二重积分练习题

二重积分练习题（一）选择题1.设D是由直线x=0，y=0，x+y=3，x+y=5所围成的闭区域，记：I1=∬ln(x+y)dσ，I2=∬ln2(x+y)dσ，则（）A.I1<I2B.I1>I2C.I2=2I1D.无法比较2.设D是由x轴和y=sinx(x∈[0，π])所围成，则积分A.∬ydσ=π/2B.∬ydσ=π/4C.∬ydσ=πD.∬ydσ=2π3.设积分区域D由y=x和y=x+2围成，则∬f(x,y)dσ=（）A.∫1-2dx∫x-2x-1f(x,y)dyB.∫2-1dx∫x-2x-1f(x,y)dyC.∫1-2dx∫x-2x+2f(x,y)dyD.∫2-1dx∫x-2x+2f(x,y)dy4.设f(x,y)是连续函数，则累次积分A.∫4dx∫2x-4yf(x,y)dy=4∫0dx∫0y/2f(x,y)dyB.∫4dx∫2x-4yf(x,y)dy=∫0^4dy∫y/2^2yf(x,y)dxC.∫4dx∫-4y-2yf(x,y)dx=∫0^4dy∫-y/2^0yf(x,y)dxD.∫4dx∫-4y-2yf(x,y)dx=∫0^4dy∫0y/2f(x,y)dx5.累次积分A.∫2dx∫e-ydy=1-e-2B.∫2dx∫e-ydy=1-e-4C.∫2dx∫e-ydy=1-e-4D.∫2dx∫e-ydy=1-e-26.设D由|x+y|≤1确定，若I1=∬(x+y)2dσ，I2=∬ln(x2+y2)dσ，则I1，I2之间的大小顺序为（）A.I1<I2B.I1<I2C.I2<I1D.I2<I17.设D由|x|≤1，|y|≤1确定，则∬cosxyesinxydxdy=（）A.0B.eC.2D.e-28.若积分区域D由x+y≤1，x≥0，y≥0确定，且∫1xf(x)dx=∫0^1xf(x)dx，则∬f(x,y)dxdy=（）A.2B.1C.1/2D.1/49.若∫-1^1dx∫1+x1/(1-x)f(x,y)dy+∫1^2dx∫1-x^21/(1-x)f(x,y)dy=∬f(x,y)dxdy，则（）A.x1(y)=y-1，x2(y)=y-1B.x1(y)=y-1，x2(y)=y+1C.x1(y)=y+1，x2(y)=y-1D.x1(y)=y+1，x2(y)=y+3(一)改写后的文章：1.对于函数$f(x,y)=1-yC/x$，求其在$D$区域内的二重积分，其中$D$是由直线$y=x$，$y=1/x$，$y=2$围成的区域。2.已知$D$是由$a\leqx\leqb$，$0\leqy\leq1$所围成的区域，且$\iint_Dyf(x)dxdy=1$，求$\int_a^bf(x)dx$。3.若$D$是由直线$x+y=1$和两坐标轴围成的区域，且$\iint_Df(x)dxdy=\int_0^1\phi(x)dx$，求$\phi(x)$。4.交换积分次序：$\int_{-1}^2dy\int_{\sqrt{1-y^2}}^{\sqrt{1-y^2}}f(x,y)dx$。5.设$D$是由圆心在原点，半径为$1$，在第一象限内的圆所围成的区域，求$\iint_Ddxdy$。6.交换积分次序：$\int_0^1dx\int_{x^2}^{\sqrt{x}}f(x,y)dy$。7.交换积分次序：$\int_0^1dy\int_{y}^{\sqrt{y}}f(x,y)dx$。8.交换积分次序：$\int_0^{\pi}dx\int_0^{\sinx}f(x,y)dy$。(二)填空题：1.设$D$是由直线$y=x$，$y=1/x$，$y=2$所围成的区域，则$\iint_Ddxdy=\frac{3}{2}$。2.已知$D$是由$a\leqx\leqb$，$0\leqy\leq1$所围成的区域，且$\iint_Dyf(x)dxdy=1$，则$\int_a^bf(x)dx=\frac{1}{2}$。3.若$D$是由直线$x+y=1$和两坐标轴围成的区域，且$\iint_Df(x)dxdy=\int_0^1\phi(x)dx$，则$\phi(x)=1-x$。4.交换积分次序：$\int_{-1}^2dy\int_{\sqrt{1-y^2}}^{\sqrt{1-y^2}}f(x,y)dx=\int_{-\sqrt{2}}^{\sqrt{2}}dx\int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}}f(x,y)dy$。5.设$D$是由圆心在原点，半径为$1$，在第一象限内的圆所围成的区域，则$\iint_Ddxdy=\frac{\pi}{4}$。6.交换积分次序：$\int_0^1dx\int_{x^2}^{\sqrt{x}}f(x,y)dy=\int_0^1dy\int_{y^{\frac{1}{2}}}^{y^2}f(x,y)dx$。7.交换积分次序：$\int_0^1dy\int_{y}^{\sqrt{y}}f(x,y)dx=\int_0^1dx\int_{x^2}^{x}f(x,y)dy$。8.交换积分次序：$\int_0^{\pi}dx\int_0^{\sinx}f(x,y)dy=\int_0^1dy\int_{0}^{\arcsiny}f(x,y)dx$。(三)计算题：1.(1)选择$y$为外层积分变量，$x$为内层积分变量，则原式为：$$\int_1^2dy\int_y^{1/y}x\cosy\dx=\int_1^2dy\left[\frac{x^2}{2}\cosy\right]_y^{1/y}=\int_1^2\frac{\cosy}{2y^2}-\frac{y\cosy}{2}dy=\frac{\sin1}{2}-\frac{5\sin2}{8}$$(2)选择$y$为外层积分变量，$x$为内层积分变量，则原式为：$$\int_0^2dy\int_0^{\sqrt{2y-y^2}}(x+y)dx=\int_0^{\pi}d\theta\int_0^{2\cos\theta}(r\cos\theta+r\sin\theta)rdr=\frac{8}{3}$$(3)选择极坐标系，则原式为：$$\int_1^2dr\int_0^{2\pi}\frac{r}{2}e^{-r^2/2}d\theta=\pi(1-e^{-2})$$(4)选择$y$为外层积分变量，$x$为内层积分变量，则原式为：$$\int_0^1dy\int_{-\sqrt{1-y^2}}^{\sqrt{1-y^2}}\frac{1}{\sqrt{x^2+y^2}}dx=\int_0^{\pi/2}d\theta\int_0^1\frac{r}{r^2}dr=\frac{\pi}{2}$$2.(1)选择极坐标系，则原式为：$$\int_0^{\pi/4}d\theta\int_{\sqrt{\tan\theta}}^{2\sec\theta}r\cos\thetadr=\frac{3\sqrt{2}-2\ln(1+\sqrt{2})}{6}$$(2)选择$y$为外层积分变量，$x$为内层积分变量，则原式为：$$\int_0^{\pi/4}dy\int_{y^2}^{2y}(x^2+y^2)dx=\frac{5\pi}{96}$$(3)选择极坐标系，则原式为：$$\int_1^2dr\int_0^{2\pi}r^3d\theta=\frac{15}{2}\pi$$3.选择$y=x^2$作为积分区域的下界，$y=x$作为积分区域的上界，则原式为：$$\int_0^1dx\int_{x^2}^xf(x,y)dy=\int_0^1dx\int_{x^2}^x\frac{f(x,y)}{f(x,x)}f(x,x)dy=\int_0^1dx\int_{x^2}^x\frac{f(x,y)}{f(x,x)}\phi(x)dy=\int_0^1\phi(x)\int_{x^2}^x\frac{f(x,y)}{f(x,x)}dydx$$由于$D$是由直线$x+y=1$和两坐标轴围成的区域，所以$f(x,y)=f(x,x+y-1)$。因此：$$\int_{x^2}^x\frac{f(x,y)}{f(x,x)}dy=\int_{x^2}^x\frac{f(x,x+y-1)}{f(x,x)}dy=\int_{x-1}^0\frac{f(x,y)}{f(x,x)}dy=\int_0^{1-x}\frac{f(x,y)}{f(x,x)}dy$$所以原式变为：$$\int_0^1\phi(x)\int_0^{1-x}\frac{f(x,y)}{f(x,x)}dydx=\int_0^1\phi(x)dx=\int_0^1(1-x)dx=\frac{1}{2}$$4.选择$x$为外层积分变量，$y$为内层积分变量，则原式为：$$\int_{-1}^2dx\int_{y=-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dyf(x,y)=\int_{-\sqrt{2}}^{\sqrt{2}}dx\int_{y=-\sqrt{1-x^2}}^{\sqrt{1-x^2}}dyf(x,y)$$选择$y$为外层积分变量，$x$为内层积分变量，则原式为：$$\int_{-\sqrt{2}}^{\sqrt{2}}dy\int_{x=-\sqrt{2-y^2}}^{\sqrt{2-y^2}}dxf(x,y)$$5.选择极坐标系，则原式为：$$\int_0^{\pi/2}d\theta\int_0^1rdr=\frac{\pi}{4}$$6.选择$y$为外层积分变量，$x$为内层积分变量，则原式为：$$\int_0^1dy\int_{x=y^2}^{\sqrt{y}}f(x,y)dx=\int_0^1dy\int_{x=0}^{\sqrt{y}}f(x,y)dx-\int_0^1dy\int_{x=0}^{y^2}f(x,y)dx=\int_0^1\sqrt{y}f(\sqrt{y},y)dy-\int_0^1y^2f(y^2,y)dy$$7.选择$x$为外层积分变量，$y$为内层积分变量，则原式为：$$\int_0^1dx\int_{y=x}^{\sqrt{x}}f(x,y)dy$$选择$y$为外层积分变量，$x$为内层积分变量，则原