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2015-2016第二学期数学分析3-2期末考试(含解答)

一、讨论f(x,y)=\sqrt{|xy|}(0,0)的可微性.

二、设u=u(x)为方程组

\left\{\begin{aligned}u=f(x,y,z)\\g(x,y,z)=0\\h(x,y,z)=0\end{aligned}\right.

确定的隐函数,求\displaystyle \frac{\mathrm{d}u}{\mathrm{d}x}.

三、求F(x,y,z)=x+y+z在条件xy+yz+xz=1下的条件极值.

四、求\displaystyle \iint_{x^2+y^2 \leqslant 1} [y-x] \mathrm{d}x \mathrm{d}y.(其中[x]表示不超过x的最大整数)

五、求\displaystyle \iint_{|x|+|y| \leqslant 4} \frac{|x^3+xy^2-2x^2-2xy|}{|x|+|y|} \mathrm{d}x \mathrm{d}y.

六、求\displaystyle \iiint_D (x^2+z^2) \mathrm{d}x \mathrm{d}y \mathrm{d}z.其中区域D为由曲面x^2+y^2=2-zz=\sqrt{x^2+y^2}所围成的区域.

感谢Chokie给出了该套题的详细解答,在其授权下将其给出的详解附在本文中。


一、讨论f(x,y)=\sqrt{|xy|}(0,0)处的可微性.

解:令

A=\dfrac{\partial f}{\partial x}(0,0)=\displaystyle\lim_{x\rightarrow0}\frac{f(x,0)-f(0,0)}{x}=0
B=\dfrac{\partial f}{\partial y}(0,0)=\displaystyle\lim_{x\rightarrow0}\frac{f(0,y)-f(0,0)}{y}=0

现求极限\displaystyle\lim_{\substack{x\rightarrow0\\ y\rightarrow0}}\frac{f(x,y)-f(0,0)-Ax-By}{\sqrt{x^{2}+y^{2}}}=\lim_{\substack{x\rightarrow0\\ y\rightarrow0}}\frac{\sqrt{|xy|}}{\sqrt{x^{2}+y^{2}}}.

x=r\cos\theta,y=r\sin\theta, 则上述极限=\displaystyle\lim_{r\rightarrow0^{+}}\frac{r\sqrt{|\cos\theta\sin\theta|}}{\sqrt{r^{2}}}=\sqrt{|\cos\theta\sin\theta|}, 可见上述极限显然不存在,则由可微性的定义即有f(x,y)(0,0)处不可微.

二、设u=u(x)为方程组\displaystyle \left\{\begin{aligned}u=f(x,y,z)\\g(x,y,z)=0\\h(x,y,z)=0\end{aligned}\right.确定的隐函数,求\displaystyle \frac{\mathrm{d}u}{\mathrm{d}x}.

解:对题目中三个等式两端求全微分,得到以下三个等式:

\begin{cases}\dfrac{\partial f}{\partial x}\mathrm{d}x+\dfrac{\partial f}{\partial y}\mathrm{d}y+\dfrac{\partial f}{\partial z}\mathrm{d}z=\mathrm{d}u\\ \dfrac{\partial g}{\partial x}\mathrm{d}x+\dfrac{\partial g}{\partial y}\mathrm{d}y+\dfrac{\partial g}{\partial z}\mathrm{d}z=0\\ \dfrac{\partial h}{\partial x}\mathrm{d}x+\dfrac{\partial h}{\partial y}\mathrm{d}y+\dfrac{\partial h}{\partial z}\mathrm{d}z=0\end{cases}

将以上三个等式看作以\mathrm{d}x,\mathrm{d}y,\mathrm{d}z为变量的线性方程组,即解得

\mathrm{d}x=\dfrac{\mathrm{d}u\dfrac{D(g,h)}{D(y,z)}}{\dfrac{D(f,g,h)}{D(x,y,z)}},即\dfrac{\mathrm{d}u}{\mathrm{d}x}=\dfrac{\dfrac{D(f,g,h)}{D(x,y,z)}}{\dfrac{D(g,h)}{D(y,z)}}.

三、求F(x,y,z)=x+y+z在条件xy+yz+xz=1下的条件极值.

解:令L(x,y,z)=f(x,y,z)+\lambda(xy+yz+xz-1)(\lambda待定).

解方程组

\begin{cases}\dfrac{\partial L}{\partial x}=1+\lambda(y+z)=0\\ \dfrac{\partial L}{\partial y}=1+\lambda(x+z)=0\\ \dfrac{\partial L}{\partial z}=1+\lambda(x+y)=0\\xy+yz+xz-1=0\end{cases}

即可解得x=y=z=\dfrac{\sqrt{3}}{3},\lambda=-\dfrac{\sqrt{3}}{2}x=y=z=-\dfrac{\sqrt{3}}{3},\lambda=\dfrac{\sqrt{3}}{2}.

\mathrm{d}L=(1+\lambda(y+z))\mathrm{d}x+(1+\lambda(x+z))\mathrm{d}y+(1+\lambda(x+y))\mathrm{d}z, \mathrm{d}^{2}L=\lambda(2\mathrm{d}x\mathrm{d}y+2\mathrm{d}y\mathrm{d}z+2\mathrm{d}x\mathrm{d}z).

xy+yz+xz=1两边取微分得

(y+z)\mathrm{d}x+(x+z)\mathrm{d}y+(x+y)\mathrm{d}z=0

由于对上述两解均有x=y=z, 于是\mathrm{d}x+\mathrm{d}y+\mathrm{d}z=0,两边取平方即有

2\mathrm{d}x\mathrm{d}y+2\mathrm{d}x\mathrm{d}z+2\mathrm{d}y\mathrm{d}z=-(\mathrm{d}^{2}x+\mathrm{d}^{2}y+\mathrm{d}^{2}z)

因此\mathrm{d}^{2}L=-\lambda(\mathrm{d}^{2}x+\mathrm{d}^{2}y+\mathrm{d}^{2}z).

\lambda=-\dfrac{\sqrt{3}}{2}时,\mathrm{d}^{2}L>0,f(x,y,z)取极小值,为f(\dfrac{\sqrt{3}}{3},\dfrac{\sqrt{3}}{3},\dfrac{\sqrt{3}}{3})=\sqrt{3}.

\lambda=\dfrac{\sqrt{3}}{2}时,\mathrm{d}^{2}L<0,f(x,y,z)取极大值,为f(-\dfrac{\sqrt{3}}{3},-\dfrac{\sqrt{3}}{3},-\dfrac{\sqrt{3}}{3})=-\sqrt{3}.

四、求\displaystyle \iint_{x^2+y^2 \leqslant 1} [y-x] \mathrm{d}x \mathrm{d}y.(其中[x]表示不超过x的最大整数)

解:记D=x^{2}+y^{2}\leqslant1.易知当(x,y)\in D,y-x\in[-\sqrt{2},\sqrt{2}].

把区域D分成四个区域, D_{1}={(x,y)|-\sqrt{2}\leqslant y-x<-1}\bigcap D D_{2}={(x,y)|-1\leqslant y-x<0}\bigcap D D_{3}={(x,y)|0\leqslant y-x<1}\bigcap D D_{4}={(x,y)|1\leqslant y-x\leqslant\sqrt{2}}\bigcap D

I=-2\displaystyle\iint_{D_{1}}\mathrm{d}x\mathrm{d}y-\iint_{D_{2}}\mathrm{d}x\mathrm{d}y+\iint_{D_{4}}\mathrm{d}x\mathrm{d}y=-2|D_{1}|-|D_{2}|+|D_{4}|.

显然有|D_{1}|=|D_{4}||D_{1}|+|D_{2}|=\dfrac{\pi}{2},因此I=-\dfrac{\pi}{2}.

五、求\displaystyle \iint_{|x|+|y| \leqslant 4} \frac{|x^3+xy^2-2x^2-2xy|}{|x|+|y|} \mathrm{d}x \mathrm{d}y.

解:记D=|x|+|y|\leqslant4,|D|=32,有|x^{3}+xy^{2}-2x^{2}-2xy|=|x||x^{2}+y^{2}-2x-2y|.

区域D关于x,y是对称的,且根据变量的对称性就有

I=\displaystyle\iint_{D}\dfrac{|x||x^{2}+y^{2}-2x-2y|}{|x|+|y|}\mathrm{d}x\mathrm{d}y=\iint_{D}\dfrac{|y||x^{2}+y^{2}-2x-2y|}{|x|+|y|}\mathrm{d}x\mathrm{d}y

可得

I=\displaystyle\dfrac{1}{2}\iint_{D}|x^{2}+y^{2}-2x-2y|\mathrm{d}x\mathrm{d}y=I=\frac{1}{2}\iint_{D}|(x-1)^{2}+(y-1)^{2}-2|\mathrm{d}x\mathrm{d}y

D_{1}=(x-1)^{2}+(y-1)^{2}\leqslant2.由于D_{1}\subseteq D,将区域D分为两个区域,D_{1}D\setminus D_{1}.

再记 I_{1}=\displaystyle\frac{1}{2}\iint_{D_{1}}(2x+2y-x^{2}-y^{2})\mathrm{d}x\mathrm{d}y I_{2}=\displaystyle\frac{1}{2}\iint_{D\setminus D_{1}}(x^{2}+y^{2}-2x-2y)\mathrm{d}x\mathrm{d}y=\frac{1}{2}\iint_{D}(x^{2}+y^{2}-2x-2y)\mathrm{d}x\mathrm{d}y+I_{1} I_{3}=\displaystyle\frac{1}{2}\iint_{D}(x^{2}+y^{2}-2x-2y)\mathrm{d}x\mathrm{d}yI=I_{1}+I_{2}=I_{3}+2I_{1}.

根据D_{1}关于x,y的对称性,I_{1}=\displaystyle\iint_{D_{1}}(1-(x-1)^{2})\mathrm{d}x\mathrm{d}y,经过极坐标代换容易解出I_{1}=\pi.

同理,有I_{3}=\displaystyle\iint_{D}((x-1)^{2}-1)\mathrm{d}x\mathrm{d}y=\iint_{D}(x-1)^{2}\mathrm{d}x\mathrm{d}y-32.

作变换u=x+y,v=x-y,得x=\dfrac{u+v}{2},y=\dfrac{u-v}{2}.则\Big{|}\dfrac{D(x,y)}{D(u,v)}\Big{|}=\dfrac{1}{2}.

区域D\rightarrow D',D'=\{(u,v)|u\in[-4,4],v\in[-4,4]\}.

I_{3}=\displaystyle\dfrac{1}{2}\iint_{D'}(\dfrac{u+v}{2}-1)^{2}\mathrm{d}u\mathrm{d}v-32=\frac{1}{2}\int_{-4}^{4}\mathrm{d}v\int_{-4}^{4}(\dfrac{u+v}{2}-1)^{2}\mathrm{d}u-32=\frac{256}{3}.

I=\dfrac{256}{3}+2\pi.

六、求\displaystyle \iiint_D (x^2+z^2) \mathrm{d}x \mathrm{d}y \mathrm{d}z.其中区域D为由曲面x^2+y^2=2-zz=\sqrt{x^2+y^2}所围成的区域.

解:联立两曲面方程,消去z,即得DxOy上的投影为x^{2}+y^{2}\leqslant1.

因此D=\{(x,y,z)|x^{2}+y^{2}\leqslant1,\sqrt{x^{2}+y^{2}}\leqslant z\leqslant2-x^{2}-y^{2}\}.

I=\displaystyle\iint_{x^{2}+y^{2}\leqslant1}\mathrm{d}x\mathrm{d}y\int_{\sqrt{x^{2}+y^{2}}}^{2-x^{2}-y^{2}}(x^{2}+z^{2})\mathrm{d}z 经过坐标变换x=r\cos\theta,y=r\sin\theta,z=z,有

I=\displaystyle\int_{0}^{2\pi}\mathrm{d}\theta\int_{0}^{1}r\mathrm{d}r\int_{r}^{2-r^{2}}(r^{2}\cos^{2}\theta+z^{2})\mathrm{d}z

=\displaystyle\int_{0}^{2\pi}cos^{2}\theta\mathrm{d}\theta\int_{0}^{1}r^{3}(2-r^{2}-r)\mathrm{d}r+\frac{2\pi}{3}\int_{0}^{1}r((2-r^{2})^{3}-r^{3})\mathrm{d}r

=\displaystyle\pi(\frac{1}{2}-\frac{1}{6}-\frac{1}{5})-\frac{\pi}{3}\int_{0}^{1}(2-r^{2})^{3}\mathrm{d}(2-r^{2})-\frac{2\pi}{15}

=-\dfrac{\pi}{12}(2-r^{2})^{4}\Big{|}_{0}^{1}=\dfrac{5\pi}{4}.