高等數(shù)學(xué)（二）多元函數(shù)考試及答案

一、填空題1.設(shè)函數(shù)\(z=\ln(x+y^2)\)，則\(\frac{\partialz}{\partialx}\)在點(diǎn)\((1,1)\)處的值為______。2.函數(shù)\(z=x^2+y^2\)在點(diǎn)\((1,2)\)處沿向量\(\vec{a}=(1,1)\)的方向?qū)?shù)為______。3.設(shè)\(z=e^{xy}\)，則\(dz\)=______。4.設(shè)\(f(x,y)\)在區(qū)域\(D\)：\(x^2+y^2\leqslant1\)上連續(xù)，將二重積分\(\iint_{D}f(x,y)dxdy\)化為極坐標(biāo)形式為______。5.設(shè)\(z=x^3y^2\)，則\(\frac{\partial^2z}{\partialx\partialy}\)=______。6.已知函數(shù)\(z=f(x,y)\)由方程\(x^2+y^2+z^2=1\)確定，則\(\frac{\partialz}{\partialy}\)=______。7.設(shè)\(D\)是由\(x=0\)，\(y=0\)，\(x+y=1\)所圍成的區(qū)域，則\(\iint_{D}dxdy\)=______。8.函數(shù)\(f(x,y)=x^2-y^2\)的駐點(diǎn)是______。9.設(shè)\(z=\sin(xy)\)，則\(\frac{\partialz}{\partialx}\)=______。10.交換積分次序\(\int_{0}^{1}dx\int_{x}^{1}f(x,y)dy\)=______。二、單項(xiàng)選擇題1.函數(shù)\(z=\frac{1}{\sqrt{x^2+y^2-1}}\)的定義域是（）A.\(x^2+y^2\geqslant1\)B.\(x^2+y^2\gt1\)C.\(x^2+y^2\leqslant1\)D.\(x^2+y^2\lt1\)2.設(shè)\(z=x^y\)，則\(\frac{\partialz}{\partialy}\big|_{(e,1)}\)=（）A.\(0\)B.\(1\)C.\(e\)D.\(\frac{1}{e}\)3.設(shè)\(D\)是由\(x=0\)，\(y=0\)，\(x+y=1\)所圍成的區(qū)域，則\(\iint_{D}(x+y)dxdy\)=（）A.\(\frac{1}{6}\)B.\(\frac{1}{3}\)C.\(\frac{1}{2}\)D.\(1\)4.函數(shù)\(z=x^2+y^2\)在點(diǎn)\((0,0)\)處（）A.有極大值B.有極小值C.無(wú)極值D.無(wú)法判斷5.設(shè)\(z=f(x,y)\)在點(diǎn)\((x_0,y_0)\)處的兩個(gè)偏導(dǎo)數(shù)\(f_x(x_0,y_0)\)，\(f_y(x_0,y_0)\)都存在，則（）A.\(z=f(x,y)\)在點(diǎn)\((x_0,y_0)\)處連續(xù)B.\(z=f(x,y)\)在點(diǎn)\((x_0,y_0)\)處可微C.\(z=f(x,y)\)在點(diǎn)\((x_0,y_0)\)處沿任何方向的方向?qū)?shù)都存在D.以上結(jié)論都不對(duì)6.設(shè)\(D\)是由\(x^2+y^2\leqslant4\)所圍成的區(qū)域，則\(\iint_{D}\sqrt{x^2+y^2}dxdy\)=（）A.\(\frac{16\pi}{3}\)B.\(\frac{8\pi}{3}\)C.\(\frac{4\pi}{3}\)D.\(\frac{2\pi}{3}\)7.已知\(z=e^{x+y}\)，則\(\frac{\partial^2z}{\partialx\partialy}\)=（）A.\(e^{x+y}\)B.\(xe^{x+y}\)C.\(ye^{x+y}\)D.\(e^{x}\)8.函數(shù)\(z=xy\)在點(diǎn)\((1,1)\)處沿向量\(\vec{l}=(1,-1)\)的方向?qū)?shù)為（）A.\(\sqrt{2}\)B.\(-\sqrt{2}\)C.\(0\)D.\(1\)9.設(shè)\(z=f(x,y)\)由方程\(x+y+z=e^{z}\)確定，則\(\frac{\partialz}{\partialx}\)=（）A.\(\frac{1}{e^{z}-1}\)B.\(\frac{1}{1-e^{z}}\)C.\(\frac{e^{z}}{e^{z}-1}\)D.\(\frac{e^{z}}{1-e^{z}}\)10.設(shè)\(D\)是由\(y=x\)，\(y=2x\)，\(x=1\)所圍成的區(qū)域，則\(\iint_{D}xydxdy\)=（）A.\(\frac{3}{8}\)B.\(\frac{1}{8}\)C.\(\frac{5}{8}\)D.\(\frac{7}{8}\)三、多項(xiàng)選擇題1.下列函數(shù)在點(diǎn)\((0,0)\)處連續(xù)的有（）A.\(f(x,y)=\begin{cases}\frac{xy}{x^2+y^2},(x,y)\neq(0,0)\\0,(x,y)=(0,0)\end{cases}\)B.\(f(x,y)=x^2+y^2\)C.\(f(x,y)=\sqrt{x^2+y^2}\)D.\(f(x,y)=\sin(x^2+y^2)\)2.設(shè)\(z=f(x,y)\)，則下列說(shuō)法正確的有（）A.若\(z=f(x,y)\)在點(diǎn)\((x_0,y_0)\)處可微，則\(z=f(x,y)\)在點(diǎn)\((x_0,y_0)\)處連續(xù)B.若\(z=f(x,y)\)在點(diǎn)\((x_0,y_0)\)處可微，則\(z=f(x,y)\)在點(diǎn)\((x_0,y_0)\)處的兩個(gè)偏導(dǎo)數(shù)都存在C.若\(z=f(x,y)\)在點(diǎn)\((x_0,y_0)\)處的兩個(gè)偏導(dǎo)數(shù)都存在，則\(z=f(x,y)\)在點(diǎn)\((x_0,y_0)\)處可微D.若\(z=f(x,y)\)在點(diǎn)\((x_0,y_0)\)處的兩個(gè)偏導(dǎo)數(shù)都連續(xù)，則\(z=f(x,y)\)在點(diǎn)\((x_0,y_0)\)處可微3.設(shè)\(D\)是由\(x^2+y^2\leqslant1\)所圍成的區(qū)域，下列二重積分值為\(0\)的有（）A.\(\iint_{D}xdxdy\)B.\(\iint_{D}ydxdy\)C.\(\iint_{D}xydxdy\)D.\(\iint_{D}(x+y)dxdy\)4.函數(shù)\(z=f(x,y)\)的駐點(diǎn)可能是（）A.極值點(diǎn)B.鞍點(diǎn)C.最值點(diǎn)D.非極值點(diǎn)5.設(shè)\(z=e^{xy}\)，則（）A.\(\frac{\partialz}{\partialx}=ye^{xy}\)B.\(\frac{\partialz}{\partialy}=xe^{xy}\)C.\(\frac{\partial^2z}{\partialx\partialy}=(1+xy)e^{xy}\)D.\(\frac{\partial^2z}{\partialy\partialx}=(1+xy)e^{xy}\)6.設(shè)\(D\)是由\(x=0\)，\(y=0\)，\(x+y=1\)所圍成的區(qū)域，將\(\iint_{D}f(x,y)dxdy\)化為二次積分，正確的有（）A.\(\int_{0}^{1}dx\int_{0}^{1-x}f(x,y)dy\)B.\(\int_{0}^{1}dy\int_{0}^{1-y}f(x,y)dx\)C.\(\int_{0}^{1}dx\int_{0}^{1}f(x,y)dy\)D.\(\int_{0}^{1}dy\int_{0}^{1}f(x,y)dx\)7.設(shè)\(z=f(x,y)\)由方程\(F(x,y,z)=0\)確定，則（）A.\(\frac{\partialz}{\partialx}=-\frac{F_x}{F_z}\)B.\(\frac{\partialz}{\partialy}=-\frac{F_y}{F_z}\)C.\(dz=\frac{\partialz}{\partialx}dx+\frac{\partialz}{\partialy}dy\)D.當(dāng)\(F_z\neq0\)時(shí)，上述公式成立8.下列關(guān)于方向?qū)?shù)的說(shuō)法正確的有（）A.函數(shù)在某點(diǎn)沿梯度方向的方向?qū)?shù)最大B.函數(shù)在某點(diǎn)沿負(fù)梯度方向的方向?qū)?shù)最小C.函數(shù)在某點(diǎn)沿任意方向的方向?qū)?shù)都存在D.函數(shù)在某點(diǎn)的方向?qū)?shù)與梯度有關(guān)9.設(shè)\(D\)是由\(y=x^2\)，\(y=1\)所圍成的區(qū)域，則\(\iint_{D}f(x,y)dxdy\)化為二次積分可以是（）A.\(\int_{-1}^{1}dx\int_{x^2}^{1}f(x,y)dy\)B.\(\int_{0}^{1}dy\int_{-\sqrt{y}}^{\sqrt{y}}f(x,y)dx\)C.\(\int_{-1}^{1}dx\int_{0}^{1}f(x,y)dy\)D.\(\int_{0}^{1}dy\int_{0}^{\sqrt{y}}f(x,y)dx\)10.設(shè)\(z=x^2-2xy+y^2\)，則（）A.\(z\)在點(diǎn)\((1,1)\)處的駐點(diǎn)B.\(z\)在點(diǎn)\((1,1)\)處的極小值為\(0\)C.\(z\)在點(diǎn)\((1,1)\)處的極大值為\(0\)D.\(z\)在點(diǎn)\((1,1)\)處無(wú)極值四、判斷題1.若函數(shù)\(z=f(x,y)\)在點(diǎn)\((x_0,y_0)\)處的兩個(gè)偏導(dǎo)數(shù)都存在，則函數(shù)在該點(diǎn)一定連續(xù)。（）2.二重積分\(\iint_{D}f(x,y)dxdy\)的幾何意義是區(qū)域\(D\)上以\(z=f(x,y)\)為曲頂?shù)那斨w的體積。（）3.函數(shù)\(z=f(x,y)\)的駐點(diǎn)一定是極值點(diǎn)。（）4.若\(z=f(x,y)\)在點(diǎn)\((x_0,y_0)\)處可微，則\(z=f(x,y)\)在該點(diǎn)沿任意方向的方向?qū)?shù)都存在。（）5.交換積分次序時(shí)，積分限不需要改變。（）6.設(shè)\(z=f(x,y)\)由方程\(F(x,y,z)=0\)確定，則\(\frac{\partialz}{\partialx}\cdot\frac{\partialx}{\partialy}\cdot\frac{\partialy}{\partialz}=-1\)。（）7.函數(shù)\(z=x^2+y^2\)在點(diǎn)\((0,0)\)處沿任何方向的方向?qū)?shù)都為\(0\)。（）8.若\(z=f(x,y)\)在區(qū)域\(D\)上連續(xù)，則\(\iint_{D}f(x,y)dxdy\)一定存在。（）9.設(shè)\(D\)是由\(x^2+y^2\leqslant1\)所圍成的區(qū)域，則\(\iint_{D}\sqrt{1-x^2-y^2}dxdy=\frac{2\pi}{3}\)。（）10.設(shè)\(z=f(x,y)\)，則\(\frac{\partial^2z}{\partialx\partialy}=\frac{\partial^2z}{\partialy\partialx}\)一定成立。（）五、簡(jiǎn)答題1.簡(jiǎn)述函數(shù)\(z=f(x,y)\)在點(diǎn)\((x_0,y_0)\)處可微、偏導(dǎo)數(shù)存在、連續(xù)之間的關(guān)系。2.如何將二重積分\(\iint_{D}f(x,y)dxdy\)化為極坐標(biāo)形式？3.求函數(shù)\(z=f(x,y)\)駐點(diǎn)的步驟是什么？4.簡(jiǎn)述方向?qū)?shù)與梯度的關(guān)系。六、討論題1.討論函數(shù)\(z=x^2+y^2\)的極值情況。2.討論二重積分\(\iint_{D}f(x,y)dxdy\)在不同積分區(qū)域\(D\)下的計(jì)算方法。3.討論函數(shù)\(z=f(x,y)\)在某點(diǎn)處偏導(dǎo)數(shù)存在與可微的關(guān)系，并舉例說(shuō)明。4.討論如何利用二重積分求平面圖形的面積。答案一、填空題1.\(\frac{1}{2}\)2.\(3\sqrt{2}\)3.\(e^{xy}(ydx+xdy)\)4.\(\int_{0}^{2\pi}d\theta\int_{0}^{1}f(r\cos\theta,r\sin\theta)rdr\)5.\(6x^2y\)6.\(-\frac{y}{z}(z\neq0)\)7.\(\frac{1}{2}\)8.\((0,0)\)9.\(y\cos(xy)\)10.\(\int_{0}^{1}dy\int_{0}^{y}f(x,y)