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

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\(x^2+y^2=2-z\)\(z=\sqrt{x^2+y^2}\)所围成的区域.

解:联立两曲面方程,消去\(z\),即得\(D\)\(xOy\)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

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

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

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

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


Last update: July 26, 2020