## 张祖锦2023年数学专业真题分类70天之第35天
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783、 (2)、 求
\begin\{aligned\} \iint\_\varSigma (x-y+z)\mathrm\{ d\} y\mathrm\{ d\} z+(y-z+x)\mathrm\{ d\} z\mathrm\{ d\} x+(z-x+y)\mathrm\{ d\} x\mathrm\{ d\} y, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle \varSigma$ 为闭曲面 $\displaystyle |x-y+z|+|y-z+x|+|z-x+y|=1$, 方向取外侧. (上海财经大学2023年数学分析考研试题) [曲线曲面积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle \varSigma$ 所围区域为 $\displaystyle \varOmega$, 则
\begin\{aligned\} I\xlongequal\{\tiny\mbox\{Gauss\}\} \iiint\_\varOmega (1+1+1)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令
\begin\{aligned\} \left.\begin\{array\}\{rrrr\}x-y+z=u\\\\ y-z+x=v\\\\ z-x+y=w\end\{array\}\right\\}\Rightarrow \frac\{\partial(u,v,w)\}\{\partial(x,y,z)\}=\left|\begin\{array\}\{cccccccccc\}1&-1&1\\\\ 1&1&-1\\\\ -1&1&1\end\{array\}\right|=4. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} I&=3\iiint\_\{|u|+|v|+|w|\leq 1\} \frac\{1\}\{4\}\mathrm\{ d\} u\mathrm\{ d\} v\mathrm\{ d\} w \xlongequal\{\tiny\mbox\{对称性\}\} \frac\{3\}\{4\}\cdot 8\iiint\_\{u,v,w\geq 0\atop u+v+w\leq 1\} \mathrm\{ d\} u\mathrm\{ d\} v\mathrm\{ d\} w\\\\ &=6 \int\_0^1 \mathrm\{ d\} w\iint\_\{u\geq 0, v\geq 0\atop u+v\leq 1-w\}\mathrm\{ d\} u\mathrm\{ d\} v =6\int\_0^1 \frac\{1\}\{2\}(1-w)^2\mathrm\{ d\} w=1. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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784、 14、 解答如下问题: (1)、 计算 $\displaystyle \int\_L(x^2+y^2)\mathrm\{ d\} s$, 其中 $\displaystyle L$ 是球面 $\displaystyle x^2+y^2+z^2=1$ 与平面 $\displaystyle x+y+z=0$ 的交线; (上海大学2023年数学分析考研试题) [曲线曲面积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}\xlongequal\{\tiny\mbox\{对称性\}\} \frac\{2\}\{3\}\int\_L (x^2+y^2+z^2)\mathrm\{ d\} s=\frac\{2\}\{3\}\int\_L \mathrm\{ d\} s =\frac\{2\}\{3\}\cdot 2\pi =\frac\{4\pi\}\{3\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
785、 (2)、 求曲面积分
\begin\{aligned\} I=\iint\_\varSigma 2x^3\mathrm\{ d\} y\mathrm\{ d\} z+2y^3\mathrm\{ d\} z\mathrm\{ d\} x+2z^3\mathrm\{ d\} x\mathrm\{ d\} y, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle \varSigma$ 是曲面 $\displaystyle z=1-x^2-y^2\ (z\geq 0)$ 的上侧. (上海大学2023年数学分析考研试题) [曲线曲面积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle S: x^2+y^2\leq 1, z=0$, 方向指向负 $\displaystyle z$ 轴, 则 $\displaystyle \iint\_S\cdots=0$. 据 Gauss 公式,
\begin\{aligned\} \mbox\{原式\}&=\mbox\{原式\}+\iint\_S \cdots\\\\ &=\iiint\_\{x^2+y^2+z^2\leq 1\atop z\geq 0\} (6x^2+6y^2+6z^2)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z\\\\ &=6\int\_0^1 r^2\cdot\frac\{4\pi r^2\}\{2\}\mathrm\{ d\} r=\frac\{12\pi\}\{5\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
786、 8、 $\displaystyle Q(x,y)$ 在全平面上偏导数存在且连续,
\begin\{aligned\} \int\_L (3x^2y+8xy^2)\mathrm\{ d\} x+Q(x,y)\mathrm\{ d\} y, \int\_L Q(x,y)\mathrm\{ d\} x+\frac\{8\}\{3\}x^3+12x(y+1)\mathrm\{e\}^y\mathrm\{ d\} y \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
与路径无关. (1)、 求 $\displaystyle Q(x,y)$, 使得 $\displaystyle Q(0,0)=0$. (2)、 就 (1) 的 $\displaystyle Q(x,y)$, 求 $\displaystyle \int\_L (3x^2y+8xy^2)\mathrm\{ d\} x+Q(x,y)\mathrm\{ d\} y$. (首都师范大学2023年数学分析考研试题) [曲线曲面积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / (1)、 由题设,
\begin\{aligned\} Q\_x=3x^2+16xy,\qquad(I)\\\\ Q\_y=8x^2+12(y+1)\mathrm\{e\}^y,\qquad(II) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} &(I)\Rightarrow Q=x^3+8x^2y+f(y)\stackrel\{(II)\}\{\Rightarrow\}8x^2+f'(y)=8x^2+12(y+1)\mathrm\{e\}^y\\\\ \Rightarrow&f'(y)=12(y+1)\mathrm\{e\}^y\Rightarrow f(y)=12y\mathrm\{e\}^y+C\\\\ \Rightarrow&Q=x^3+8x^2y+12y\mathrm\{e\}^y+C \stackrel\{Q(0,0)=0\}\{\Rightarrow\}C=0\\\\ \Rightarrow& Q=x^3+8x^2y+12y\mathrm\{e\}^y. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 设 $\displaystyle L$ 的起点为 $\displaystyle (x\_0,y\_0)$, 终点为 $\displaystyle (x\_1,y\_1)$, 则
\begin\{aligned\} \mbox\{原式\}=&\int\_L (3x^2y+8xy^2)\mathrm\{ d\} x+(x^3+8x^2y+12y\mathrm\{e\}^y)\mathrm\{ d\} y\\\\ =&\int\_L\mathrm\{ d\} \left(x^3y+4x^2y^2+12(y-1)\mathrm\{e\}^y\right)\\\\ =&\left(x^3y+4x^2y^2+12(y-1)\mathrm\{e\}^y\right)|\_\{(x\_0,y\_0)\}^\{(x\_1,y\_1)\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
787、 9、 (1)、 求 $\displaystyle \int\_\{(1,1,1)\}^\{(2,3,4)\}x\mathrm\{ d\} x+2y\mathrm\{ d\} y+3z\mathrm\{ d\} z$. (首都师范大学2023年数学分析考研试题) [曲线曲面积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \mbox\{原式\}=\left.\left(\frac\{x^2\}\{2\}+y^2+\frac\{3z^2\}\{2\}\right)\right|\_\{(1,1,1)\}^\{(2,3,4)\} =32$. 当然你可由 Stokes 公式知题中曲线积分与路径无关, 而可取定一条路径后化为定积分的形式再求解.跟锦数学微信公众号. [在线资料](
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---
788、 (2)、 求 $\displaystyle \iint\_S x^5\mathrm\{ d\} x\mathrm\{ d\} y$, 其中 $\displaystyle S$ 为 $\displaystyle x^2+\frac\{y^2\}\{4\}+\frac\{z^2\}\{16\}=1, z\geq 0$, 取上侧. (首都师范大学2023年数学分析考研试题) [曲线曲面积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle \varSigma: x^2+\frac\{y^2\}\{4\}\leq 1, z=0$, 取上侧, 则
\begin\{aligned\} \iint\_\varSigma\cdots =\iint\_\{x^2+\frac\{y^2\}\{4\}\leq 1\}x^5\mathrm\{ d\} x\mathrm\{ d\} y\xlongequal\{\tiny\mbox\{对称性\}\} 0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} \mbox\{原式\}=&\mbox\{原式\}-\iint\_\varSigma\cdots\xlongequal\{\tiny\mbox\{Gauss\}\} \iiint\_\{x^2+\frac\{y^2\}\{4\}+\frac\{z^2\}\{16\}\leq 1\atop z\geq 0\}5x^4\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z\\\\ =&5\int\_\{-1\}^1 x^4\mathrm\{ d\} x\iint\_\{\frac\{y^2\}\{4\}+\frac\{z^2\}\{16\}\leq 1-x^2\atop z\geq 0\}\mathrm\{ d\} y\mathrm\{ d\} z =5\int\_\{-1\}^1 x^4\cdot \frac\{\pi \cdot 2\sqrt\{1-x^2\}\cdot 4\sqrt\{1-x^2\}\}\{2\}\mathrm\{ d\} x\\\\ \xlongequal\{\tiny\mbox\{对称性\}\}& 40\pi \int\_0^1 x^4(1-x^2)\mathrm\{ d\} x=\frac\{16\pi\}\{7\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
789、 (2)、 $\displaystyle \oint\_L (x+z)\mathrm\{ d\} x+x^4\mathrm\{ d\} z$, 其中 $\displaystyle L$ 为曲面
\begin\{aligned\} x^2+y^2+z^2=1, x^2+y^2=3z^2\left(z\geq 0\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
的交线, 从 $\displaystyle z$ 轴正向看是逆时针方向. (四川大学2023年数学分析考研试题) [曲线曲面积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 联合两方程知
\begin\{aligned\} z=\frac\{1\}\{2\}, x^2+y^2=\frac\{3\}\{4\}\Rightarrow x=\frac\{\sqrt\{3\}\}\{2\}\cos\theta, y=\frac\{\sqrt\{3\}\}\{2\}\sin\theta, \theta: 0\to 2\pi. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} \mbox\{原式\}=\int\_0^\{2\pi\} \left(\frac\{\sqrt\{3\}\}\{2\}\cos\theta+\frac\{1\}\{2\}\right)\left(\frac\{\sqrt\{3\}\}\{2\}\cos\theta\right)'\mathrm\{ d\} \theta=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
790、 (3)、 $\displaystyle \iint\_\varSigma(y-2z)\mathrm\{ d\} y\mathrm\{ d\} z+z\mathrm\{ d\} x\mathrm\{ d\} y$, 其中 $\displaystyle \varSigma$ 为曲面
\begin\{aligned\} x^2+y^2+z^2=2z\left(0\leq z\leq 1\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
法向量与 $\displaystyle z$ 轴正向的夹角为锐角. (四川大学2023年数学分析考研试题) [曲线曲面积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 取 $\displaystyle S: x^2+y^2\leq 1, z=1$, 取上侧, 则
\begin\{aligned\} \iint\_S \cdots=\iint\_\{x^2+y^2\leq 1\} \mathrm\{ d\} x\mathrm\{ d\} y=\pi, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而
\begin\{aligned\} \mbox\{原式\}=&-\left(-\mbox\{原式\}+\iint\_S\cdots\right)+\iint\_S\cdots\\\\ \xlongequal\{\tiny\mbox\{Gauss\}\}&-\iiint\_\{x^2+y^2+z^2\leq 2z\atop 0\leq z\leq 1\}(0+1)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z+\pi\\\\ =&-\frac\{1\}\{2\}\cdot\frac\{4\pi\}\{3\}+\pi=\frac\{\pi\}\{3\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
791、 9、 (15 分) 设 $\displaystyle \varOmega$ 为 $\displaystyle \mathbb\{R\}^3$ 中一闭区域, $\displaystyle \varSigma$ 为其边界, 且分段光滑, $\displaystyle u,v$ 有连续的二阶偏导数, 证明:
\begin\{aligned\} \iiint\_\varOmega v\Delta u\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z=\iint\_\varSigma v\frac\{\partial u\}\{\partial \vec\{n\}\}\mathrm\{ d\} S -\iiint\_\varOmega \nabla u\cdot \nabla v \mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle \Delta u=\frac\{\partial^2u\}\{\partial x^2\}+\frac\{\partial^2u\}\{\partial y^2\}+\frac\{\partial^2u\}\{\partial z^2\}$, $\displaystyle \frac\{\partial u\}\{\partial \vec\{n\}\}$ 为 $\displaystyle u$ 沿曲面 $\displaystyle \varSigma$ 外法线方向的方向导数, $\displaystyle \nabla u, \nabla v$ 分别为 $\displaystyle u,v$ 的的梯度. (苏州大学2023年数学分析考研试题) [曲线曲面积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} &\iiint\_\varOmega v\Delta u\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z+\iiint\_\varOmega \nabla u\cdot \nabla v \mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z\\\\ =&\iiint\_\varOmega \left\[\frac\{\partial\}\{\partial x\}\left(v\frac\{\partial u\}\{\partial x\}\right) +\frac\{\partial\}\{\partial y\}\left(v\frac\{\partial u\}\{\partial y\}\right) +\frac\{\partial\}\{\partial z\}\left(v\frac\{\partial u\}\{\partial z\}\right)\right\]\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z\\\\ \xlongequal\{\tiny\mbox\{Gauss\}\}&\iint\_\varSigma v\frac\{\partial u\}\{\partial x\}\mathrm\{ d\} y\mathrm\{ d\} z +v\frac\{\partial u\}\{\partial y\}\mathrm\{ d\} z\mathrm\{ d\} x +v\frac\{\partial u\}\{\partial z\}\mathrm\{ d\} x\mathrm\{ d\} y =\iint\_\varSigma v\frac\{\partial u\}\{\partial \vec\{n\}\}\mathrm\{ d\} S. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
792、 5、 计算曲线积分
\begin\{aligned\} I=\int\_L (yz^2-\cos z)\mathrm\{ d\} x+2xz^2\mathrm\{ d\} y+(2xyz+x\sin z)\mathrm\{ d\} z, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle L$ 为取上侧的曲面 $\displaystyle S: 4x^2+y^2+z^2=1\ (x,y,z\geq 0)$ 的边界曲线, 其正向与曲面 $\displaystyle S$ 的正法向量满足右手法则. (太原理工大学2023年数学分析考研试题) [曲线曲面积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle F=4x^2+y^2+z^2$, 则 $\displaystyle S$ 上 $\displaystyle (x,y,z)$ 处的法向量为
\begin\{aligned\} \frac\{1\}\{2\}\left\\{F\_x,F\_y,F\_z\right\\}=\left\\{4x,y,z\right\\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其模长为 $\displaystyle \sqrt\{16x^2+y^2+z^2\}=\sqrt\{12x^2+1\}$. 故 $\displaystyle \vec\{n\}=\frac\{\left\\{4x,y,z\right\\}\}\{\sqrt\{12x^2+1\}\}$. 于是
\begin\{aligned\} I&\xlongequal\{\tiny\mbox\{Stokes\}\} \iint\_S \frac\{1\}\{\sqrt\{12x^2+1\}\}\left|\begin\{array\}\{cccccccccc\}4x&y&z\\\\ \frac\{\partial\}\{\partial x\}&\frac\{\partial\}\{\partial y\}&\frac\{\partial\}\{\partial z\}\\\\ yz^2-\cos z&2xz^2&2xyz+x\sin z\end\{array\}\right|\mathrm\{ d\} S\\\\ &=\iint\_S \frac\{\left\\{4x[2xz-4xz]-y[2yz+\sin z-(2yz+\sin z)] +z[2z^2-z^2]\right\\}\}\{\sqrt\{12x^2+1\}\}\mathrm\{ d\} S\\\\ &=\iint\_S\frac\{z^3-8x^2z\}\{\sqrt\{12x^2+1\}\}\mathrm\{ d\} S\\\\ &\stackrel\{z=\sqrt\{1-4x^2-y^2\}\}\{=\} \iint\_\{4x^2+y^2\leq 1\atop x\geq 0, y\geq 0\} \frac\{\sqrt\{1-4x^2-y^2\}(1-12x^2-y^2)\}\{\sqrt\{12x^2-1\}\}\cdot \sqrt\{\frac\{12x^2+1\}\{1-4x^2-y^2\}\}\mathrm\{ d\} x\mathrm\{ d\} y\\\\ &=\iint\_\{4x^2+y^2\leq 1\atop x\geq 0, y\geq 0\}(1-12x^2-y^2)\mathrm\{ d\} x\mathrm\{ d\} y \stackrel\{2x=u, y=v\}\{=\}\frac\{1\}\{2\}\iint\_\{u^2+v^2\leq 1\atop u\geq 0, v\geq 0\}(1-3u^2-v^2)\mathrm\{ d\} u\mathrm\{ d\} v\\\\ &\xlongequal\{\tiny\mbox\{对称性\}\} \frac\{1\}\{2\}\frac\{1\}\{2\}\iint\_\{u^2+v^2\leq 1\atop u\geq 0, v\geq 0\}[1-2(u^2+v^2)]\mathrm\{ d\} u\mathrm\{ d\} v =\frac\{1\}\{2\}\int\_0^1 (1-2r^2)\cdot\frac\{2\pi r\}\{4\}\mathrm\{ d\} r=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
793、 9、 (1)、 求
\begin\{aligned\} \iint\_S (x^2+?) \mathrm\{ d\} y\mathrm\{ d\} z+(2y^2+?)\mathrm\{ d\} x\mathrm\{ d\} z+(3z^2+?)\mathrm\{ d\} x\mathrm\{ d\} y, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle S: (x-1)^2+(y-2)^2+(z-3)^3=1$, 取外侧. [张祖锦注: ? 中的部分构成的向量的散度为 $\displaystyle 0$] (天津大学2023年数学分析考研试题) [曲线曲面积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 注意 $\displaystyle \mathrm\{ d\} x\mathrm\{ d\} z=-\mathrm\{ d\} z\mathrm\{ d\} x$,
\begin\{aligned\} \mbox\{原式\}&\xlongequal\{\tiny\mbox\{Gauss\}\} \iiint\_\{(x-1)^2+(y-2)^2+(z-3)^3\leq 1\}(2x-4y+6xz)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z\\\\ &\stackrel\{x-1=u, y-2=v, z-3=w\}\{=\}\iiint\_\{u^2+v^2+w^2\leq 1\}\left\[\begin\{array\}\{c\}2(u+1)-4(v+2)\\\\ +6(w+3)\end\{array\}\right\]\mathrm\{ d\} u\mathrm\{ d\} v\mathrm\{ d\} w\\\\ &\xlongequal\{\tiny\mbox\{对称性\}\} 12\iiint\_\{u^2+v^2+w^2\leq 1\}\mathrm\{ d\} u\mathrm\{ d\} v\mathrm\{ d\} w=12\cdot \frac\{4\pi\}\{3\}=16\pi. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
794、 (2)、 求 $\displaystyle \iint\_\varSigma \sqrt\{\frac\{x^2\}\{a^4\}+\frac\{y^2\}\{b^4\}+\frac\{z^2\}\{c^4\}\}\mathrm\{ d\} S$, 其中 $\displaystyle \varSigma: \frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}+\frac\{z^2\}\{c^2\}=1$. (天津大学2023年数学分析考研试题) [曲线曲面积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} &\mbox\{原式\}=2\iint\_\{\frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}\leq 1\} \sqrt\{\frac\{x^2\}\{a^4\}+\frac\{y^2\}\{b^4\} +\frac\{z^2\}\{c^4\}\}\cdot \frac\{c\}\{\sqrt\{1-\frac\{x^2\}\{a^2\}-\frac\{y^2\}\{b^2\}\}\} \sqrt\{\frac\{x^2\}\{a^4\}+\frac\{y^2\}\{b^4\}+\frac\{z^2\}\{c^4\}\}\mathrm\{ d\} x\mathrm\{ d\} y\\\\ =&2c \iint\_\{\frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}\leq 1\} \frac\{\frac\{x^2\}\{a^4\}+\frac\{y^2\}\{b^4\} +\frac\{z^2\}\{c^4\}\}\{ \sqrt\{1-\frac\{x^2\}\{a^2\}-\frac\{y^2\}\{b^2\}\} \}\mathrm\{ d\} x\mathrm\{ d\} y\\\\ =&2c \iint\_\{\frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}\leq 1\} \frac\{\frac\{x^2\}\{a^4\}+\frac\{y^2\}\{b^4\} +\frac\{1\}\{c^2\}\left(1-\frac\{x^2\}\{a^2\}-\frac\{y^2\}\{b^2\}\right)\}\{ \sqrt\{1-\frac\{x^2\}\{a^2\}-\frac\{y^2\}\{b^2\}\} \}\mathrm\{ d\} x\mathrm\{ d\} y\\\\ =&2abc\iint\_\{u^2+v^2\leq 1\} \frac\{\frac\{u^2\}\{a^2\}+\frac\{v^2\}\{b^2\} +\frac\{1\}\{c^2\}\left(1-u^2-v^2\right)\}\{\sqrt\{1-u^2-v^2\}\}\mathrm\{ d\} u\mathrm\{ d\} v\left(x=au,\ y=bv\right)\\\\ =&2abc\left\[\begin\{array\}\{c\}\frac\{1\}\{a^2\} \iint\_\{u^2+v^2\leq 1\} \frac\{u^2\}\{\sqrt\{1-u^2-v^2\}\}\mathrm\{ d\} u\mathrm\{ d\} v +\frac\{1\}\{b^2\}\iint\_\{u^2+v^2\leq 1\} \frac\{v^2\}\{\sqrt\{1-u^2-v^2\}\}\mathrm\{ d\} u\mathrm\{ d\} v\\\\ +\frac\{1\}\{c^2\}\iint\_\{u^2+v^2\leq 1\}\sqrt\{1-u^2-v^2\}\mathrm\{ d\} u\mathrm\{ d\} v\end\{array\}\right\]\\\\ =&2abc\left\[\frac\{1\}\{2\} \left(\frac\{1\}\{a^2\}+\frac\{1\}\{b^2\}\right) 2\pi \int\_0^1 \frac\{r^2\}\{\sqrt\{1-r^2\}\}r\mathrm\{ d\} r +\frac\{1\}\{c^2\} 2\pi \int\_0^1 \sqrt\{1-r^2\}r\mathrm\{ d\} r\right\]\\\\ =&2abc\left\[\frac\{1\}\{2\} \left(\frac\{1\}\{a^2\}+\frac\{1\}\{b^2\}\right) 2\pi \cdot\frac\{2\}\{3\} +\frac\{1\}\{c^2\} 2\pi \cdot \frac\{1\}\{3\}\right\] =\frac\{4\pi abc\}\{3\}\left(\frac\{1\}\{a^2\}+\frac\{1\}\{b^2\}+\frac\{1\}\{c^2\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
795、 6、 求 $\displaystyle u(x,y)=\oint\_C \frac\{\cos\angle(\vec\{r\},\vec\{n\})\}\{r\}\mathrm\{ d\} s$, 其中
\begin\{aligned\} r=\sqrt\{(x-\xi)^2+(y-\eta)^2\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
$\displaystyle \vec\{r\}$ 为 $\displaystyle M(x,y)$ 到 $\displaystyle C$ 上点 $\displaystyle (\xi,\eta)$ 的向量, $\displaystyle M(x,y)$ 不在 $\displaystyle C$ 上, $\displaystyle \vec\{n\}$ 为 $\displaystyle C$ 在 $\displaystyle (\xi,\eta)$ 的单位外法向量. (同济大学2023年数学分析考研试题) [曲线曲面积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
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\begin\{aligned\} u(x,y)=&\oint\_C \frac\{\vec\{r\}\cdot\vec\{n\}\}\{r^2\}\mathrm\{ d\} s =\oint\_C \left\[\frac\{\xi-x\}\{r^2\}\cos\angle(\vec\{n\},x)+\frac\{\eta-y\}\{r^2\}\cos\angle(\vec\{n\},y)\right\]\mathrm\{ d\} s\\\\ =&\oint\_C \left\\{\frac\{\xi-x\}\{r^2\}\cos\angle(\vec\{t\},y) +\frac\{\eta\}\{r^2\}\cos\left\[\pi-(\vec\{t\},y)\right\]\right\\}\mathrm\{ d\} s\\\\ =&\oint\_C \frac\{-(\eta-y)\mathrm\{ d\} \xi+(\xi-x)\mathrm\{ d\} \eta\}\{r^2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 若 $\displaystyle M$ 在 $\displaystyle C$ 的外部, 则 $\displaystyle u(x,y)\xlongequal\{\tiny\mbox\{Green\}\} 0$. (2)、 若 $\displaystyle M$ 在 $\displaystyle C$ 的内部, 则
\begin\{aligned\} &u(x,y)=\oint\_\{(\xi-x)^2+(\eta-y)^2=\varepsilon^2\}\cdots\\\\ =&\frac\{1\}\{\varepsilon^2\}\oint\_\{(\xi-x)^2+(\eta-y)^2=\varepsilon^2\}[-(\eta-y)\mathrm\{ d\} \xi+(\xi-x)\mathrm\{ d\} \eta]\\\\ \xlongequal\{\tiny\mbox\{Green\}\}&\frac\{1\}\{\varepsilon^2\}\iint\_\{(\xi-x)^2+(\eta-y)^2\leq \varepsilon^2\}2\mathrm\{ d\} \xi\mathrm\{ d\} \eta=2\pi. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
796、 (2)、 已知 $\displaystyle f$ 连续可微, 且 $\displaystyle f(1)=2, f(4)=3$, 求 $\displaystyle \oint\_L\frac\{f(xy)\}\{y\}\mathrm\{ d\} y$, 其中 $\displaystyle L$ 是 $\displaystyle y=x, y=4x, xy=1, xy=4$ 所围区域的边界, 取逆时针方向. (武汉大学2023年数学分析考研试题) [曲线曲面积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle L$ 围成的区域为 $\displaystyle \varOmega$, 则
\begin\{aligned\} \mbox\{原式\}\xlongequal\{\tiny\mbox\{Green\}\} \iint\_\varOmega \frac\{1\}\{y\}f'(xy)\cdot x\mathrm\{ d\} x\mathrm\{ d\} y =\iint\_\varOmega f'(xy)\mathrm\{ d\} x\mathrm\{ d\} y. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle \frac\{y\}\{x\}=u, xy=v$, 则 $\displaystyle \left|\frac\{\partial(u,v)\}\{\partial(x,y)\}\right|=2u$, 而
\begin\{aligned\} \mbox\{原式\}=&\iint\_\{[1,4]^2\}f'(v)\frac\{1\}\{2u\}\mathrm\{ d\} u\mathrm\{ d\} v =\frac\{1\}\{2\}\int\_1^4\frac\{\mathrm\{ d\} u\}\{u\}\int\_1^4 f'(v)\mathrm\{ d\} v\\\\ =&\frac\{1\}\{2\}\ln 4[f(4)-f(1)]=\ln 2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
797、 (3)、 求曲面积分 $\displaystyle \iint\_S z\left(\frac\{\alpha x\}\{a^2\}+\frac\{\beta y\}\{b^2\}+\frac\{\gamma z\}\{c^2\}\right)\mathrm\{ d\} S$, 其中 $\displaystyle S$ 为 $\displaystyle \frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}+\frac\{z^2\}\{c^2\}=1$ 的上半部分, $\displaystyle \alpha,\beta,\gamma$ 为 $\displaystyle S$ 的外方向余弦. (武汉大学2023年数学分析考研试题) [曲线曲面积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle \frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}+\frac\{z^2\}\{c^2\}=1$ 知
\begin\{aligned\} &\frac\{x\}\{a^2\}+\frac\{zz\_x\}\{c^2\}=0, \frac\{y\}\{b^2\}+\frac\{zz\_y\}\{c^2\}=0\\\\ \Rightarrow& \frac\{z^2\}\{c^4\}(1+z\_x^2+z\_y^2)=\frac\{x^2\}\{a^4\}+\frac\{y^2\}\{b^4\}+\frac\{z^2\}\{c^4\}\\\\ \Rightarrow& z\sqrt\{1+z\_x^2+z\_y^2\}=c^2\sqrt\{\frac\{x^2\}\{a^4\}+\frac\{y^2\}\{b^4\}+\frac\{z^2\}\{c^4\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
又设 $\displaystyle A=\sqrt\{\frac\{x^2\}\{a^4\}+\frac\{y^2\}\{b^4\}+\frac\{z^2\}\{c^4\}\}$, 则
\begin\{aligned\} \mbox\{原式\}&=\iint\_S z\left(\frac\{x\}\{a^2\}\cdot\frac\{\frac\{x\}\{a^2\}\}\{A\} +\frac\{y\}\{b^2\}\cdot\frac\{\frac\{y\}\{b^2\}\}\{A\} +\frac\{z\}\{c^2\}\cdot\frac\{\frac\{z\}\{c^2\}\}\{A\}\right)\mathrm\{ d\} S\\\\ &=\iint\_S z\sqrt\{\frac\{x^2\}\{a^4\}+\frac\{y^2\}\{b^4\}+\frac\{z^2\}\{c^4\}\}\mathrm\{ d\} S =\iint\_\{\frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}\leq 1\} c^2\left(\frac\{x^2\}\{a^4\}+\frac\{y^2\}\{b^4\}+\frac\{z^2\}\{c^4\}\right)\mathrm\{ d\} x\mathrm\{ d\} y\\\\ &=\iint\_\{\frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}\leq 1\} \left(\frac\{c^2\}\{a^4\}x^2 +\frac\{c^2\}\{b^4\}y^2+1-\frac\{x^2\}\{a^2\}-\frac\{y^2\}\{b^2\}\right)\mathrm\{ d\} x\mathrm\{ d\} y\\\\ &=ab\iint\_\{u^2+v^2\leq 1\} \left(\frac\{c^2\}\{a^2\}u^2+\frac\{c^2\}\{b^2\}v^2+1-u^2-v^2\right)\mathrm\{ d\} u\mathrm\{ d\} v\\\\ &\xlongequal\{\tiny\mbox\{对称性\}\} ab\left\[\frac\{c^2\}\{2\}\left(\frac\{1\}\{a^2\}+\frac\{1\}\{b^2\}\right)-1\right\] \iint\_\{u^2+v^2\leq 1\}(u^2+v^2)\mathrm\{ d\} u\mathrm\{ d\} v +\pi ab\\\\ &=ab\left\[\frac\{c^2\}\{2\}\left(\frac\{1\}\{a^2\}+\frac\{1\}\{b^2\}\right)-1\right\] \int\_0^1 r^2\cdot 2\pi r\mathrm\{ d\} r +\pi ab\\\\ &=\frac\{\pi ab\}\{2\}\left\[\frac\{c^2\}\{2\}\left(\frac\{1\}\{a^2\}+\frac\{1\}\{b^2\}\right)-1\right\] +\pi ab =\frac\{\pi abc^2\}\{4\}\left(\frac\{1\}\{a^2\}+\frac\{1\}\{b^2\}+\frac\{1\}\{c^2\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
798、 (2)、 求曲线积分 $\displaystyle \int\_L [(x^2+y^2)^2+y^2]\mathrm\{ d\} s$, 其中 $\displaystyle L: x^2+y^2=x$. (武汉理工大学2023年数学分析考研试题) [曲线曲面积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}&\stackrel\{x^2+y^2=x\}\{=\}\int\_L (x^2+y^2)\mathrm\{ d\} s \stackrel\{x^2+y^2=x\}\{=\}\int\_L x\mathrm\{ d\} s\\\\ &\stackrel\{x=\frac\{1\}\{2\}+\frac\{1\}\{2\}\cos\theta, y=\frac\{1\}\{2\}\sin \theta\}\{=\}\int\_0^\{2\pi\}\left(\frac\{1\}\{2\}+\frac\{1\}\{2\}\cos\theta\right)\cdot \frac\{1\}\{2\}\mathrm\{ d\} \theta\\\\ &\xlongequal\{\tiny\mbox\{对称性\}\} \frac\{1\}\{4\}\cdot 2\pi=\frac\{\pi\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
799、 (6)、 求曲面积分
\begin\{aligned\} \iint\_S xz\mathrm\{ d\} y\mathrm\{ d\} z-x^2y\mathrm\{ d\} z\mathrm\{ d\} x+y^2z\mathrm\{ d\} x\mathrm\{ d\} y, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle S$ 由 $\displaystyle z=x^2+y^2$, 柱面 $\displaystyle x^2+y^2=1$ 以及三个坐标面在第一个卦限所围曲面外侧. (武汉理工大学2023年数学分析考研试题) [曲线曲面积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}\xlongequal\{\tiny\mbox\{Gauss\}\}&\iiint\_\{z\leq x^2+y^2\leq 1\atop x,y\geq 0\} (z-x^2+y^2)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z \xlongequal\{\tiny\mbox\{对称性\}\} \iiint\_\{z\leq x^2+y^2\leq 1\atop x,y\geq 0\} z\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z\\\\ =&\int\_0^1 z\mathrm\{ d\} z\iint\_\{z\leq x^2+y^2\leq 1\atop x,y\geq 0\}s\mathrm\{ d\} x\mathrm\{ d\} y =\int\_0^1 z\frac\{\pi(1-z)\}\{4\}\mathrm\{ d\} z=\frac\{\pi\}\{24\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
800、
(10)、 $\displaystyle \int\_\{x^2+y^2=1\}\ln \left\[(x-4)^2+y^2\right\]\mathrm\{ d\} s=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (西安交通大学2023年数学分析考研试题) [曲线曲面积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) /
\begin\{aligned\} \mbox\{原式\}=&\int\_\{x^2+y^2=1\}\ln(17-8x)\mathrm\{ d\} s=\int\_0^\{2\pi\} \ln (17-8\cos\theta)\mathrm\{ d\} \theta\\\\ \stackrel\{\theta-\pi=\tau\}\{=\}&\int\_\{-\pi\}^\pi \ln (17+8\cos\tau)\mathrm\{ d\} \tau \xlongequal\{\tiny\mbox\{对称性\}\} 2\int\_0^\pi \ln (17+8\cos\tau)\mathrm\{ d\} \tau\\\\ =&2\pi \ln 17+2\int\_0^\pi \ln \left(1+\frac\{8\}\{17\}\cos\tau\right)\mathrm\{ d\} \tau. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle I(s)=\int\_0^\pi \ln\left(1+s\cos\tau\right)\mathrm\{ d\} \tau, 0 < s < 1$, 则
\begin\{aligned\} I'(s)=&\int\_0^\pi \frac\{\cos\tau\}\{1+s\cos\tau\}\mathrm\{ d\} \tau \stackrel\{\tan\frac\{\tau\}\{2\}=u\}\{=\}\int\_0^\infty \frac\{\frac\{1-u^2\}\{1+u^2\}\}\{1+s\frac\{1-u^2\}\{1+u^2\}\}\frac\{2\}\{1+u^2\}\mathrm\{ d\} u\\\\ =&2\int\_0^\infty \frac\{1-u^2\}\{(1+u^2)[(1+s)+(1-s)u^2]\}\mathrm\{ d\} u\\\\ =&\frac\{2\}\{s\}\int\_0^\infty \left\[\frac\{1\}\{1+u^2\}-\frac\{1\}\{(1+s)+(1-s)u^2\}\right\]\mathrm\{ d\} u\\\\ =&\frac\{2\}\{s\}\left(\frac\{\pi\}\{2\}-\frac\{\pi\}\{2\sqrt\{1-s^2\}\}\right) =\frac\{\pi\}\{s\}\left(1-\frac\{1\}\{\sqrt\{1-s^2\}\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} I(s)=&\int\_0^s \frac\{\pi\}\{\tau\}\left(1-\frac\{1\}\{\sqrt\{1-\tau^2\}\}\right)\mathrm\{ d\} \tau\stackrel\{\tau=\sin \theta\}\{=\}\cdots=\pi \ln \frac\{1+\sqrt\{1-s^2\}\}\{2\},\\\\ \mbox\{原式\}=&2\pi \ln 17+2\pi \ln \frac\{1+\frac\{15\}\{17\}\}\{2\}=8\pi \ln 2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
801、 6、 (15 分) 求曲面积分
\begin\{aligned\} I=\iint\_\varSigma 4xz\mathrm\{ d\} y\mathrm\{ d\} z-2yz\mathrm\{ d\} z\mathrm\{ d\} x+(1-z^2)\mathrm\{ d\} x\mathrm\{ d\} y, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle \varSigma$ 是由 $\displaystyle z=\mathrm\{e\}^y\ (0\leq y\leq 1)$ 绕 $\displaystyle z$ 轴旋转一周所得的曲面, 方向取下侧. (西北大学2023年数学分析考研试题) [曲线曲面积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle \varSigma: x^2+y^2\leq \mathrm\{e\}^\{2a\}, z=\mathrm\{e\}^a$, 取上侧, 则
\begin\{aligned\} \iint\_\varSigma \cdots =(1-\mathrm\{e\}^\{2a\}) \iint\_\varSigma \mathrm\{ d\} x\mathrm\{ d\} y =(1-\mathrm\{e\}^\{-2a\})\cdot \pi \mathrm\{e\}^\{2a\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} \mbox\{原式\}=&-\left\[\iint\_\varSigma-\iint\_S\cdots\right\]+\iint\_\varSigma\cdots\\\\ \xlongequal\{\tiny\mbox\{Gauss\}\}& -\iiint\_\{\mathrm\{e\}^\{\sqrt\{x^2+y^2\}\}\leq z\atop 0\leq z\leq \mathrm\{e\}^a\} (4z-2z-2z)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z +\iint\_\varSigma\cdots\\\\ =&0+(1-\mathrm\{e\}^\{-2a\})\cdot \pi \mathrm\{e\}^\{2a\} =(1-\mathrm\{e\}^\{-2a\})\cdot \pi \mathrm\{e\}^\{2a\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
802、 2、 计算积分. (1)、 (10 分) 求
\begin\{aligned\} I=\int\_L(\mathrm\{e\}^x+1)\cos y\mathrm\{ d\} x-[(\mathrm\{e\}^x+x)\sin y-x]\mathrm\{ d\} y, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle L$ 为由点 $\displaystyle A(2,0)$ 沿曲线 $\displaystyle y=\sqrt\{4-x^2\}$ 到点 $\displaystyle B(-2,0)$ 的有限曲线段. (西南大学2023年数学分析考研试题) [曲线曲面积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle P=(\mathrm\{e\}^x+1)\cos y, Q=-[(\mathrm\{e\}^x+x)\sin y-x]$, 则 $\displaystyle Q\_x-P\_y=1$. 记 $\displaystyle \ell: (-2,0)\xrightarrow\{y=0\}\{\to\}(2,0)$, 则
\begin\{aligned\} \int\_\ell\cdots =\int\_\{-2\}^2 (\mathrm\{e\}^x+1)\mathrm\{ d\} x=\mathrm\{e\}^2-\mathrm\{e\}^\{-2\}+4. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} I=&\left(I+\int\_\ell\cdots\right)-\iint\_\ell\cdots\\\\ \xlongequal\{\tiny\mbox\{Green\}\}&\iint\_\{0\leq y\leq \sqrt\{4-x^2\}\atop 0\leq x\leq 2\}1\mathrm\{ d\} x\mathrm\{ d\} y-(\mathrm\{e\}^2-\mathrm\{e\}^\{-2\}+4)\\\\ =&\frac\{\pi \cdot 2^2\}\{2\}-(\mathrm\{e\}^2-\mathrm\{e\}^\{-2\}+4) =2\pi+\mathrm\{e\}^\{-2\}-\mathrm\{e\}^2-4. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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---
803、 11、 (10 分)设 $\displaystyle S$ 是锥面 $\displaystyle z^2=x^2+y^2$ 与平面 $\displaystyle z=\sqrt\{2\}\left(\frac\{x\}\{2\}+1\right)$ 所围区域 $\displaystyle V$ 的全表面, 求 $\displaystyle S$ 的面积. (西南交通大学2023年数学分析考研试题) [曲线曲面积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由
\begin\{aligned\} \sqrt\{1+z\_x^2+z\_y^2\}=\left\\{\begin\{array\}\{llllllllllll\}\sqrt\{2\},&z=\sqrt\{x^2+y^2\},\\\\ \sqrt\{\frac\{3\}\{2\}\},&z=\sqrt\{2\}\left(\frac\{x\}\{2\}+1\right)\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及两曲面的交线在 $\displaystyle xOy$ 面上的投影为 $\displaystyle x^2+y^2=\sqrt\{2\}\left(\frac\{x\}\{2\}+1\right)\Leftrightarrow \left(x-\frac\{\sqrt\{2\}\}\{4\}\right)^2+y^2=\sqrt\{2\}+\frac\{1\}\{8\}$ 知所求
\begin\{aligned\} =&\iint\_\{\left(x-\frac\{\sqrt\{2\}\}\{4\}\right)^2+y^2\leq\sqrt\{2\}+\frac\{1\}\{8\}\} \left(\sqrt\{2\}+\sqrt\{\frac\{3\}\{2\}\}\right)\mathrm\{ d\} x\mathrm\{ d\} y\\\\ =&\left(\sqrt\{2\}+\sqrt\{\frac\{3\}\{2\}\}\right)\cdot \pi \left(\sqrt\{2\}+\frac\{1\}\{8\}\right) =\frac\{\pi(16+\sqrt\{2\})(2+\sqrt\{3\})\}\{16\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
804、 12、 (10 分) 设 $\displaystyle S$ 是单位球面 $\displaystyle x^2+y^2+z^2=1$ 的外侧, 求曲面积分
\begin\{aligned\} \iint\_S \frac\{x\mathrm\{ d\} y\mathrm\{ d\} z+y\mathrm\{ d\} z\mathrm\{ d\} x+z\mathrm\{ d\} x\mathrm\{ d\} y\}\{\left(2x^2+2y^2+z^2\right)^\frac\{3\}\{2\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(西南交通大学2023年数学分析考研试题) [曲线曲面积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle (P,Q,R)=\frac\{(x,y,z)\}\{(2x^2+2y^2+z^2)^\frac\{3\}\{2\}\}$, 则 $\displaystyle P\_x+Q\_y+R\_z=0$, 而
\begin\{aligned\} \mbox\{原式\}&\xlongequal\{\tiny\mbox\{Gauss\}\} \iint\_\{2x^2+2y^2+z^2=\varepsilon^2\ll 1\}P\mathrm\{ d\} y\mathrm\{ d\} z+Q\mathrm\{ d\} z\mathrm\{ d\} x+R\mathrm\{ d\} x\mathrm\{ d\} y\\\\ &=\frac\{1\}\{\varepsilon^3\}\iint\_\{2x^2+2y^2+z^2=\varepsilon^2\ll 1\}x\mathrm\{ d\} y\mathrm\{ d\} z+y\mathrm\{ d\} z\mathrm\{ d\} x+z\mathrm\{ d\} x\mathrm\{ d\} y\\\\ &\xlongequal\{\tiny\mbox\{Gauss\}\}\frac\{1\}\{\varepsilon^3\}\iiint\_\{2x^2+2y^2+z^2\leq \varepsilon^2\} 3\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z =\frac\{3\}\{\varepsilon^3\} \cdot \frac\{4\pi\}\{3\}\cdot \frac\{1\}\{\sqrt\{2\}\}\cdot\frac\{1\}\{\sqrt\{2\}\}\cdot 1 =2\pi. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
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---
805、 3、 设 $\displaystyle a > 0$, 计算曲线积分
\begin\{aligned\} I=\int\_L (\mathrm\{e\}^x\sin y-y^2)\mathrm\{ d\} x+\mathrm\{e\}^x\cos y\mathrm\{ d\} y, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle L$ 是从 $\displaystyle A (a,0)$ 到 $\displaystyle O (0,0)$ 的上半圆周 $\displaystyle x^2+y^2=ax$. (湘潭大学2023年数学分析考研试题) [曲线曲面积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle \ell: y=0, x: 0\to a$, 则 $\displaystyle \int\_\ell \cdots=0$, 而
\begin\{aligned\} I&=I+\int\_\{\ell\}\cdots\xlongequal\{\tiny\mbox\{Green\}\} \iint\_\{x^2+y^2\leq ax\atop y\geq 0\}2y\mathrm\{ d\} x\mathrm\{ d\} y\\\\ &\stackrel\{x=r\cos\theta, y=r\sin\theta\}\{=\}\int\_0^\frac\{\pi\}\{2\}\mathrm\{ d\} \theta\int\_0^\{a\cos \theta\}2r\sin\theta\cdot r\mathrm\{ d\} r =\frac\{a^3\}\{6\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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发表于 2023-3-5 09:13:35
