# MA127 高等数学下 2019年春 期末试卷 答案

一 Determine whether the following statements are true or false?

F T F

二 Multiple Choice Questions

B D D C D

三 Please fill in the blank for the questions below

3-1

$\frac{1}{2}$

3-2

$(0,1)$

3-3

$\ln (\sec 1)$

四

4-1

$\underset{n\to \mathrm{\infty }}{lim}\sqrt[n]{\left|a_n\right|}=(x-1{)}^2<1,x\in (0,2)$

当 $x=0$ 时, $a_n=\frac{(-1{)}^{n+1}}{\sqrt{n+9012}\ln n}$

$\left|a_n\right|$ 单减.

且 $\frac{\left|a_n\right|}{\frac{1}{n\ln x}}=\mathrm{\infty }\sum \frac{1}{n\ln n}$ 发散

$\therefore$ 条件收敛.

$x=2$ 时, $a_n=\frac{(-1{)}^n}{\sqrt{n+9912}\ln n}$

同上

4-2

绝对收敛$(0,2)$

条件收敛 $x=0,x=2$

五

5-1

$${\int }_0^{2\pi }{\int }_0^{\frac{\sqrt{2}}{2}}{\int }_r^{\sqrt{1-r^2}}(r\cos \theta +z)rdzdrd\theta$$

5-2

$${\int }_0^{2\pi }{\int }_0^{\frac{\pi }{4}}{\int }_0^1(\rho \sin \phi \cos \theta +\rho \cos \phi ){\rho }^2\sin \phi d\rho d\phi d\theta$$

六

$V=8xyz$

$$\begin{array}{rl}& g(x,y,z)=\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1\\ & \mathrm{\nabla }\overrightarrow{V}=\lambda \mathrm{\nabla }\overrightarrow{g}\\ & \{\begin{array}{ll}8yz=\lambda \frac{2x}{a^2}& \\ 8xz=\lambda \frac{2y}{b^2}& \\ 8xy=\lambda \frac{2z}{c^2}& \end{array}\end{array}$$

when $\lambda =0$ 不成立

if $\lambda \ne 0$

$\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}\Rightarrow x=\frac{a}{\sqrt{3}}y=\frac{b}{\sqrt{3}}z=\frac{c}{\sqrt{3}}$

$\therefore V=\frac{8abc}{3\sqrt{3}}$

七

$$\begin{array}{rl}& y=x^{2}x=1\\ & \overrightarrow{r}(t)=(t,t^2)\\ & \overrightarrow{V}(t)=(1,2t)\\ & |\overrightarrow{V}(t)|=\sqrt{4t^2+1}\\ & \overrightarrow{T}=(\frac{1}{\sqrt{4t^2+1}},\frac{2t}{\sqrt{4t^2+1}})\end{array}$$$$\begin{array}{c}\frac{d\overrightarrow{T}}{dt}=(-\frac{4t}{(1+4{)}^{3/2}},\frac{2}{{(1+4t^2)}^{\frac{3}{2}}})\\ k(1)=\frac{2}{5\sqrt{5}}r=\frac{5\sqrt{5}}{2}\\ (1,1)+\frac{5\sqrt{5}}{2}\cdot \frac{1}{\sqrt{5}}(-2,1)=(-4,\frac{7}{2})\\ \therefore (x+4{)}^2+{(y-\frac{7}{2})}^2=\frac{125}{4}\end{array}$$

八

$$\begin{array}{r}M\\ & ={\int }_0^2{\int }_0^{\sqrt{x}}{\int }_0^{4-x^2}xydzdydx=\frac{32}{15}\\ & M_{xy}=\frac{256}{105}\\ & M_{yz}=\frac{8}{3}\\ & M_{xz}=\frac{256\sqrt{2}}{231}\\ & \therefore \overline{x}=\frac{M_{yz}}{M}=\frac{5}{4}\overline{y}=\frac{M_{xz}}{M}=\frac{40\sqrt{2}}{77}\overline{z}=\frac{M_{xy}}{M}=\frac{8}{7}\end{array}$$

九

$$\begin{array}{l}u=y-x\\ v=y+x\\ J=\left|\left|\begin{array}{cc}-\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& \frac{1}{2}\end{array}\right|\right|=\frac{1}{2}\end{array}$$$$\begin{array}{l}x=0\\ y=0\\ x+y=2\\ v=u\\ u=-v\\ v=2\end{array}$$$$\therefore I={\int }_0^2{\int }_{-v}^v{e}^{\frac{u}{v}}\cdot \frac{1}{2}dudv=e-e^{-1}$$

十

10-1

$\frac{\partial P}{\partial y}=\frac{\partial N}{\partial z};\frac{\partial P}{\partial x}=\frac{\partial M}{\partial z};\frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}$

$$\therefore \text{ exact }$$

10-2

$$\begin{array}{rl}& \frac{\partial f}{\partial x}=e^x\ln y\\ & f=e^x\ln y+g(y,z)\\ & \frac{\partial f}{\partial z}=y\cos z\therefore f=e^x\ln y+y\sin z+h(y)\\ & \frac{\partial f}{\partial y}=\frac{e^x}{y}+\sin z+\frac{\partial h}{\partial z}\therefore \frac{\partial h}{\partial z}=c\\ & \therefore f=e^x\ln y+y\sin z+c\\ & f(1,3,\pi )-f(1,1,1)=e\ln 3-\sin 1\end{array}$$

十一

$$\begin{array}{rl}& \mathrm{\nabla }\cdot (4\times \overrightarrow{j})=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ \frac{\partial }{\partial x}& \frac{j}{\partial y}& \frac{\delta }{dz}\\ 0\cdot & 4x& 0\end{array}\right|=4\overrightarrow{k}\\ & \therefore {\iint }_s\mathrm{\nabla }\times (4\times \overrightarrow{j})\cdot \overrightarrow{k}d\sigma ={\iint }_{s}4\overrightarrow{k}\cdot \overrightarrow{k}d\sigma =4{\iint }_{s}d\sigma =64\pi \end{array}$$

十二

$\mathrm{\nabla }\cdot \overrightarrow{F}=6+4y$, Stokes定理

$$\begin{array}{rl}{\iiint }_v(6+4y)dv& ={\int }_0^1{\int }_0^{\sqrt{1-x^2}}{\int }_0^{\sqrt{x^2+y^2}}(6+4y)dzdydx\\ & ={\int }_0^{\frac{\pi }{2}}{\int }_0^1{\int }_0^r(6+4r\sin \theta )rdzdrd\theta \\ & =\pi +1\end{array}$$

十三

$$\begin{array}{rl}& \cos a_n-a_n=\cos b_n\\ & \cos a_n-\cos b_n=a_n>0\\ & \Rightarrow 0<a_n<b_n\\ & \sum _{n=1}^{\mathrm{\infty }}{b}_n\text{ 收敛 }\Rightarrow \underset{n\to \mathrm{\infty }}{lim}{b}_n=0\\ & 0\le \underset{n\to \mathrm{\infty }}{lim}{a}_n\le b_n\therefore \underset{n\to \mathrm{\infty }}{lim}{a}_n=0\end{array}$$