\sqrt{1-y^2}}^{\sqrt{4-y^2}} \sqrt{x^2+y^2} f\left(x^2+y^2\right) \mathrm{d} x$
$\text{D.}$ $\int_0^2 \mathrm{~d} y \int_{1+\sqrt{1-y^2}}^{\sqrt{4-y^2}} f\left(x^2+y^2\right) \mathrm{d} x$
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 $D_k$ 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 $f(x, y)$ 是连续函数，则 $\int_0^1 \mathrm{~d} y \int_{-\sqrt{1-y^2}}^{1-y} f(x, y) \mathrm{d} x=$
$\text{A.}$ $\int_0^1 \mathrm{~d} x \int_0^{x-1} f(x, y) \mathrm{d} y+\int_{-1}^0 \mathrm{~d} x \int_0^{\sqrt{1-x^2}} f(x, y) \mathrm{d} y$
$\text{B.}$ $\int_0^1 \mathrm{~d} x \int_0^{1-x} f(x, y) \mathrm{d} y+\int_{-1}^0 \mathrm{~d} x \int_{-\sqrt{1-x^2}}^0 f(x, y) \mathrm{d} y$
$\text{C.}$ $\int_0^{\frac{\pi}{2}} \mathrm{~d} \theta \int_0^{\frac{1}{\cos \theta+\sin \theta}} f(r \cos \theta, r \sin \theta) \mathrm{d} r$+ $ \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) \mathrm{d} r $+ $\int_{\frac{\pi}{2}}^\pi \mathrm{d} \theta \int_0^1 f(r \cos \theta, r \sin \theta) r \mathrm{~d} r$
设 $D$ 是第一象限中曲线 $2 x y=1,4 x y=1$ 与直线 $y=x$ ， $y=\sqrt{3} x$ 围成的平面区域，函数 $f(x, y)$ 在 $D$ 上连续，则
$$
\iint_D f(x, y) \mathrm{d} x \mathrm{~d} y=(\quad)
$$
$\text{A.}$ $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \mathrm{~d} \theta \int_{\frac{1}{2 \sin 2 \theta}}^{\frac{1}{\sin 2 \theta}} f(r \cos \theta, r \sin \theta) r \mathrm{~d} r$
$\text{B.}$ $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \mathrm{~d} \theta \int_{\frac{1}{\sqrt{2 \sin 2 \theta}}}^{\frac{1}{\sqrt{\sin 2 \theta}}} f(r \cos \theta, r \sin \theta) r \mathrm{~d} r$
$\text{C.}$ $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \mathrm{~d} \theta \int_{\frac{1}{2 \sin 2 \theta}}^{\frac{1}{\sin 2 \theta}} f(r \cos \theta, r \sin \theta) \mathrm{d} r$
$\text{D.}$ $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \mathrm{~d} \theta \int_{\frac{1}{\sqrt{2 \sin 2 \theta}}}^{\frac{1}{\sqrt{\sin 2 \theta}}} f(r \cos \theta, r \sin \theta) \mathrm{d} r$
设 $D$ 是第一象限中曲线 $2 x y=1,4 x y=1$ 与直线 $y=x$ ， $y=\sqrt{3} x$ 围成的平面区域，函数 $f(x, y)$ 在 $D$ 上连续，则 $\iint_D f(x, y) \mathrm{d} x \mathrm{~d} y=$
$\text{A.}$ $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \mathrm{~d} \theta \int_{\frac{1}{2 \sin 2 \theta}}^{\frac{1}{\sin 2 \theta}} f(r \cos \theta, r \sin \theta) r \mathrm{~d} r$
$\text{B.}$ $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \mathrm{~d} \theta \int_{\frac{1}{\sqrt{2 \sin 2 \theta}}}^{\frac{1}{\sqrt{\sin 2 \theta}}} f(r \cos \theta, r \sin \theta) r \mathrm{~d} r$
$\text{C.}$ $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \mathrm{~d} \theta \int_{\frac{1}{2 \sin 2 \theta}}^{\frac{1}{\sin 2 \theta}} f(r \cos \theta, r \sin \theta) \mathrm{d} r$
$\text{D.}$ $\int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \mathrm{~d} \theta \int_{\frac{1}{\sqrt{2 \sin 2 \theta}}}^{\frac{1}{\sqrt{\sin 2 \theta}}} f(r \cos \theta, r \sin \theta) \mathrm{d} r$
设 $D=\left\{(x, y) \mid x^2+y^2 \leq 2 x, x^2+y^2 \leq 2 y\right\}$ ，函数 $f(x, y)$ 在 $D$ 上连续，则 $\iint_D f(x, y) \mathrm{d} x \mathrm{~d} y=(\quad)$
$\text{A.}$ $\int_0^{\frac{\pi}{4}} \mathrm{~d} \theta \int_0^{2 \cos \theta} f( r \cos \theta, r \sin \theta) r \mathrm{~d} r +\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_0^{2 \sin \theta} f(r \cos \theta, r \sin \theta) r \mathrm{~d} r $
$\text{B.}$ $ \int_0^{\frac{\pi}{4}} \mathrm{~d} \theta \int_0^{2 \sin \theta} f(r \cos \theta, r \sin \theta) r \mathrm{~d} r \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_0^{2 \cos \theta} f(r \cos \theta, r \sin \theta) r \mathrm{~d} r $
$\text{C.}$ $\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \mathrm{~d} \theta \int_0^{2 \cos \theta} f(r \cos \theta, r \sin \theta) r \mathrm{~d} r$
$\text{D.}$ $2 \int_0^1 \mathrm{~d} x \int_{1-\sqrt{1-x^2}}^x f(x, y) \mathrm{d} y$
设 $f(x, y)$ 是连续函数，则 $\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \mathrm{~d} x \int_{\sin x}^1 f(x, y) \mathrm{d} y=(\quad)$
$\text{A.}$ $\int_{\frac{1}{2}}^1 \mathrm{~d} y \int_{\frac{\pi}{6}}^{\arcsin y} f(x, y) \mathrm{d} x$
$\text{B.}$ $\int_{\frac{1}{2}}^1 \mathrm{~d} y \int_{\arcsin y}^{\frac{\pi}{2}} f(x, y) \mathrm{d} x$
$\text{C.}$ $\int_0^{\frac{1}{2}} \mathrm{~d} y \int_{\frac{\pi}{6}}^{\arcsin y} f(x, y) \mathrm{d} x$
$\text{D.}$ $\int_0^{\frac{1}{2}} \mathrm{~d} y \int_{\arcsin y}^{\frac{\pi}{2}} f(x, y) \mathrm{d} x$