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一、对二重积分概念及性质的考查

小题

  1. **【1999-34-3 分】**设f(x,y)\displaystyle f(x,y)连续,且 f(x,y)=xy+Df(u,v)dudv\displaystyle f(x, y)=x y+\iint_{D} f(u, v) d u d v,其中 D\displaystyle D 是由y=0, y=x2, x=1\displaystyle y=0,\ y=x^{2},\ x=1 所围区域,则f(x,y)\displaystyle f(x,y)等于( ) A. xy\displaystyle xy    B. 2xy\displaystyle 2xy    C. xy+18\displaystyle xy+\dfrac18    D. xy+1\displaystyle xy+1

  2. **【2005-34-4 分】**设 I1=Dcosx2+y2dσ,I2=Dcos(x2+y2)dσ,I3=Dcos(x2+y2)2dσI_{1}=\iint_{D} \cos \sqrt{x^{2}+y^{2}} d \sigma,\quad I_{2}=\iint_{D} \cos (x^{2}+y^{2}) d \sigma,\quad I_{3}=\iint_{D} \cos (x^{2}+y^{2})^{2} d \sigma 其中 D={(x,y)x2+y21}\displaystyle D=\{(x, y) | x^{2}+y^{2} \leq 1\},则( ) A. I3>I2>I1\displaystyle I_3>I_2>I_1    B. I1>I2>I3\displaystyle I_1>I_2>I_3 C. I2>I1>I3\displaystyle I_2>I_1>I_3    D. I3>I1>I2\displaystyle I_3>I_1>I_2

  3. **【2009-1-4 分】**正方形{(x,y)x1,y1}\displaystyle \{(x,y)||x|\leq 1,|y|\leq 1\}被对角线划分为四个区域Dk(k=1,2,3,4)\displaystyle D_k(k=1,2,3,4)Ik=Dkycosxdxdy\displaystyle I_{k}=\iint_{D_{k}} y \cos x d x d y,则max1k4{Ik}=\displaystyle \max\limits_{1 \le k \le4}\{I_k\}=( ) A. I1\displaystyle I_1    B. I2\displaystyle I_2    C. I3\displaystyle I_3    D. I4\displaystyle I_4

  4. **【2013-23-4 分】**设Dk\displaystyle D_k是圆域D={(x,y)x2+y21}\displaystyle D=\{(x,y)|x^2+y^2\le1\}在第k\displaystyle k象限的部分,Ik=Dk(yx)dxdy(k=1,2,3,4)\displaystyle I_{k}=\iint_{D_{k}}(y-x) d x d y(k=1,2,3,4),则( ) A. I1>0\displaystyle I_1>0    B. I2>0\displaystyle I_2>0    C. I3>0\displaystyle I_3>0    D. I4>0\displaystyle I_4>0

  5. **【2016-3-4 分】**记 D1={(x,y)0x1,0y1},D_{1}=\{(x, y) | 0 \leq x \leq 1,0 \leq y \leq 1\}, D2={(x,y)0x1,0yx},D_{2}=\{(x, y) | 0 \leq x \leq 1,0 \leq y \leq \sqrt{x}\}, D3={(x,y)0x1,x2y1},D_{3}=\{(x, y) | 0 \leq x \leq 1, x^{2} \leq y \leq 1\}, Ji=Dixy3dxdy(i=1,2,3)\displaystyle J_{i}=\iint_{D_{i}} \sqrt[3]{x-y} d x d y(i=1,2,3),则( ) A. J2<J3<J1\displaystyle J_2<J_3<J_1    B. J3<J1<J2\displaystyle J_3<J_1<J_2 C. J2<J1<J3\displaystyle J_2<J_1<J_3    D. J1<J2<J3\displaystyle J_1<J_2<J_3

  6. **【2019-2-4 分】**已知平面区域D={(x,y)x+yπ2}\displaystyle D=\{(x, y)||x|+|y| \leq \dfrac{\pi}{2}\},记 I1=D(1x2+y2)dxdy,I2=D(1cosx2+y2)dxdy,I3=D(1cosx2+y2)dxdyI_{1}=\iint_{D}(1-\sqrt{x^{2}+y^{2}}) d x d y,\quad I_{2}=\iint_{D}(1-\cos\sqrt{x^{2}+y^{2}}) d x d y,\quad I_{3}=\iint_{D}(1-\cos \sqrt{x^{2}+y^{2}}) d x d y 则( ) A. I3<I2<I1\displaystyle I_3<I_2<I_1    B. I2<I1<I3\displaystyle I_2<I_1<I_3 C. I1<I2<I3\displaystyle I_1<I_2<I_3    D. I2<I3<I1\displaystyle I_2<I_3<I_1