习题答案与提示

习题答案与提示

第八章

习题8-1（第13页）

1. 5a - 11b + 7c.

1. 略.

1. $ \overrightarrow{D_1A} = -\left(c + \frac{1}{5}a\right) $, $ \overrightarrow{D_2A} = -\left(c + \frac{2}{5}a\right) $, $ \overrightarrow{D_3A} = -\left(c + \frac{3}{5}a\right) $, $ \overrightarrow{D_4A} = -\left(c + \frac{4}{5}a\right) $.

4.（1，-2，-2），(-2，4，4）.

1. $ \left(\frac{6}{11},\frac{7}{11},-\frac{6}{11}\right) $ 或 $ \left(-\frac{6}{11},-\frac{7}{11},\frac{6}{11}\right) $.

1. $A: \mathrm{IV}, B: \mathrm{V}, C: \mathrm{VIII}, D: \mathrm{III}$.

1. A 在 xOy 面上，B 在 yOz 面上，C 在 x 轴上，D 在 y 轴上.

1. (1) $ (a, b, -c) $, $ (-a, b, c) $, $ (a, -b, c) $;

(2) $ (a, -b, -c) $, $ (-a, b, -c) $, $ (-a, -b, c) $;

(3) $ (-a, -b, -c) $.

1. xOy 面： $ (x_{0},y_{0},0) $，yOz 面： $ (0,y_{0},z_{0}) $，xOz 面： $ (x_{0},0,z_{0}) $；x 轴： $ (x_{0},0,0) $，y 轴： $ (0,y_{0},0) $，z 轴： $ (0,0,z_{0}) $.

1. 略.

1. $ \left(\frac{\sqrt{2}}{2}a,0,0\right),\quad\left(-\frac{\sqrt{2}}{2}a,0,0\right),\quad\left(0,\frac{\sqrt{2}}{2}a,0\right),\quad\left(0,-\frac{\sqrt{2}}{2}a,0\right),\quad\left(\frac{\sqrt{2}}{2}a,0,a\right), $

$ \left(-\frac{\sqrt{2}}{2}a,0,a\right),\left(0,\frac{\sqrt{2}}{2}a,a\right),\left(0,-\frac{\sqrt{2}}{2}a,a\right). $

1. x 轴： $ \sqrt{34} $，y 轴： $ \sqrt{41} $，z 轴：5.

1. (0,1, -2).

1. 略.

1. 模:2;方向余弦: $ -\frac{1}{2} $, $ -\frac{\sqrt{2}}{2} $, $ \frac{1}{2} $; 方向角 $ \frac{2\pi}{3}, \frac{3\pi}{4}, \frac{\pi}{3} $.

16 平面；

（2）指向与 y 轴正向一致，垂直于 xOz 平面；

（3）平行于z轴，垂直于xOy平面.

17.2.

1. $ A(-2,3,0) $.

1. 13,7j.

习题8-2（第23页）

1. (1) $ 3,5i + j + 7k $; (2) $ -18,10i + 2j + 14k $; (3) $ \cos(\widehat{a,b}) = \frac{3}{2\sqrt{21}} $.

1. $ -\frac{3}{2} $

1. $ \pm\frac{1}{\sqrt{17}}(3i-2j-2k) $.

4.5 880 J.

1. $ \left|F_{1}\right|x_{1}\sin\theta_{1}=\left|F_{2}\right|x_{2}\sin\theta_{2}. $

6.2.

1. $ \lambda = 2\mu $.

1. 略.

1. (1) $ -8j - 24k $; (2) $ -j - k $;

1. $ \frac{1}{2}\sqrt{19} $

$ ^{*} $11—12. 略.

习题8-3（第29页）

1. $ 3x - 7y + 5z - 4 = 0 $.

1. $ 2x + 9y - 6z - 121 = 0 $.

1. x - 3y - 2z = 0.

4.（1）yOz面；

（2）平行于 xOz 面的平面；

（3）平行于z轴的平面；（4）通过z轴的平面；

（5）平行于 x 轴的平面；（6）通过 y 轴的平面；

（7）通过原点的平面.

1. $ \frac{1}{3}, \frac{2}{3}, \frac{2}{3} $.

1. $ x + y - 3z - 4 = 0 $. 公众号：考研讲课

1. (1, -1, 3).

1. (1) $ y + 5 = 0 $; (2) $ x + 3y = 0 $; (3) 9y - z - 2 = 0.

1. 1.

习题8-4（第36页）

1. $ \frac{x-4}{2}=\frac{y+1}{1}=\frac{z-3}{5} $.

1. $ \frac{x-3}{-4}=\frac{y+2}{2}=\frac{z-1}{1} $

1. $ \frac{x-1}{-2}=\frac{y-1}{1}=\frac{z-1}{3} $, $ \{\begin{aligned}x&=1-2t,\\ y&=1+t,\\ z&=1+3t\end{aligned}. $ (t为任意常数).

1. $ 16x - 14y - 11z - 65 = 0 $.

1. $ \cos \varphi = 0 $.

1. 略.

1. $ \frac{x}{-2} = \frac{y-2}{3} = \frac{z-4}{1} $.

1. 8x - 9y - 22z - 59 = 0.

1. $ \varphi = 0 $.

10.（1）平行；（2）垂直；（3）直线在平面上.

1. $ x - y + z = 0 $.

1. $ \left(-\frac{5}{3},\frac{2}{3},\frac{2}{3}\right) $

$$ \frac{3\sqrt{2}}{2}. $$

1. 略.

1. $ \begin{cases} 17x + 31y - 37z - 117 = 0, \\ 4x - y + z - 1 = 0. \end{cases} $

1. 略.

习题8-5（第44页）

1. $ x^{2}+y^{2}+z^{2}-4x-2y+4z=0 $，球心为 $ (2,1,-2) $，R=3.

1. $ x^{2} + y^{2} + z^{2} - 2x - 6y + 4z = 0. $

1. 以点(1, -2, -1)为球心，半径为 $ \sqrt{6} $的球面开

1. $ \left(x+\frac{2}{3}\right)^{2}+\left(y+1\right)^{2}+\left(z+\frac{4}{3}\right)^{2}=\frac{116}{9} $，它表示一球面，球心为 $ \left(-\frac{2}{3},-1,-\frac{4}{3}\right) $，半径为 $ \frac{2}{3}\sqrt{29} $.

1. $ y^{2} + z^{2} = 5x $

1. $ x^{2} + y^{2} + z^{2} = 9 $

1. 绕 x 轴: $ 4x^{2}-9(y^{2}+z^{2})=36 $, 绕 y 轴: $ 4(x^{2}+z^{2})-9y^{2}=36 $.

8 —9. 略

1. (1) xOy 平面上的椭圆 $ \frac{x^{2}}{4} + \frac{y^{2}}{9} = 1 $ 绕 x 轴旋转一周；

(2) xOy 平面上的双曲线 $ x^{2}-\frac{y^{2}}{4}=1 $ 绕 y 轴旋转一周；

（3）xOy 平面上的双曲线 $ x^{2}-y^{2}=1 $ 绕 x 轴旋转一周；

（4）yOz 平面上的直线 z = y + a 绕 z 轴旋转一周.

注：本题各小题均有多个答案，以上给出的均是其中一个答案。11—12. 略.

习题8-6（第51页）

1—2. 略.

1. 母线平行于 x 轴的柱面方程为 $ 3y^{2}-z^{2}=16 $，

母线平行于 y 轴的柱面方程为 $ 3x^{2}+2z^{2}=16 $.

1. $ \{\begin{array}{l}2x^{2}-2x+y^{2}=8,\\z=0.\end{array}. $

1. (1) $ \begin{cases} x = \dfrac{3}{\sqrt{2}} \cos t, \\ y = \dfrac{3}{\sqrt{2}} \cos t, & (0 \leqslant t \leqslant 2\pi) \\ z = 3 \sin t \end{cases} $; (2) $ \begin{cases} x = 1 + \sqrt{3} \cos \theta, \\ y = \sqrt{3} \sin \theta, & (0 \leqslant \theta \leqslant 2\pi) \\ z = 0 \end{cases} $

1. $ \{\begin{array}{l}x^{2}+y^{2}=a^{2},\\z=0,\end{array}.\quad\{\begin{array}{l}y=a\sin\frac{z}{b},\\x=0,\end{array}.\quad\{\begin{array}{l}x=a\cos\frac{z}{b},\\y=0.\end{array}. $

1. $ x^{2} + y^{2} \leqslant ax; x^{2} + z^{2} \leqslant a^{2}, x \geqslant 0, z \geqslant 0. $

1. $ x^{2} + y^{2} \leqslant 4 $, $ x^{2} \leqslant z \leqslant 4 $, $ y^{2} \leqslant z \leqslant 4 $.

总习题八（第51页）

1. (1) $ M(x-x_0,y-y_0,z-z_0) $, $ \overrightarrow{OM}=(x,z,0) $ 共面；(3) 3；(4) 36.

2.（1）（A）;

1. (0,2,0).

4.

$$ \sqrt{30}. $$

1. $ \overrightarrow{AD} = \boldsymbol{c} + \frac{1}{2}\boldsymbol{a} $, $ \overrightarrow{BE} = \boldsymbol{a} + \frac{1}{2}\boldsymbol{b} $, $ \overrightarrow{CF} = \boldsymbol{b} + \frac{1}{2}\boldsymbol{c} $.

1. 略.

7.1.

$$ \operatorname{arccos}\frac{2}{\sqrt{7}}. $$

9.

$$ \frac{\pi}{3}. $$

$$ z=-4,\theta_{\min}=\frac{\pi}{4}. $$

1. 30.

1. (14,10,2).

1. c = 5a + b.

1. $ 4(z-1)=(x-1)^2+(y+1)^2 $.

$$ \{\begin{aligned}x=0,\\ z=2y^{2},\end{aligned}.z $$

(2）$\{\begin{aligned}x&=0,\\ \frac{y^{2}}{9}+\frac{z^{2}}{36}&=1,\end{aligned}.$ y轴；

(3) $ \{\begin{aligned}x=0,\\ z=\sqrt{3}y,\end{aligned}. $z轴；

(4) $ \{\begin{aligned}&z=0,\\&x^{2}-\frac{y^{2}}{4}=1,\end{aligned}. $ x轴.

1. $ x + \sqrt{26}y + 3z - 3 = 0 $ 或 $ x - \sqrt{26}y + 3z - 3 = 0 $.

1. $ x + 2y + 1 = 0 $

18.

$$ \frac{x+1}{16}=\frac{y}{19}=\frac{z-4}{28}. $$

19.

$$ \left(0,0,\frac{1}{5}\right). $$

$$ z=0,x^{2}+y^{2}=x+y;x=0,2y^{2}+2yz+z^{2}-4y-3z+2=0; $$

$$ y=0,2x^{2}+2xz+z^{2}-4x-3z+2=0. $$

$$ z=0,(x-1)^{2}+y^{2}\leqslant1;x=0,\left(\frac{z^{2}}{2}-1\right)^{2}+y^{2}\leqslant1,z\geqslant0;y=0,x\leqslant z\leqslant\sqrt{2x}. $$

1. 略.

第 九 章

习题9-1（第64页）

1. (1) 开集，无界集，导集： $ \mathbb{R}^2 $，边界： $ \{(x,y)\mid x=0 \text{ 或 } y=0\} $；

（2）既非开集，又非闭集，有界集，导集： $ \{(x,y)\mid1\leqslant x^{2}+y^{2}\leqslant4\} $，边界： $ \{(x,y)\mid x^{2}+y^{2}=1\}\cup\{(x,y)\mid x^{2}+y^{2}=4\} $；

（3）开集，区域，无界集，导集： $ \{(x,y)\mid y\geqslant x^{2}\} $，边界： $ \{(x,y)\mid y=x^{2}\} $；

(4) 闭集，有界集，导集：集合本身，

边界： $ \{(x,y)\mid x^{2}+(y-1)^{2}=1\}\cup\{(x,y)\mid x^{2}+(y-2)^{2}=4\} $

1. $ t^{2}f(x,y) $.

1. 略.

1. $ (x + y)^{xy} + (xy)^{2x} $

1. (1) $ \{(x,y)\mid y^{2}-2x+1>0\} $;

(2) $ \{(x,y)\mid x+y>0,x-y>0\} $;

(3) $ \{(x,y) \mid x \geqslant 0, y \geqslant 0, x^{2} \geqslant y\} $;

(4) $ \{(x,y)\mid y-x>0,x\geqslant0,x^{2}+y^{2}<1\} $;

(5) $ \{ (x,y,z) \mid r^{2} < x^{2} + y^{2} + z^{2} \leqslant R^{2} \} $;

(6) $ \{(x,y,z)\mid x^{2}+y^{2}-z^{2}\geqslant0,x^{2}+y^{2}\neq0\} $

1. (1)1; (2) $ \ln 2 $; (3) $ -\frac{1}{4} $; (4)-2; (5)2; (6)0.

1. 略.

1. $ \{(x,y)\mid y^{2}-2x=0\} $

*9. 提示： $ |xy| \leqslant \frac{x^2 + y^2}{2} $.

$ ^{*} $10. 略.

习题9-2（第71页）

1. (1) $ \frac{\partial z}{\partial x} = 3x^{2}y - y^{3} $, $ \frac{\partial z}{\partial y} = x^{3} - 3xy^{2} $;

(2) $ \frac{\partial s}{\partial u} = \frac{1}{v} - \frac{v}{u^2} $, $ \frac{\partial s}{\partial v} = \frac{1}{u} - \frac{u}{v^2} $;

(3) $ \frac{\partial z}{\partial x} = \frac{1}{2x \sqrt{\ln(xy)}} $, $ \frac{\partial z}{\partial y} = \frac{1}{2y \sqrt{\ln(xy)}} $

(4)

$$ \frac{\partial z}{\partial x}=y\left[\cos\left(xy\right)-\sin\left(2xy\right)\right],\quad\frac{\partial z}{\partial y}=x\left[\cos\left(xy\right)-\sin\left(2xy\right)\right]; $$

(5)

$$ \frac{\partial z}{\partial x}=\frac{2}{y}\csc\frac{2x}{y},\frac{\partial z}{\partial y}=-\frac{2x}{y^{2}}\csc\frac{2x}{y}; $$

(6)

$$ \frac{\partial z}{\partial x}=y^{2}\left(1+xy\right)^{y-1},\quad\frac{\partial z}{\partial y}=\left(1+xy\right)^{y}\left[\ln\left(1+xy\right)+\frac{xy}{1+xy}\right]; $$

(7)

$$ \frac{\partial u}{\partial x}=\frac{y}{z}x^{\frac{y}{z}-1},\quad\frac{\partial u}{\partial y}=\frac{1}{z}x^{\frac{y}{z}}\cdot\ln x,\quad\frac{\partial u}{\partial z}=-\frac{y}{z^{2}}x^{\frac{y}{z}}\cdot\ln x; $$

(8)

$$ \frac{\partial u}{\partial x}=\frac{z(x-y)^{z-1}}{1+(x-y)^{2z}},\quad\frac{\partial u}{\partial y}=-\frac{z(x-y)^{z-1}}{1+(x-y)^{2z}}, $$

$$ \frac{\partial u}{\partial z}=\frac{\left(x-y\right)^{z}\ln\left(x-y\right)}{1+\left(x-y\right)^{2z}}. $$

2—3. 略.

1. $ f_{x}(x,1) = 1 $

1. $ \frac{\pi}{4} $.

6.

$$ \frac{\partial^{2}z}{\partial x^{2}}=12x^{2}-8y^{2},\frac{\partial^{2}z}{\partial y^{2}}=12y^{2}-8x^{2},\frac{\partial^{2}z}{\partial x\partial y}=-16xy; $$

(2)

$$ \frac{\partial^{2}z}{\partial x^{2}}=\frac{2xy}{\left(x^{2}+y^{2}\right)^{2}},\frac{\partial^{2}z}{\partial y^{2}}=-\frac{2xy}{\left(x^{2}+y^{2}\right)^{2}},\frac{\partial^{2}z}{\partial x\partial y}=\frac{y^{2}-x^{2}}{\left(x^{2}+y^{2}\right)^{2}} $$

(3)

$$ \frac{\partial^{2}z}{\partial x^{2}}=y^{x}\cdot\ln^{2}y,\frac{\partial^{2}z}{\partial y^{2}}=x(x-1)y^{x-2},\frac{\partial^{2}z}{\partial x\partial y}=y^{x-1}(1+x\ln y). $$

$$ 7.f_{xx}(0,0,1)=2,f_{xz}(1,0,2)=2,f_{yz}(0,-1,0)=0,f_{zzx}(2,0,1)=0. $$

1. $ \frac{\partial^{3}z}{\partial x^{2}\partial y}=0 $, $ \frac{\partial^{3}z}{\partial x\partial y^{2}}=-\frac{1}{y^{2}} $.

1. 略.

习题9-3（第77页）

$$ \left(y+\frac{1}{y}\right)\mathrm{d}x+x\left(1-\frac{1}{y^{2}}\right)\mathrm{d}y; $$

$$ -\frac{1}{x}\mathrm{e}^{\frac{x}{x}}\left(\frac{y}{x}\mathrm{d}x-\mathrm{d}y\right) $$

(3)

$$ -\frac{x}{\left(x^{2}+y^{2}\right)^{3/2}}(y\mathrm{d}x-x\mathrm{d}y) $$

(4) $ yzx^{yz-1}dx + zx^{yz} \cdot \ln xdy + yx^{yz} \cdot \ln xdz $.

1. $ \frac{1}{3}dx + \frac{2}{3}dy $

1. $ \Delta z = -0.119 $, dz = -0.125.

1. 0.25e.

1. (A).

$ ^{*}6 $. 2.95.

$ ^{7} $. 2.039.

$ ^{*} $8. -5 cm.

$ ^{*} $9.55.3 cm $ ^{3} $.

$ 10.0.124 \, cm. $

$$ 11.2128\mathrm{~m}^{2},27.6\mathrm{~m}^{2},1.30\% $$

$ ^{*}12\text{—}^{*}13. $ 略.

习题9-4（第84页）

1. $ \frac{\partial z}{\partial x}=4x $, $ \frac{\partial z}{\partial y}=4y $.

$$ 2.\ \frac{\partial z}{\partial x}=\frac{2x}{y^{2}}\ln\left(3x-2y\right)+\frac{3x^{2}}{\left(3x-2y\right)y^{2}},\ \frac{\partial z}{\partial y}=-\frac{2x^{2}}{y^{3}}\ln\left(3x-2y\right)-\frac{2x^{2}}{\left(3x-2y\right)y^{2}}. $$

1. $ \mathrm{e}^{\sin t-2t^{3}}(\cos t-6t^{2}) $.

1. $ \frac{3(1-4t^{2})}{\sqrt{1-(3t-4t^{3})^{2}}} $

1. $ \frac{e^{x}(1+x)}{1+x^{2}e^{2x}} $

1. $ e^{ax}\sin x $

1. 略.

1. (1) $ \frac{\partial u}{\partial x} = 2xf'_{1} + ye^{xy}f'_{2} $, $ \frac{\partial u}{\partial y} = -2yf'_{1} + xe^{xy}f'_{2} $;

(2) $ \frac{\partial u}{\partial x} = \frac{1}{y} f'_{1} $, $ \frac{\partial u}{\partial y} = -\frac{x}{y^{2}} f'_{1} + \frac{1}{z} f'_{2} $, $ \frac{\partial u}{\partial z} = -\frac{y}{z^{2}} f'_{2} $;

(3) $ \frac{\partial u}{\partial x} = f'_1 + yf'_2 + yzf'_3 $, $ \frac{\partial u}{\partial y} = xf'_2 + xzf'_3 $, $ \frac{\partial u}{\partial z} = xyf'_3 $.

9 —10. 略

$$ 11.\frac{\partial^{2}z}{\partial x^{2}}=2f^{\prime}+4x^{2}f^{\prime\prime},\frac{\partial^{2}z}{\partial x\partial y}=4xyf^{\prime\prime},\frac{\partial^{2}z}{\partial y^{2}}=2f^{\prime}+4y^{2}f^{\prime\prime}. $$

$ ^{*} $12. (1) $ \frac{\partial^2 z}{\partial x^2} = y^2 f_{11}'' $, $ \frac{\partial^2 z}{\partial x \partial y} = f_1' + y(xf_{11}'' + f_{12}'') $, $ \frac{\partial^2 z}{\partial y^2} = x^2 f_{11}'' + 2xf_{12}'' + f_{22}'' $;

(2)

$$ \frac{\partial^{2}z}{\partial x^{2}}=f_{11}^{\prime\prime}+\frac{2}{y}f_{12}^{\prime\prime}+\frac{1}{y^{2}}f_{22}^{\prime\prime},\quad\frac{\partial^{2}z}{\partial x\partial y}=-\frac{x}{y^{2}}\left(f_{12}^{\prime\prime}+\frac{1}{y}f_{22}^{\prime\prime}\right)-\frac{1}{y^{2}}f_{2}^{\prime}, $$

$$ \frac{\partial^{2}z}{\partial y^{2}}=\frac{2x}{y^{3}}f_{2}^{\prime}+\frac{x^{2}}{y^{4}}f_{22}^{\prime \prime}; 众众号 \vdots 考研讲课 $$

(3)

$$ \frac{\partial^{2}z}{\partial x^{2}}=2y f_{2}^{\prime}+y^{4}f_{11}^{\prime\prime}+4x y^{3}f_{12}^{\prime\prime}+4x^{2}y^{2}f_{22}^{\prime\prime}, $$

$$ \frac{\partial^{2}z}{\partial x\partial y}=2y f_{1}^{\prime}+2x f_{2}^{\prime}+2x y^{3}f_{11}^{\prime\prime}+2x^{3}y f_{22}^{\prime\prime}+5x^{2}y^{2}f_{12}^{\prime\prime} $$

$$ \frac{\partial^{2}z}{\partial y^{2}}=2x f_{1}^{\prime}+4x^{2}y^{2}f_{11}^{\prime\prime}+4x^{3}y f_{12}^{\prime\prime}+x^{4}f_{22}^{\prime\prime}; $$

(4)

$$ \frac{\partial^{2}z}{\partial x^{2}}=\mathrm{e}^{x+y}f_{3}^{\prime}-\sin x f_{1}^{\prime}+\cos^{2}x f_{11}^{\prime\prime}+2\mathrm{e}^{x+y}\cos x f_{13}^{\prime\prime}+\mathrm{e}^{2(x+y)}f_{33}^{\prime\prime} $$

$$ \frac{\partial^{2}z}{\partial x\partial y}=\mathrm{e}^{x+y}f_{3}^{\prime}-\cos x\sin y f_{12}^{\prime\prime}+\mathrm{e}^{x+y}\cos x f_{13}^{\prime\prime}-\mathrm{e}^{x+y}\sin y f_{32}^{\prime\prime}+\mathrm{e}^{2(x+y)}f_{33}^{\prime\prime}, $$

$$ \frac{\partial^{2}z}{\partial y^{2}}=\mathrm{e}^{x+y}f_{3}^{\prime}-\cos y f_{2}^{\prime}+\sin^{2}y f_{22}^{\prime\prime}-2\mathrm{e}^{x+y}\sin y f_{23}^{\prime\prime}+\mathrm{e}^{2(x+y)}f_{33}^{\prime\prime}. $$

$ ^{*} $13. 略.

习题9-5（第91页）

1.

$$ \frac{y^{2}-\mathrm{e}^{x}}{\cos y-2xy} $$

2.

$$ \frac{x+y}{x-y}. $$

3.

$$ \frac{\partial z}{\partial x}=\frac{yz-\sqrt{xyz}}{\sqrt{xyz}-xy},\ \frac{\partial z}{\partial y}=\frac{xz-2\sqrt{xyz}}{\sqrt{xyz}-xy}. $$

1. $ \frac{\partial z}{\partial x} = \frac{z}{x + z} $, $ \frac{\partial z}{\partial y} = \frac{z^{2}}{y(x + z)} $.

5—7. 略.

$$ \star8.\frac{2y^{2}z\mathrm{e}^{z}-2x y^{3}z-y^{2}z^{2}\mathrm{e}^{z}}{\left(\mathrm{e}^{z}-x y\right)^{3}}. $$

$$ \cdot9.\frac{z\left(z^{4}-2xyz^{2}-x^{2}y^{2}\right)}{\left(z^{2}-xy\right)^{3}}. $$

10.

$$ \frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{x(6z+1)}{2y(3z+1)},\quad\frac{\mathrm{d}z}{\mathrm{d}x}=\frac{x}{3z+1}; $$

(2)

$$ \frac{\mathrm{d}x}{\mathrm{d}z}=\frac{y-z}{x-y},\frac{\mathrm{d}y}{\mathrm{d}z}=\frac{z-x}{x-y}; $$

(3)

$$ \frac{\partial u}{\partial x}=\frac{-u f_{_{1}}^{\prime}\left(2y v g_{_{2}}^{\prime}-1\right)\;-f_{_{2}}^{\prime}\cdot g_{_{1}}^{\prime}}{\left(x f_{_{1}}^{\prime}-1\right)\left(2y v g_{_{2}}^{\prime}-1\right)\;-f_{_{2}}^{\prime}\cdot g_{_{1}}^{\prime}}, $$

$$ \frac{\partial v}{\partial x}=\frac{g_{1}^{\prime}\left(xf_{1}^{\prime}+uf_{1}^{\prime}-1\right)}{\left(xf_{1}^{\prime}-1\right)\left(2yvg_{2}^{\prime}-1\right)\cdots\left(f_{2}^{\prime}\right)\cdots g_{1}^{\prime}} $$

(4)

$$ \frac{\partial u}{\partial x}=\frac{\sin v}{e^{u}\left(\sin v-\cos v\right)+1},\quad\frac{\partial u}{\partial y}=\frac{-\cos v}{e^{u}\left(\sin v-\cos v\right)+1}, $$

$$ \frac{\partial v}{\partial x}=\frac{\cos v-\mathrm{e}^{u}}{u\left[\mathrm{e}^{u}\left(\sin v-\cos v\right)+1\right]},\frac{\partial v}{\partial y}=\frac{\sin v+\mathrm{e}^{u}}{u\left[\mathrm{e}^{u}\left(\sin v-\cos v\right)+1\right]}. $$

1. 略.

习题9-6（第102页）

1. 略.

1. (1) $ v_{0} = i + 2j + 2k, a_{0} = 2j $, $ \left|v(t)\right| = \sqrt{5 + 4t^{2}} $;

$$ v_{0}=-2i+4k,a_{0}=-3j,|v(t)|=\sqrt{20+5\cos^{2}t}; $$

$$ \boldsymbol{v}_{0}=\boldsymbol{i}+2\boldsymbol{j}+\boldsymbol{k},\boldsymbol{a}_{0}=-\frac{1}{2}\boldsymbol{i}+2\boldsymbol{j}+\boldsymbol{k},\left|\boldsymbol{v}(t)\right|=\sqrt{5t^{2}+\frac{4}{(t+1)^{2}}}. $$

1. 切线方程： $ \frac{x - \left(\frac{\pi}{2} - 1\right)}{1} = \frac{y - 1}{1} = \frac{z - 2\sqrt{2}}{\sqrt{2}} $，

法平面方程： $ x + y + \sqrt{2}z = \frac{\pi}{2} + 4 $。

1. 切线方程： $ \frac{x-\frac{1}{2}}{1}=\frac{y-2}{-4}=\frac{z-1}{8} $，法平面方程： $ 2x-8y+16z-1=0 $。

1. 切线方程： $ \frac{x - x_0}{1} = \frac{y - y_0}{\frac{m}{y_0}} = \frac{z - z_0}{-\frac{1}{2z_0}} $,

法平面方程： $ (x-x_{0})+\frac{m}{y_{0}}(y-y_{0})-\frac{1}{2z_{0}}(z-z_{0})=0 $.

1. 切线方程： $ \frac{x-1}{16}=\frac{y-1}{9}=\frac{z-1}{-1} $，法平面方程： $ 16x+9y-z-24=0 $。

1. $ P_{1}(-1,1,-1) $ 及 $ P_{2}\left(-\frac{1}{3},\frac{1}{9},-\frac{1}{27}\right) $

1. 切平面方程： $ x+2y-4=0 $，法线方程： $ \{\begin{aligned}\frac{x-2}{1}&=\frac{y-1}{2}\\ z&=0.\end{aligned}. $

1. 切平面方程： $ ax_{0}x + by_{0}y + cz_{0}z = 1 $，法线方程： $ \frac{x - x_{0}}{ax_{0}} = \frac{y - y_{0}}{by_{0}} = \frac{z - z_{0}}{cz_{0}} $.

1. 切平面方程： $ x - y + 2z = \pm \sqrt{\frac{11}{2}} $.

1. $ \cos \gamma = \frac{3}{\sqrt{22}} $.

12—13. 略

习题9-7（第111页）

1. $ 1 + 2\sqrt{3} $.

1. $ \frac{\sqrt{2}}{3} $

1. $ \frac{1}{ab}\sqrt{2(a^{2}+b^{2})} $.

1. 5.

1. $ \frac{98}{13} $

1. $ \frac{6}{7}\sqrt{14} $.

1. $ x_{0} + y_{0} + z_{0} $

1. $ \mathrm{grad}f(0,0,0)=3i-2j-6k $, $ \mathrm{grad}f(1,1,1)=6i+3j $.

1. 略.

1. 增加最快的方向为 $ n = \frac{1}{\sqrt{21}}(2i - 4j + k) $，方向导数为 $ \sqrt{21} $;

减少最快的方向为 $ -n = \frac{1}{\sqrt{21}}(-2i + 4j - k) $，方向导数为 $ -\sqrt{21} $.

习题9-8（第121页）

1. (A).

1. 极大值: $ f(2, -2) = 8 $.

1. 极大值: $ f(3,2)=36 $.

1. 极小值: $ f\left(\frac{1}{2}, -1\right) = -\frac{e}{2} $.

1. 极大值: $ z\left(\frac{1}{2}, \frac{1}{2}\right) = \frac{1}{4} $.

1. 当两直角边都是 $ \frac{l}{\sqrt{2}} $时，可得最大的周长.

1. 当长、宽都是 $ \sqrt[3]{2k} $，而高为 $ \frac{1}{2}\sqrt[3]{2k} $ 时，水池的表面积最小.

1. $ \left(\frac{8}{5},\frac{16}{5}\right) $.

1. 当矩形的边长分别为 $ \frac{2p}{3} $及 $ \frac{p}{3} $时，绕短边旋转所得圆柱体的体积最大.

1. 当长、宽、高都是 $ \frac{2a}{\sqrt{3}} $时，可得最大的体积.

1. 最大值为 $ \sqrt{9 + 5\sqrt{3}} $，最小值为 $ \sqrt{9 - 5\sqrt{3}} $.

1. 最热点在 $ \left(-\frac{1}{2}, \pm\frac{\sqrt{3}}{2}\right) $，最冷点在 $ \left(\frac{1}{2}, 0\right) $

1. 最热点在 $ \left(\pm\frac{4}{3},-\frac{4}{3},-\frac{4}{3}\right) $

习题9-9（第127页）

$$ f(x,y)=5+2(x-1)^{2}-(x-1)(y+2)-(y+2)^{2}. $$

$$ \begin{aligned}&\therefore\quad\mathrm{e}^{x}\ln\left(1+y\right)=y+\frac{1}{2!}(2xy-y^{2})+\frac{1}{3!}(3x^{2}y-3xy^{2}+2y^{3})+R_{3}, 其中 R_{3}=\\&\quad\frac{\mathrm{e}^{\theta x}}{24}\Big[x^{4}\ln\left(1+\theta y\right)+\frac{4x^{3}y}{1+\theta y}-\frac{6x^{2}y^{2}}{\left(1+\theta y\right)^{2}}+\frac{8xy^{3}}{\left(1+\theta y\right)^{3}}-\frac{6y^{4}}{\left(1+\theta y\right)^{4}}\Big](0<\theta<1).\end{aligned} $$

1. $ \sin x \sin y = \frac{1}{2} + \frac{1}{2} \left( x - \frac{\pi}{4} \right) + \frac{1}{2} \left( y - \frac{\pi}{4} \right) - $

$$ \frac{1}{4}\left[\left(x-\frac{\pi}{4}\right)^{2}-2\left(x-\frac{\pi}{4}\right)\left(y-\frac{\pi}{4}\right)+\left(y-\frac{\pi}{4}\right)^{2}\right]+R_{2} $$

其中 $ R_{2} = -\frac{1}{6}\left[\cos \xi \sin \eta \left(x - \frac{\pi}{4}\right)^{3} + 3 \sin \xi \cos \eta \left(x - \frac{\pi}{4}\right)^{2} \left(y - \frac{\pi}{4}\right) + \right] $

$$ 3\cos\xi\sin\eta\left(x-\frac{\pi}{4}\right)\left(y-\frac{\pi}{4}\right)^{2}+\sin\xi\cos\eta\left(y-\frac{\pi}{4}\right)^{3}, $$

且 $ \xi = \frac{\pi}{4} + \theta \left( x - \frac{\pi}{4} \right), \eta = \frac{\pi}{4} + \theta \left( y - \frac{\pi}{4} \right) \quad (0 < \theta < 1) $.

$$ x^{y}=1+\left(x-1\right)+\left(x-1\right)\left(y-1\right)+\frac{1}{2}\left(x-1\right)^{2}\left(y-1\right)+R_{3},\\1.1^{1.02}\approx1.1021. $$

$$ \mathrm{e}^{x+y}=1+(x+y)+\frac{1}{2!}(x^{2}+2xy+y^{2})+\cdots+\frac{1}{n!}(x^{n}+C_{n}^{1}x^{n-1}y+\cdots+y^{n})+R_{n}, $$

其中 $ R_{n}=\frac{e^{\theta(x+y)}}{(n+1)!}(x^{n+1}+C_{n+1}^{1}x^{n}y+\cdots+C_{n}^{n+1}y^{n+1}) $， $ 0<\theta<1 $。

$ ^{*} $习题9-10（第132页）

1. $ \theta = 2.234p + 95.33 $

2.

$$ \{\begin{array}{l}a\displaystyle\sum_{i=1}^{n}x_{i}^{4}+b\displaystyle\sum_{i=1}^{n}x_{i}^{3}+c\displaystyle\sum_{i=1}^{n}x_{i}^{2}=\sum_{i=1}^{n}x_{i}^{2}y_{i},\\a\displaystyle\sum_{i=1}^{n}x_{i}^{3}+b\displaystyle\sum_{i=1}^{n}x_{i}^{2}+c\displaystyle\sum_{i=1}^{n}x_{i}=\sum_{i=1}^{n}x_{i}y_{i},\\a\displaystyle\sum_{i=1}^{n}x_{i}^{2}+b\displaystyle\sum_{i=1}^{n}x_{i}+nc=\sum_{i=1}^{n}y_{i}.\end{array}. $$

总习题九（第132页）

1.（1）充分，必要；（2）必要，充分；（3）充分；（4）充分.

1. (C).

3.

$$ \{(x,y)\mid0

$ ^{*}4. $ 略.

5.

$$ f_{x}(x,y)=\{\begin{array}{ll}\displaystyle\frac{2x y^{3}}{(x^{2}+y^{2})^{2}},&\quad x^{2}+y^{2}\neq0,\\0,&\quad x^{2}+y^{2}=0;\end{array}. $$

$$ f_{y}(x,y)=\{\begin{array}{cc}\frac{x^{2}(x^{2}-y^{2})}{(x^{2}+y^{2})^{2}},&\quad x^{2}+y^{2}\neq0,\\0,&\quad x^{2}+y^{2}=0.\end{array}. $$

6.(1)

$$ \begin{aligned}\frac{\partial z}{\partial x}&=\frac{1}{x+y^{2}},\frac{\partial z}{\partial y}=\frac{2y}{x+y^{2}},\frac{\partial^{2}z}{\partial x^{2}}=-\frac{1}{\left(x+y^{2}\right)^{2}}\\\frac{\partial^{2}z}{\partial x\partial y}&=-\frac{2y}{\left(x+y^{2}\right)^{2}},\frac{\partial^{2}z}{\partial y^{2}}=\frac{2\left(x-y^{2}\right)}{\left(x+y^{2}\right)^{2}};\end{aligned} $$

(2)

$$ \frac{\partial z}{\partial x}=yx^{y-1},\quad\frac{\partial z}{\partial y}=x^{y}\ln x,\quad\frac{\partial^{2}z}{\partial x^{2}}=y(y-1)x^{y-2}, $$

$$ \frac{\partial^{2}z}{\partial x\partial y}=x^{y-1}\left(1+y\ln x\right),\quad\frac{\partial^{2}z}{\partial y^{2}}=x^{y}\left(\ln x\right)^{2}. $$

1. $ \Delta z = 0.02 $, $ \mathrm{d}z = 0.03 $.

1. 略.

1. $ \frac{\mathrm{d}u}{\mathrm{d}t} = y x^{y-1} \varphi'(t) + x^y \ln x \psi'(t) $.

公众号：考研讲课

1. $ \frac{\partial z}{\partial \xi} = -\frac{\partial z}{\partial v} + \frac{\partial z}{\partial w} $, $ \frac{\partial z}{\partial \eta} = \frac{\partial z}{\partial u} - \frac{\partial z}{\partial w} $, $ \frac{\partial z}{\partial \zeta} = -\frac{\partial z}{\partial u} + \frac{\partial z}{\partial v} $.

1. $ \frac{\partial^{2}z}{\partial x\partial y}=xe^{2y}f_{uu}^{\prime\prime}+e^{y}f_{uy}^{\prime\prime}+xe^{y}f_{xu}^{\prime\prime}+f_{xy}^{\prime\prime}+e^{y}f_{u}^{\prime} $.

$$ \frac{\partial z}{\partial x}=\left(v\cos v-u\sin v\right)\mathrm{e}^{-u},\quad\frac{\partial z}{\partial y}=\left(u\cos v+v\sin v\right)\mathrm{e}^{-u}. $$

1. 切线方程 $ \{\begin{array}{l}x=a,\\ by-az=0\end{array}. $；法平面方程 $ ay+bz=0 $.

1. $ (-3, -1, 3) $, $ \frac{x + 3}{1} = \frac{y + 1}{3} = \frac{z - 3}{1} $.

1. $ \frac{\partial f}{\partial l} = \cos \theta + \sin \theta $, (1) $ \theta = \frac{\pi}{4} $, (2) $ \theta = \frac{5\pi}{4} $, (3) $ \theta = \frac{3\pi}{4} $及 $ \frac{7\pi}{4} $.

1. $ \frac{\partial u}{\partial n} = \frac{2}{\sqrt{\frac{x_0^2}{a^4} + \frac{y_0^2}{b^4} + \frac{z_0^2}{c^4}}} $,

1. $ \left(\frac{4}{5},\frac{3}{5},\frac{35}{12}\right) $.

1. 切点 $ \left(\frac{a}{\sqrt{3}},\frac{b}{\sqrt{3}},\frac{c}{\sqrt{3}}\right) $， $ V_{\min}=\frac{\sqrt{3}}{2}abc $.

1. 当 $ p_{1}=80, p_{2}=120 $ 时，总利润最大，最大总利润为605.

1. (1) $ g(x_{0}, y_{0}) = \sqrt{5x_{0}^{2} + 5y_{0}^{2} - 8x_{0}y_{0}} $;

（2）攀岩的起点可取为 $ M_{1}(5,-5) $或 $ M_{2}(-5,5) $

第 1 十 章

习题10-1（第139页）

1. $ \iint_{D}\mu(x,y)d\sigma. $

1. $ I_{1}=4I_{2} $

1. 略.

$$ D=\{(x,y)\mid2x^{2}+y^{2}\leqslant1\} $$

1. (1) $ \iint_{D}(x+y)^{2}\mathrm{d}\sigma\geqslant\iint_{D}(x+y)^{3}\mathrm{d}\sigma;\quad(2)\iint_{D}(x+y)^{3}\mathrm{d}\sigma\geqslant\iint_{D}(x+y)^{2}\mathrm{d}\sigma; $

(3) $ \iint_{D} \ln (x + y) \, \mathrm{d}\sigma \geqslant \iint_{D} [\ln (x + y)]^2 \, \mathrm{d}\sigma $.

(4) $ \iint_{D}\left[\ln\left(x+y\right)\right]^{2}\mathrm{d}\sigma\geqslant\iint_{D}\ln(x+y)\mathrm{d}\sigma. $

1. (1) $ 0 \leqslant I \leqslant 2 $; (2) $ 0 \leqslant I \leqslant \pi^{2} $; (3) $ 2 \leqslant I \leqslant 8 $; (4) $ 36\pi \leqslant I \leqslant 100\pi $.

习题10-2（第156页）

1. (1) $ \frac{8}{3} $; (2) $ \frac{20}{3} $; (3) 1; (4) $ -\frac{3\pi}{2} $

1. (1) $ \frac{6}{55} $; (2) $ \frac{64}{15} $; (3) $ e - e^{-1} $; (4) $ \frac{13}{6} $.

1. 略.

1. (1) $ \int_{0}^{4} \mathrm{d}x \int_{x}^{2\sqrt{x}} f(x,y) \, \mathrm{d}y $ 或 $ \int_{0}^{4} \mathrm{d}y \int_{\frac{y}{4}}^{y} f(x,y) \, \mathrm{d}x $;

(2) $ \int_{-r}^{r} \mathrm{d}x \int_{0}^{\sqrt{r^{2}-x^{2}}} f(x,y) \, \mathrm{d}y $ 或 $ \int_{0}^{r} \mathrm{d}y \int_{-\sqrt{r^{2}-y^{2}}}^{\sqrt{r^{2}-y^{2}}} f(x,y) \, \mathrm{d}x $;

(3) $ \int_{1}^{2} \mathrm{d}x \int_{\frac{1}{x}}^{x} f(x,y) \, \mathrm{d}y $ 或 $ \int_{\frac{1}{2}}^{1} \mathrm{d}y \int_{\frac{1}{y}}^{2} f(x,y) \, \mathrm{d}x + \int_{1}^{2} \mathrm{d}y \int_{y}^{2} f(x,y) \, \mathrm{d}x $;

(4) $ \int_{-1}^{1} \mathrm{d}x \int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} f(x,y) \, \mathrm{d}y + \int_{-1}^{1} \mathrm{d}x \int_{-\sqrt{4-x^{2}}}^{-\sqrt{1-x^{2}}} f(x,y) \, \mathrm{d}y + \int_{-2}^{-1} \mathrm{d}x \int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} f(x,y) \, \mathrm{d}y + \int_{1}^{2} \mathrm{d}x \int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} f(x,y) \, \mathrm{d}y $

或 $ \int_{1}^{2} \mathrm{d}y \int_{1}^{-\sqrt{4-y^{2}}} f(x,y) \, \mathrm{d}x + \int_{-2}^{-1} \mathrm{d}y \int_{-\sqrt{4-y^{2}}}^{\sqrt{4-y^{2}}} f(x,y) \, \mathrm{d}x + \int_{-1}^{1} \mathrm{d}y \int_{-\sqrt{4-y^{2}}}^{-\sqrt{1-y^{2}}} f(x,y) \, \mathrm{d}x + \int_{-1}^{1} \mathrm{d}y \int_{-\sqrt{4-y^{2}}}^{-\sqrt{1-y^{2}}} f(x,y) \, \mathrm{d}x + \int_{-1}^{1} \mathrm{d}y \int_{-\sqrt{4-y^{2}}}^{\sqrt{4-y^{2}}} f(x,y) \, \mathrm{d}x $.

1. 略.

1. (1) $\int_{0}^{1} \mathrm{d}x \int_{x}^{1} f(x,y) \mathrm{d}y$; (2) $\int_{0}^{4} \mathrm{d}x \int_{\frac{x}{2}}^{\sqrt{x}} f(x,y) \mathrm{d}y$;

(3) $\int_{-1}^{1} \mathrm{d}x \int_{0}^{\sqrt{1-x^{2}}} f(x,y) \mathrm{d}y$; (4) $\int_{0}^{1} \mathrm{d}y \int_{2-y}^{1+\sqrt{1-y^{2}}} f(x,y) \mathrm{d}x$;

(5) $\int_{0}^{1} \mathrm{d}y \int_{x^{2}}^{e} f(x,y) \mathrm{d}x$;

(6) $\int_{-1}^{0} \mathrm{d}y \int_{-2\arcsin y}^{\pi} f(x,y) \mathrm{d}x + \int_{0}^{1} \mathrm{d}y \int_{\arcsin y}^{\pi-\arcsin y} f(x,y) \mathrm{d}x$.

1. $ \frac{4}{3} $

1. $ \frac{7}{2} $

1. $ \frac{17}{6} $

1. $ 6\pi $.

1. (1) $ \int_{0}^{2\pi} \mathrm{d}\theta \int_{0}^{a} f(\rho\cos\theta, \rho\sin\theta) \rho\mathrm{d}\rho $;

(2) $ \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \mathrm{d}\theta \int_{0}^{2\cos\theta} f(\rho\cos\theta, \rho\sin\theta) \rho\mathrm{d}\rho; $

(3) $ \int_{0}^{2\pi} \mathrm{d}\theta \int_{a}^{\delta} f(\rho\cos\theta, \rho\sin\theta) \rho\mathrm{d}\rho; $

(4) $ \int_{0}^{\frac{\pi}{2}} \mathrm{d}\theta \int_{0}^{\left(\cos\theta + \sin\theta\right)^{-1}} f(\rho\cos\theta, \rho\sin\theta) \rho \mathrm{d}\rho. $

1. (1) $ \int_{0}^{\frac{\pi}{4}} \mathrm{d}\theta \int_{0}^{\sec\theta} f(\rho\cos\theta,\rho\sin\theta)\rho\mathrm{d}\rho + \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \mathrm{d}\theta \int_{0}^{\csc\theta} f(\rho\cos\theta,\rho\sin\theta)\rho\mathrm{d}\rho $;

(2) $ \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} \mathrm{d}\theta \int_{0}^{2\sec\theta} f(\rho)\rho\mathrm{d}\rho; $ (3) $ \int_{0}^{\frac{\pi}{2}} \mathrm{d}\theta \int_{(\cos\theta+\sin\theta)^{-1}}^{1} f(\rho\cos\theta,\rho\sin\theta)\rho\mathrm{d}\rho; $

(4) $ \int_{0}^{\frac{\pi}{4}} \mathrm{d}\theta \int_{\sec\theta\tan\theta}^{\sec\theta} f(\rho\cos\theta,\rho\sin\theta)\rho\mathrm{d}\rho. $

1. (1) $ \frac{3}{4}\pi a^{4} $; (2) $ \frac{1}{6}a^{3}\left[\sqrt{2}+\ln(1+\sqrt{2})\right] $; (3) $ \sqrt{2}-1 $; (4) $ \frac{1}{8}\pi a^{4} $.

1. (1) $ \pi(e^4 - 1) $; (2) $ \frac{\pi}{4}(2\ln 2 - 1) $; (3) $ \frac{3}{64}\pi^2 $.

1. (1) $ \frac{9}{4} $; (2) $ \frac{\pi}{8}(\pi - 2) $; (3) $ 14a^{4} $; (4) $ \frac{2}{3}\pi(b^{3} - a^{3}) $.

1. $ \frac{1}{40}\pi^{5} $

1. $ \frac{1}{3}R^{3}\arctan k $.

1. $ \frac{3}{32}\pi a^{4} $.

$ ^{*} $19. (1) $ \frac{\pi^{4}}{3} $; (2) $ \frac{7}{3}\ln 2 $; (3) $ \frac{e-1}{2} $;

(4) $ \frac{1}{2}\pi ab $. 提示: 作变换 $ x = a\rho\cos\theta $, $ y = b\rho\sin\theta $.

$ ^{*}20.\ (1)\ 2\ln 3 $; (2) $ \frac{1}{8} $.

1. 略.

*22.（1）略；（2）提示：作变换 $ x=\frac{au-bv}{\sqrt{a^{2}+b^{2}}} $， $ y=\frac{bu+av}{\sqrt{a^{2}+b^{2}}} $

习题10-3（第166页）

1. (1) $ \int_{0}^{1} \mathrm{d}x \int_{0}^{1-x} \mathrm{d}y \int_{0}^{xy} f(x,y,z) \, \mathrm{d}z; $ (2) $ \int_{-1}^{1} \mathrm{d}x \int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}} \mathrm{d}y \int_{x^{2}+y^{2}}^{1} f(x,y,z) \, \mathrm{d}z; $

(3) $ \int_{-1}^{1} \mathrm{d}x \int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}} \mathrm{d}y \int_{x^{2}+2y^{2}}^{2-x^{2}} f(x,y,z) \, \mathrm{d}z; $

(4) $ \int_{0}^{a} \mathrm{d}x \int_{0}^{b} \int_{-x^{2}}^{1-x^{2}/a^{2}} \mathrm{d}y \int_{0}^{xy/e} f(x,y,z) \, \mathrm{d}z. $

1. $ \frac{3}{2} $

1. 略.

1. $ \frac{1}{364} $

1. $ \frac{1}{2}\left(\ln2-\frac{5}{8}\right) $

1. $ \frac{1}{48} $.

1. 0.

1. $ \frac{\pi}{4}h^{2}R^{2} $

1. (1) $ \frac{7\pi}{12} $; (2) $ \frac{16}{3}\pi $.

$ 10.\ (1)\ \frac{4\pi}{5};\quad(2)\ \frac{7}{6}\pi a^{4}. $

1. (1) $ \frac{1}{8} $; * (2) $ \frac{\pi}{10} $; (3) $ 8\pi $; * (4) $ \frac{4\pi}{15}(A^{5} - a^{5}) $.

1. (1) $ \frac{32}{3}\pi $; $ *(2) $ $ \pi a^{3} $; (3) $ \frac{\pi}{6} $; (4) $ \frac{2}{3}\pi(5\sqrt{5}-4) $.

$ 13. \frac{2}{3}\pi a^3 $.

1. $ \frac{8\sqrt{2}-7}{6}\pi $

$$ ^{*}15.~k\pi R^{4}. $$

习题10-4（第177页）

1. $ 2a^{2}(\pi-2) $.

1. $ \sqrt{2}\pi $

1. $ 16R^{2} $

1. (1) $ \bar{x} = \frac{3}{5}x_0 $; $ \bar{y} = \frac{3}{8}y_0 $; (2) $ \bar{x} = 0 $, $ \bar{y} = \frac{4b}{3\pi} $; (3) $ \bar{x} = \frac{b^2 + ab + a^2}{2(a + b)} $, $ \bar{y} = 0 $.

1. $ \bar{x} = \frac{35}{48} $, $ \bar{y} = \frac{35}{54} $.

1. $ \bar{x} = \frac{2}{5}a $, $ \bar{y} = \frac{2}{5}a $.

1. (1) $ \left(0,0,\frac{3}{4}\right) $; $ *(2)\left(0,0,\frac{3(A^{4}-a^{4})}{8(A^{3}-a^{3})}\right) $; (3) $ \left(\frac{2}{5}a,\frac{2}{5}a,\frac{7}{30}a^{2}\right) $.

1. $ \left(0,0,\frac{5}{4}R\right) $.

1. (1) $ I_{y} = \frac{1}{4}\pi a^{3}b $; (2) $ I_{x} = \frac{72}{5} $, $ I_{y} = \frac{96}{7} $; (3) $ I_{x} = \frac{1}{3}ab^{3} $, $ I_{y} = \frac{1}{3}ba^{3} $.

1. $ \frac{1}{12}M h^{2} $, $ \frac{1}{12}M b^{2} $ (M = bh $ \mu $ 为矩形板的质量).

1. (1) $ \frac{8}{3}a^{4} $; (2) $ \bar{x} = \bar{y} = 0, \bar{z} = \frac{7}{15}a^{2} $; (3) $ \frac{112}{45}a^{6}\rho $.

1. $ \frac{1}{2}a^{2}M $ ( $ M=\pi a^{2}h\rho $ 为圆柱体的质量).

$$ \begin{aligned}{\mathbf{F}}&{{}=\left(2G\mu\left(\ln\frac{R_{2}+\sqrt{R_{2}^{2}+a^{2}}}{R_{1}+\sqrt{R_{1}^{2}+a^{2}}}-\frac{R_{2}}{\sqrt{R_{2}^{2}+a^{2}}}+\frac{R_{1}}{\sqrt{R_{1}^{2}+a^{2}}}\right)\right.,}\\ {}&{{}\quad\left.\vphantom{\frac{R_{2}}{\sqrt{R_{2}^{2}+a^{2}}}}\pi G a\mu\left(\frac{1}{\sqrt{R_{2}^{2}+a^{2}}}-\frac{1}{\sqrt{R_{1}^{2}+a^{2}}}\right)\right).}\\ \end{aligned} $$

1. $ F_{x}=F_{y}=0 $, $ F_{z}=-2\pi G\rho[\sqrt{(h-a)^{2}+R^{2}}-\sqrt{R^{2}+a^{2}}+h] $.

$ ^{*} $习题10-5（第184页）

1. (1) $ \frac{\pi}{4} $; (2) 1; (3) $ \frac{8}{3} $.

1. (1) $ \frac{1}{3}\cos x(\cos x - \sin x)(1 + 2\sin 2x) $; (2) $ \frac{2}{x}\ln(1 + x^2) $;

(3) $ \ln\sqrt{\frac{x^{2}+1}{x^{4}+1}}+3x^{2}\arctan x^{2}-2x\arctan x $; (4) $ 2xe^{-x^{5}}-e^{-x^{3}}-\int_{x}^{x^{2}}y^{2}e^{-xy^{2}}dy $.

1. $ 3f(x) + 2xf'(x) $

1. (1) $ \pi\arcsin a $; 公众号：考研讲课

(2) $ \pi\ln\frac{1+a}{2} $. 提示: 设 $ \varphi(\alpha)=\int_{0}^{\frac{\pi}{2}}\ln(\cos^{2}x+\alpha^{2}\sin^{2}x)\,\mathrm{d}x $, $ I=\varphi(a) $.

1. (1) $ \frac{\pi}{2}\ln(1+\sqrt{2}) $; 提示: 利用公式 $ \frac{\arctan x}{x} = \int_{0}^{1} \frac{dy}{1+x^2y^2} $.

(2) $ \arctan(1+b)-\arctan(1+a) $. 提示: 利用公式 $ \frac{x^{b}-x^{a}}{\ln x}=\int_{a}^{b}x^{y}dy $.

总习题十（第185页）

1. (1) $ \frac{1}{2}(1 - e^{-4}) $; (2) $ \frac{\pi}{4}R^{4}\left(\frac{1}{a^{2}} + \frac{1}{b^{2}}\right) $.

2.（1）（C）；（2）（A）；（3）（B）.

1. (1) $ \frac{3}{2} + \cos 1 + \sin 1 - \cos 2 - 2\sin 2 $; (2) $ \pi^2 - \frac{40}{9} $;

(3) $ \frac{1}{3}R^3\left(\pi - \frac{4}{3}\right) $; (4) $ \frac{\pi}{4}R^4 + 9\pi R^2 $.

1. (1) $ \int_{-2}^{0} \mathrm{d}x \int_{2x+4}^{4-x^2} f(x,y) \mathrm{d}y $; (2) $ \int_{0}^{2} \mathrm{d}x \int_{\frac{1}{2}x}^{3-x} f(x,y) \mathrm{d}y $;

(3) $ \int_{0}^{1} \mathrm{d}y \int_{0}^{y^{2}} f(x,y) \, \mathrm{d}x + \int_{1}^{2} \mathrm{d}y \int_{0}^{\sqrt{2y-y^{2}}} f(x,y) \, \mathrm{d}x. $

1. 略.

1. $ \int_{0}^{\frac{\pi}{4}}\mathrm{d}\theta\int_{0}^{\sec\theta\tan\theta}f(\rho\cos\theta,\rho\sin\theta)\rho\mathrm{d}\rho+\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\mathrm{d}\theta\int_{0}^{\csc\theta}f(\rho\cos\theta,\rho\sin\theta)\rho\mathrm{d}\rho+ $

$ \int_{\frac{3\pi}{4}}^{\pi}\mathrm{d}\theta\int_{0}^{\sec\theta\tan\theta}f(\rho\cos\theta,\rho\sin\theta)\rho\mathrm{d}\rho. $

1. $ f(x,y) = \sqrt{1 - x^{2} - y^{2}} + \frac{8}{9\pi} - \frac{2}{3} $.

1. $ \int_{-1}^{1} \mathrm{d}x \int_{x^{2}}^{1} \mathrm{d}y \int_{0}^{x^{2}+y^{2}} f(x,y,z) \, \mathrm{d}z. $

1. (1) $ \frac{59}{480}\pi R^{5} $; (2) 0; (3) $ \frac{250}{3}\pi $.

$ ^{*} $10. (1) $ F(t) $ 在 $ (0, +\infty) $ 内单调增加；（2）略.

1. $ \frac{1}{2}\sqrt{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}} $

1. $ \sqrt{\frac{2}{3}}R $ (R为圆的半径).

1. $ I = \frac{368}{105}\mu $.

1. $ F = (F_{x}, F_{y}, F_{z}) $，其中 $ F_{x} = 0, F_{y} = \frac{4GmM}{\pi R^{2}} \left( \ln \frac{R + \sqrt{R^{2} + a^{2}}}{a} - \frac{R}{\sqrt{R^{2} + a^{2}}} \right) $

$$ F_{z}=-\frac{2G m M}{R^{2}}\left(1-\frac{a}{\sqrt{R^{2}+a^{2}}}\right). $$

1. $ (0,0,\frac{3}{8}b) $.

$$ 16.~\mu~\mid_{r=0}=\frac{3M}{\pi R^{3}}. $$

第十一章

习题11-1（第193页）

1. (1) $ I_{x} = \int_{L} y^{2} \mu(x, y) \, \mathrm{d}s $, $ I_{y} = \int_{L} x^{2} \mu(x, y) \, \mathrm{d}s $;

$$ \bar{x}=\frac{\int_{L}x\mu(x,y)\mathrm{d}s}{\int_{L}\mu(x,y)\mathrm{d}s},\quad\bar{y}=\frac{\int_{L}y\mu(x,y)\mathrm{d}s}{\int_{L}\mu(x,y)\mathrm{d}s} $$

1. 略.

1. (1) $ 2\pi a^{2n+1} $; (2) $ \sqrt{2} $; (3) $ \frac{1}{12}(5\sqrt{5}+6\sqrt{2}-1) $; (4) $ \mathrm{e}^{a}\left(2+\frac{\pi}{4}a\right)-2 $; (5) $ \frac{\sqrt{3}}{2}(1-\mathrm{e}^{-2}) $; (6) 9; (7) $ \frac{256}{15}a^{3} $; (8) $ 2\pi^{2}a^{3}(1+2\pi^{2}) $.

1. 质心在扇形的对称轴上且与圆心距离 $ \frac{a\sin\varphi}{\varphi} $处.

1. (1) $ I_{z} = \frac{2}{3} \pi a^{2} \sqrt{a^{2} + k^{2}} (3a^{2} + 4\pi^{2}k^{2}) $;

(2) $ \bar{x} = \frac{6ak^2}{3a^2 + 4\pi^2k^2} $, $ \bar{y} = \frac{-6\pi a k^2}{3a^2 + 4\pi^2k^2} $, $ \bar{z} = \frac{3k(\pi a^2 + 2\pi^3k^2)}{3a^2 + 4\pi^2k^2} $.

习题11-2（第203页）

1—2. 略.

1. (1) $ -\frac{56}{15} $; (2) $ -\frac{\pi}{2}a^3 $; (3) 0; (4) $ -2\pi $;

(5) $ \frac{k^3\pi^3}{3} - a^2\pi $; (6) 13; (7) $ -\frac{1}{2} $; (8) $ -\frac{14}{15} $.

1. (1) $ \frac{34}{3} $; (2) 11; (3) 14; (4) $ \frac{32}{3} $.

1. $ -|F|R. $

1. $ mg(z_{2}-z_{1}) $

1. (1) $ \int_{L}\frac{P(x,y)+Q(x,y)}{\sqrt{2}}\mathrm{d}s;\quad(2)\int_{L}\frac{P(x,y)+2xQ(x,y)}{\sqrt{1+4x^{2}}}\mathrm{d}s; $

(3) $ \int_{L}\left[\sqrt{2x-x^{2}}P(x,y)+(1-x)Q(x,y)\right]\mathrm{d}s. $

1. $ \int_{T}\frac{P+2xQ+3yR}{\sqrt{1+4x^{2}+9y^{2}}}ds. $

习题11-3（第216页）

1. (1) $ \frac{1}{30} $; (2) 8.

1. (1) $ \frac{3}{8}\pi a^{2} $; (2) $ 12\pi $; (3) $ \pi a^{2} $.

1. $ -\pi $.

1. C 为椭圆 $ 2x^{2} + y^{2} = 1 $，沿逆时针方向.

1. 提示: 利用面积公式 $ A = \frac{1}{2} \oint_{C} x \, dy - y \, dx $，再逐条边地计算此曲线积分.

1. (1) $ \frac{5}{2} $; (2) 236; (3) 5.

1. (1) 12; (2) 0; (3) $ \frac{\pi^{2}}{4} $; (4) $ \frac{\sin 2}{4} - \frac{7}{6} $.

1. (1) $ \frac{1}{2}x^{2} + 2xy + \frac{1}{2}y^{2} $; (2) $ x^{2}y $; (3) $ -\cos 2x \cdot \sin 3y $; (4) $ x^{3}y + 4x^{2}y^{2} - 12e^{y} + 12ye^{y} $; (5) $ y^{2}\sin x + x^{2}\cos y $.

1. 略.

*10. (1) $ x^3 + 3x^2 y^2 + \frac{4}{3} y^3 = C $; (2) $ a^2 x - x^2 y - xy^2 - \frac{1}{3} y^3 = C $; (3) $ x e^y - y^2 = C $; (4) $ x \sin y + y \cos x = C $; (5) $ xy - \frac{1}{3} x^3 = C $; (6) 不是全微分方程; (7) $ \rho (1 + e^{2\theta}) = C $; (8) 不是全微分方程.

1. $ \lambda = -1, u(x, y) = -\arctan\frac{y}{x^2} + C $.

公众号：考研讲课

习题11-4（第222页）

1. $ I_{x} = \iint_{\Sigma} (y^{2} + z^{2}) \mu(x, y, z) \, \mathrm{d}S. $

2—3. 略.

1. (1) $ \frac{13}{3}\pi $; (2) $ \frac{149}{30}\pi $; (3) $ \frac{111}{10}\pi $.

1. (1) $ \frac{1 + \sqrt{2}}{2}\pi $; (2) $ 9\pi $.

1. (1) $ 4\sqrt{61} $; (2) $ -\frac{27}{4} $; (3) $ \pi a(a^{2}-h^{2}) $; (4) $ \frac{64}{15}\sqrt{2}a^{4} $.

1. $ \frac{2\pi}{15}(6\sqrt{3}+1) $.

1. $ \frac{4}{3}\mu_0\pi a^4 $.

习题11-5（第231页）

1—2. 略.

1. (1) $ \frac{2}{105}\pi R^{7} $; (2) $ \frac{3}{2}\pi $; (3) $ \frac{1}{2} $; (4) $ \frac{1}{8} $.

1. (1) $ \iint_{\Sigma} \left( \frac{3}{5} P + \frac{2}{5} Q + \frac{2\sqrt{3}}{5} R \right) \mathrm{d}S $; (2) $ \iint_{\Sigma} \frac{2x P + 2y Q + R}{\sqrt{1 + 4x^{2} + 4y^{2}}} \mathrm{d}S $.

习题11-6（第239页）

1. (1) $ 3a^{4} $; * (2) $ \frac{12}{5}\pi a^{5} $; * (3) $ \frac{2}{5}\pi a^{5} $; (4) $ 81\pi $; (5) $ \frac{3}{2} $.

$ ^{*}2.\ (1)\ 0;\ (2)\ a^{3}\left(2-\frac{a^{2}}{6}\right);\ (3)\ 108\pi. $

*3. (1) $ \operatorname{div} A = 2x + 2y + 2z $; (2) $ \operatorname{div} A = ye^{xy} - x\sin(xy) - 2xz\sin(xz^2) $; (3) $ \operatorname{div} A = 2x $.

1. 略.

*5. 提示：取液面为 xOy 面，z 轴铅直向下。这物体表面 Σ 上点 $ (x, y, z) $ 处单位面积上所受液体的压力为 $ (-v_0 z \cos \alpha, -v_0 z \cos \beta, -v_0 z \cos \gamma) $，其中 $ v_0 $ 为液体单位体积的重力， $ \cos \alpha, \cos \beta, \cos \gamma $ 为点 $ (x, y, z) $ 处 Σ 的外法线的方向余弦。

习题11-7（第248页）

1. 略.

$ ^{*}2.\ (1)\ -\sqrt{3}\pi a^{2};\ (2)\ -2\pi a(a+b);\ (3)\ -20\pi;\ (4)\ 9\pi. $

$ ^{*} $3. (1) $ \mathrm{rot} A = 2i + 4j + 6k $; (2) $ \mathrm{rot} A = i + j $;

(3) $ \mathbf{rot} A = \left[ x \sin (\cos z) - xy^2 \cos (xz) \right] i - y \sin (\cos z) j + \left[ y^2 z \cos (xz) - x^2 \cos y \right] k $.

$ ^{*} $4. (1) 0; (2) -4.

$ ^{*} $5. (1) $ 2\pi $; (2) $ 12\pi $.

$ ^{*} $6. 略.

$ ^{*}7.\ 0. $

总习题十一（第249页）

1. (1) $ \int_{R} (P\cos\alpha + Q\cos\beta + R\cos\gamma)\,ds $，切向量；

(2) $ \iint_{\Sigma} (P\cos\alpha + Q\cos\beta + R\cos\gamma)\,\mathrm{d}S $，法向量.

2.（C）.

1. (1) $ 2a^{2} $; (2) $ \frac{(2+t_{0}^{2})^{\frac{3}{2}}-2\sqrt{2}}{3} $; (3) $ -2\pi a^{2} $; (4) $ \frac{1}{35} $; (5) $ \pi a^{2} $; (6) $ \frac{\sqrt{2}}{16}\pi $.

1. (1) $ 2\pi\arctan\frac{H}{R} $; (2) $ -\frac{\pi}{4}h^{4} $; (3) $ 2\pi R^{3} $; (4) $ \frac{2}{15} $.

1. $ \frac{1}{2}\ln\left(x^{2}+y^{2}\right) $.

1. 略.

1. (1) 略；(2) $ \frac{c}{d} - \frac{a}{b} $.

1. $ \left(0,0,\frac{a}{2}\right) $

1. 略.

$ ^{*}10.3. $

1. $ \frac{3}{2} $.

公众号：考研讲课

第 12 2 章

习题12-1（第258页）

$$ \frac{1+1}{1+1^{2}}+\frac{1+2}{1+2^{2}}+\frac{1+3}{1+3^{2}}+\frac{1+4}{1+4^{2}}+\frac{1+5}{1+5^{2}}+\cdots; $$

$$ \frac{1}{2}+\frac{1\cdot3}{2\cdot4}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8}+\frac{1\cdot3\cdot5\cdot7\cdot9}{2\cdot4\cdot6\cdot8\cdot10}+\cdots; $$

$$ \frac{1}{5}-\frac{1}{5^{2}}+\frac{1}{5^{3}}-\frac{1}{5^{4}}+\frac{1}{5^{5}}-\cdots; $$

(4) $ \frac{1!}{1^1} + \frac{2!}{2^2} + \frac{3!}{3^3} + \frac{4!}{4^4} + \frac{5!}{5^5} + \cdots $

2.（1）发散；（2）收敛；（3）发散. 提示：先乘 $ 2\sin\frac{\pi}{12} $，再将一般项分解为两个余弦函数之差；（4）发散.

3.（1）收敛；（2）发散；（3）发散；（4）发散；（5）收敛

4.（1）收敛；（2）发散；（3）收敛；（4）发散.

习题12-2（第271页）

1.（1）发散；（2）发散；（3）收敛；（4）收敛；

(5) a > 1 时收敛, $ a \leq 1 $ 时发散.

2.（1）发散；（2）收敛；（3）收敛；（4）收敛.

$ ^{*} $3.（1）收敛；（2）收敛；（3）收敛；

（4）当 b < a 时收敛，当 b > a 时发散，当 b = a 时不能肯定.

4.（1）收敛；（2）收敛；（3）发散；（4）收敛；（5）发散；（6）发散

5.（1）条件收敛；（2）绝对收敛；（3）绝对收敛；（4）条件收敛；

(5) 发散.

习题12-3（第281页）

1. (1) $ (-1,1) $; (2) $ (-1,1) $; (3) $ (-\infty,+\infty) $; (4) $ (-3,3) $;

(5) $ \left(-\frac{1}{2},\frac{1}{2}\right) $; (6)(-1,1); (7) $ \left(-\sqrt{2},\sqrt{2}\right) $; (8)(4,6).

1. (1) $ \frac{1}{(1-x)^{2}} $ (-1 < x < 1); 考研讲课

(2)

$$ \frac{1}{4}\ln\frac{1+x}{1-x}+\frac{1}{2}\arctan x-x\quad\left(-1

(3) $ \frac{1}{2}\ln\frac{1+x}{1-x} $ (-1 < x < 1).

(4) $ \frac{x^{2}}{\left(1-x\right)^{2}}-x^{2}-2x^{3}(-1

习题12-4（第289页）

$$ \cos x=\cos x_{0}+\cos\left(x_{0}+\frac{\pi}{2}\right)\left(x-x_{0}\right)+\cdots+\frac{\cos\left(x_{0}+\frac{n\pi}{2}\right)}{n!}\left(x-x_{0}\right)^{n}+\cdots $$

$$ (-\infty,+\infty). $$

1. (1) $ \frac{e^x - e^{-x}}{2} = \sum_{n=1}^{\infty} \frac{x^{2n-1}}{(2n-1)!} $, $ (-\infty, +\infty) $;

(2)

$$ \ln(a+x)=\ln a+\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n}\left(\frac{x}{a}\right)^{n},(-a,a] $$

(3)

$$ a^{x}=\sum_{n=0}^{\infty}\frac{\left(x\ln a\right)^{n}}{n!},\quad(-\infty,+\infty) $$

(4)

$$ \sin^{2}x=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{(2x)^{2n}}{2(2n)!},(-\infty,+\infty); $$

(5)

$$ (1+x)\ln(1+x)=x+\sum_{n=2}^{\infty}\frac{(-1)^{n}x^{n}}{n(n-1)},\quad(-1,1]; $$

(6)

$$ \frac{x}{\sqrt{1+x^{2}}}=x+\sum_{n=1}^{\infty}(-1)^{n}\frac{2(2n)!}{(n!)^{2}}\left(\frac{x}{2}\right)^{2n+1},[-1,1]. $$

1. (1) $ \sqrt{x^3} = 1 + \frac{3}{2}(x-1) + \sum_{n=0}^{\infty}(-1)^n \frac{(2n)!}{(n!)^2} \frac{3}{(n+1)(n+2)2^n} \left(\frac{x-1}{2}\right)^{n+2} $, [0,2];

(2) $ \lg x = \frac{1}{\ln 10} \sum_{n=1}^{\infty} (-1)^{n-1} \frac{(x-1)^n}{n} $, (0,2].

$$ \cos x=\frac{1}{2}\sum_{n=0}^{\infty}\left(-1\right)^{n}\left[\frac{\left(x+\frac{\pi}{3}\right)^{2n}}{\left(2n\right)!}+\sqrt{3}\frac{\left(x+\frac{\pi}{3}\right)^{2n+1}}{\left(2n+1\right)!}\right],\left(-\infty,+\infty\right). $$

1. $ \frac{1}{x} = \frac{1}{3} \sum_{n=0}^{\infty} (-1)^{n} \frac{(x-3)^{n}}{3^{n}}, (0,6). $

1. $ \frac{1}{x^{2}+3x+2}=\sum_{n=0}^{\infty}\left(\frac{1}{2^{n+1}}-\frac{1}{3^{n+1}}\right)(x+4)^{n} $ (6)

习题12-5（第298页）

1. (1) 1.0986; (2) 1.648; (3) 2.00430; (4) 0.9994.

2.（1）0.4940；（2）0.487.

1. (1) $ y = Ce^{\frac{x^2}{2}} + [-1 + x + \frac{1}{1 \cdot 3}x^3 + \cdots + \frac{x^{2n-1}}{1 \cdot 3 \cdot 5 \cdots (2n-1)} + \cdots] $;

$$ y=a_{0}\mathrm{e}^{-\frac{x^{2}}{2}}+a_{1}\left[x-\frac{x^{3}}{1\cdot3}+\frac{x^{5}}{1\cdot3\cdot5}-\cdots+(-1)^{n-1}\frac{x^{2n-1}}{1\cdot3\cdot5\cdots(2n-1)}+\cdots\right]; $$

$$ y=C(1-x)+x^{3}\left[\frac{1}{3}+\frac{1}{6}x+\frac{1}{10}x^{2}+\cdots+\frac{2}{(n+2)(n+3)}x^{n}+\cdots\right]. $$

1. (1) $ y = \frac{1}{2} + \frac{1}{4}x + \frac{1}{8}x^{2} + \frac{1}{16}x^{3} + \frac{9}{32}x^{4} + \cdots $;

(2) $ y = x + \frac{1}{1 \cdot 2}x^2 + \frac{1}{2 \cdot 3}x^3 + \frac{1}{3 \cdot 4}x^4 + \cdots + \frac{1}{n(n-1)}x^n + \cdots $.

1. 和函数为 $ y(x)=\frac{2}{3}\mathrm{e}^{-\frac{x}{2}}\cos\frac{\sqrt{3}}{2}x+\frac{1}{3}\mathrm{e}^{x}\quad(-\infty

1. $ \mathrm{e}^{x}\cos x = \sum_{n=0}^{\infty}2^{\frac{n}{2}}\cos\frac{n\pi}{4}\cdot\frac{x^{n}}{n!},(-\infty,+\infty) $.

提示： $ \mathrm{e}^{x}\cos x = \mathrm{Re}\mathrm{e}^{(1+\mathrm{i})x} = \mathrm{Re}\mathrm{e}^{\sqrt{2}(\cos\frac{\pi}{4}+\mathrm{i}\sin\frac{\pi}{4})x} $

习题12-6（第307页）

1. (1) 取正整数 $ N \geqslant \frac{|x|}{\varepsilon} $; (2) 略.

1. (1) $ s(x) = \begin{cases} 0, & x = 0, \\ 1 + x^2, & x \neq 0; \end{cases} $

(2) 当 $ x \neq 0 $ 时取正整数 $ N \geq \frac{\ln \frac{1}{\varepsilon}}{\ln (1 + x^2)} $，当 x = 0 时取 N = 1；

（3）在 $ [0,1] $上不一致收敛，在 $ \left[\frac{1}{2},1\right] $上一致收敛.

3.（1）一致收敛；（2）不一致收敛.

1. 略.

习题12-7（第320页）

1. (1) $ f(x) = \pi^{2} + 1 + 12 \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^{2}} \cos nx $，+ $ \infty $）；

(2)

$$ \begin{align*}f(x)&=\frac{\mathrm{e}^{2\pi}-\mathrm{e}^{-2\pi}}{\pi}\bigg[\frac{1}{4}+\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{2}+4}(2\cos nx-n\sin nx)\bigg],\ $ x\neq(2n+1)\pi,n=0,\pm1,\pm2,\cdots);\end{align*} $$

(3)

$$ \begin{align*}f(x)=&\frac{a-b}{4}\pi+\sum_{n=1}^{\infty}\{\frac{[1-(-1)^{n}](b-a)}{n^{2}\pi}\cos nx+\frac{(-1)^{n-1}(a+b)}{n}\sin nx\},\\&(x\neq(2n+1)\pi,n=0,\pm1,\pm2,\cdots).\end{align*} $$

2.（1）

$$ 2\sin\frac{x}{3}=\frac{18\sqrt{3}}{\pi}\sum_{n=1}^{\infty}(-1)^{n-1}\frac{n\sin nx}{9n^{2}-1},(-\pi,\pi); $$

(2)

$$ \begin{align*}f(x)&=\frac{1+\pi-e^{-\pi}}{2\pi}+\frac{1}{\pi}\sum_{n=1}^{\infty}\{\frac{1-(-1)^n e^{-\pi}}{1+n^2}\cos nx.\\&\quad.[\frac{-n+(-1)^n n e^{-\pi}}{1+n^2}+\frac{1}{n}(1-(-1)^n)]\sin nx\},(-\pi,\pi).\end{align*} $$

$$ \cos\frac{x}{2}=\frac{2}{\pi}+\frac{4}{\pi}\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n-1}}{4n^{2}-1}\cos nx,\left[-\pi,\pi\right]. $$

4.

$$ \begin{align*}\cdot f(x)&=\frac{2}{\pi}\sum_{n=1}^{\infty}\left[\frac{1}{n^{2}}\sin\frac{n\pi}{2}+(-1)^{n+1}\frac{\pi}{2n}\right]\sin nx\quad(x\neq(2n+1)\pi,n=0,\pm1,\\&\pm2,\cdots)\end{align*} $$

5.

$$ \frac{\pi-x}{2}=\sum_{n=1}^{\infty}\frac{1}{n}\sin nx,(0,\pi]. $$

6.

$$ 2x^{2}=\frac{4}{\pi}\sum_{n=1}^{\infty}\left[-\frac{2}{n^{3}}+\left(-1\right)^{n}\left(\frac{2}{n^{3}}-\frac{\pi^{2}}{n}\right)\right]\sin nx,\left[0,\pi\right); $$

$$ 2x^{2}=\frac{2}{3}\pi^{2}+8\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n}}{n^{2}}\cos nx,\quad[0,\pi]. $$

1. 略.

习题12-8（第327页）

1.

$$ f(x)=\frac{11}{12}+\frac{1}{\pi^{2}}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^{2}}\cos2n\pi x,(-\infty,+\infty); $$

(2)

$$ \begin{align*}2)\ f(x)&=-\frac{1}{4}+\sum_{n=1}^{\infty}\{[\frac{1-(-1)^{n}}{n^{2}\pi^{2}}+\frac{2\sin\frac{n\pi}{2}}{n\pi}]\cos n\pi x+\frac{1-2\cos\frac{n\pi}{2}}{n\pi}\sin n\pi x\},\\&\quad(x\neq2k,2k+\frac{1}{2},k=0,\pm1,\pm2,\cdots);\end{align*} $$

(3)

$$ f(x)=-\frac{1}{2}+\sum_{n=1}^{\infty}\{\frac{6}{n^{2}\pi^{2}}[1-(-1)^{n}\cos\frac{n\pi x}{3}+\frac{6}{n\pi}(-1)^{n+1}\sin\frac{n\pi x}{3}]\} $$

$$ (x\neq3(2k+1),k=0,\pm1,\pm2,\cdots). $$

1. (1) $ f(x) = \frac{4l}{\pi^2} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{(2k-1)^2} \sin \frac{(2k-1)\pi x}{l}, [0,l] $,

$$ f(x)=\frac{l}{4}-\frac{2l}{\pi^{2}}\sum_{k=1}^{\infty}\frac{1}{\left(2k-1\right)^{2}}\cos\frac{2(2k-1)\pi x}{l},\left[0,l\right]; $$

(2)

$$ f(x)=\frac{8}{\pi}\sum_{n=1}^{\infty}\{\frac{(-1)^{n+1}}{n}+\frac{2}{n^{3}\pi^{2}}[(-1)^{n}-1]\}\sin\frac{n\pi x}{2},[0,2), $$

$$ f(x)=\frac{4}{3}+\frac{16}{\pi^{2}}\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{2}}\cos\frac{n\pi x}{2},[0,2]. $$

$$ f(x)=\operatorname{sh}1\sum_{n=-\infty}^{\infty}\frac{(-1)^{n}(1-n\pi\mathrm{i})}{1+(n\pi)^{2}}\mathrm{e}^{n\pi\mathrm{x i}}\quad(x\neq2k+1,k=0,\pm1,\pm2,\cdots) $$

$$ u(t)=\frac{h\tau}{T}+\frac{2h}{\pi}\sum_{n=1}^{\infty}\frac{1}{n}\sin\frac{n\tau\pi}{T}\cos\frac{2n\pi t}{T}\quad(-\infty,+\infty). $$

总习题十二（第327页）

1.（1）必要，充分；（2）充分必要；（3）收敛，发散

1. (A).

3.（1）发散；（2）发散；（3）收敛；（4）发散；

(5) a<1 时收敛, a>1 时发散, a=1 时, s>1 收敛, $ s\leq1 $ 发散.

1. 略.

1. 不一定. 考虑级数 $ \sum_{n=1}^{\infty}(-1)^{n}\frac{1}{\sqrt{n}} $ 及 $ \sum_{n=1}^{\infty}\left((-1)^{n}\frac{1}{\sqrt{n}}+\frac{1}{n}\right) $.

6.（1）p>1时绝对收敛, $ 0

（2）绝对收敛；（3）条件收敛；（4）绝对收敛.

1. (1) 0; (2) $ \sqrt[4]{8} $. 提示: 化成 $ 2^{\frac{1}{3} + \frac{2}{3^2} + \cdots + \frac{n}{3^n} + \cdots} $.

1. (1) $ \left(-\frac{1}{5},\frac{1}{5}\right) $; (2) $ \left(-\frac{1}{e},\frac{1}{e}\right) $; (3) $ (-2,0) $; (4) $ (-\sqrt{2},\sqrt{2}) $.

1. (1) $ s(x) = \frac{2 + x^2}{(2 - x^2)^2}, (-\sqrt{2}, \sqrt{2}) $; $ ^ * (2) s(x) = \arctan x, [-1, 1]; $

(3) $ s(x) = \frac{x - 1}{(2 - x)^2} $, (0,2);

$$ \begin{aligned}*(4)s(x)=\{\begin{matrix}1+(\frac{1}{x}-1)\ln(1-x),&x\in\lbrack-1,0)\cup(0,1),\\0,&x=0\\1,& 代众号：\begin{array}{c} 考研拼课 \\x=1.\end{array}\end{array}.\end{aligned} $$

1. (1) $ 2e $; (2) $ \frac{1}{2}(\cos 1 + \sin 1) $. 提示: 利用 $ \cos 1 $ 和 $ \sin 1 $ 的展开式.

11.

$$ \ln\left(x+\sqrt{x^{2}+1}\right)=x+\sum_{n=1}^{\infty}(-1)^{n}\frac{(2n-1)!!x^{2n+1}}{(2n)!!2n+1},x\in[-1,1], $$

提示:利用积分 $ \int_{0}^{x}\frac{dt}{\sqrt{t^{2}+1}} $;

(2)

$$ \frac{1}{\left(2-x\right)^{2}}=\sum_{n=1}^{\infty}\frac{n}{2^{n+1}}x^{n-1},x\in\left(-2,2\right). $$

12.

$$ \begin{aligned}f(x)=&\frac{\mathrm{e}^{\pi}-1}{2\pi}+\frac{1}{\pi}\sum_{n=1}^{\infty}\left[\frac{\left(-1\right)^{n}\mathrm{e}^{\pi}-1}{n^{2}+1}\cos nx+\frac{n\left(\left(-1\right)^{n+1}\mathrm{e}^{\pi}+1\right)}{n^{2}+1}\sin nx\right],\\&-\infty

13.

$$ f(x)=\frac{2}{\pi}\sum_{n=1}^{\infty}\frac{1-\cos n h}{n}\sin nx,x\in(0,h)\cup(h,\pi] $$

$$ f(x)=\frac{h}{\pi}+\frac{2}{\pi}\sum_{n=1}^{\infty}\frac{\sin nh}{n}\cos nx,x\in[0,h)\cup(h,\pi]. $$

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