## 张祖锦2023年数学专业真题分类70天之第36天
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806、 4、 计算曲面积分
\begin\{aligned\} I=\iint\_S 2x^2\mathrm\{ d\} y\mathrm\{ d\} z+2y^2\mathrm\{ d\} z\mathrm\{ d\} x+2z^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 $\displaystyle S$ 是锥面 $\displaystyle x^2+y^2=z^2\ (0\leq z\leq h)$ 的外侧. (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 $\displaystyle \varSigma: x^2+y^2\leq h^2, z=h$, 取上侧, 则
\begin\{aligned\} \iint\_\varSigma \cdots=\iint\_\{x^2+y^2\leq h^2\}\mathrm\{ d\} x\mathrm\{ d\} y=2\pi h^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+\iint\_\varSigma\cdots\right)-\iint\_\varSigma\cdots\\\\ &\xlongequal\{\tiny\mbox\{Gauss\}\} \iiint\_\{x^2+y^2\leq z^2\atop 0\leq z\leq h\} (4x+4y+4z)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z-2\pi h^4\\\\ &\xlongequal\{\tiny\mbox\{对称性\}\} 4 \iiint\_\{x^2+y^2\leq z^2\atop 0\leq z\leq h\} z\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z-2\pi h^4\\\\ &=4\int\_0^h z\cdot \pi z^2\mathrm\{ d\} z-2\pi h^4=-\pi h^4. \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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---
807、 10、 (15 分) 计算
\begin\{aligned\} \int\_\{\widetilde\{AB\}\}(2x\cos y-y^2\sin x+x^2y)\mathrm\{ d\} x +(2y\cos x-x^2\sin 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\}
其中曲线 $\displaystyle \widetilde\{AB\}$ 为圆 $\displaystyle x^2+y^2=1$ 的上半部分, 起点为 $\displaystyle A=(1,0)$, 终点为 $\displaystyle B=(-1,0)$. (新疆大学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 $\displaystyle \ell: y=0, x: -1\to 1$, 则 $\displaystyle \int\_\{\ell\} \cdots=\int\_\{-1\}^1 2x\mathrm\{ d\} x=0$. 于是
\begin\{aligned\} \mbox\{原式\}=&\mbox\{原式\}+\int\_\{\ell\} \cdots\xlongequal\{\tiny\mbox\{Green\}\} \iint\_\{x^2+y^2\leq 1\atop y\geq 0\}(-x^2)\mathrm\{ d\} x\mathrm\{ d\} y\\\\ \xlongequal\{\tiny\mbox\{对称性\}\}& -\frac\{1\}\{2\}\iint\_\{x^2+y^2\leq 1\atop y\geq 0\}(x^2+y^2)\mathrm\{ d\} x\mathrm\{ d\} y =-\frac\{1\}\{2\}\int\_0^1 r^2\mathrm\{ d\} r\int\_0^\pi r\mathrm\{ d\} \theta=-\frac\{\pi\}\{8\}. \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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---
808、 7、 (15 分) 计算曲线积分 $\displaystyle I=\oint\_L \frac\{x\mathrm\{ d\} y-y\mathrm\{ d\} x\}\{x^2+y^2\}$, 其中 $\displaystyle L$ 是不经过原点的简单封闭曲线, 且取逆时针方向. (云南大学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 $\displaystyle P=\frac\{-y\}\{x^2+y^2\}, Q=\frac\{x\}\{x^2+y^2\}$, 则 $\displaystyle Q\_x=P\_y$. (1)、 若原点不在 $\displaystyle L$ 内部, 则由 Green 公式, 原式 $\displaystyle =0$. (2)、 若原点在 $\displaystyle L$ 内部, 则
\begin\{aligned\} \mbox\{原式\}&\xlongequal\{\tiny\mbox\{Green\}\}\oint\_\{x^2+y^2=\varepsilon^2\} \frac\{x\mathrm\{ d\} y-y\mathrm\{ d\} x\}\{x^2+y^2\} =\frac\{1\}\{\varepsilon^2\}\oint\_\{x^2+y^2=\varepsilon^2\} -y\mathrm\{ d\} x+x\mathrm\{ d\} y\\\\ &\xlongequal\{\tiny\mbox\{Green\}\}\frac\{1\}\{\varepsilon^2\}\iint\_\{x^2+y^2 < \varepsilon^2\}2\mathrm\{ d\} x\mathrm\{ d\} y =\frac\{2\}\{\varepsilon^2\}\cdot \pi \varepsilon^2=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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---
809、 (5)、 设 $\displaystyle \varSigma$ 是下半球面 $\displaystyle z=-\sqrt\{a^2-x^2-y^2\}$ 的下侧, $\displaystyle \varOmega$ 是 $\displaystyle \varSigma$ 与 $\displaystyle z=0$ 所围成的空间闭区域, 则曲面积分 $\displaystyle \iint\_\varSigma z\mathrm\{ d\} x\mathrm\{ d\} y=\underline\{\ \ \ \ \ \ \ \ \ \ \}$.
A. $\displaystyle -\iiint\_\varOmega \mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z$
B. $\displaystyle \int\_0^\{2\pi\}\mathrm\{ d\} \theta\int\_0^a \sqrt\{a^2-\rho^2\}\rho\mathrm\{ d\} \rho$
C. $\displaystyle -\int\_0^\{2\pi\}\mathrm\{ d\} \theta\int\_0^a \sqrt\{a^2-\rho^2\}\rho\mathrm\{ d\} \rho$
D. $\displaystyle \iint\_\varSigma (x+y+z)\mathrm\{ d\} x\mathrm\{ d\} y$ (长安大学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) / $\displaystyle B$.
\begin\{aligned\} \mbox\{原式\}=-\iint\_\{x^2+y^2\leq a^2\}z\mathrm\{ d\} x\mathrm\{ d\} y=-\int\_0^a \left(-\sqrt\{a^2-\rho^2\}\right) 2\pi \rho\mathrm\{ d\} \rho. \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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---
810、 (3)、 若 $\displaystyle L$ 是以 $\displaystyle O (0,0), A (1,0), B (0,1)$ 为顶点的三角形, 则 $\displaystyle \oint\_L (x+y)\mathrm\{ d\} s=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (长安大学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\_\{y=0, x\in [0,1]\}+\int\_\{y=x, x\in [0,1]\}+\int\_\{x=0, y\in [0,1]\}\cdots\\\\ =&\int\_0^1 x\mathrm\{ d\} x+\int\_0^1 (x+x)\sqrt\{2\}\mathrm\{ d\} x+\int\_0^1 y\mathrm\{ d\} y=\sqrt\{2\}+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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---
811、 (6)、 设 $\displaystyle S$ 表示立方体 $\displaystyle 0\leq x,y,z\leq a$ 的外表面, 则
\begin\{aligned\} \iint\_S x^2\mathrm\{ d\} y\mathrm\{ d\} z+y^2\mathrm\{ d\} z\mathrm\{ d\} x+z^2\mathrm\{ d\} x\mathrm\{ d\} y=\underline\{\ \ \ \ \ \ \ \ \ \ \}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(长安大学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\_\{[0,a]^3\} (2x+2y+2z)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z \xlongequal\{\tiny\mbox\{对称性\}\} 6\iiint\_\{[0,a]^3\} x\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z\\\\ =&3a^2\int\_0^a 2x\mathrm\{ d\} x=3a^4. \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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---
812、 (5)、 (10 分) 计算 $\displaystyle I=\oint\_L \frac\{\partial f\}\{\partial \vec\{n\}\}\mathrm\{ d\} s$, 其中 $\displaystyle f(x,y)=x^2+y^2$, $\displaystyle L$ 为椭圆 $\displaystyle 2x^2+y^2=1$, $\displaystyle \vec\{n\}$ 为 $\displaystyle L$ 的外法线向量. (长安大学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\} I&\xlongequal\{\tiny\mbox\{Green\}\} \iint\_\{2x^2+y^2\leq 1\}\Delta f\mathrm\{ d\} x\mathrm\{ d\} y =4\iint\_\{2x^2+y^2\leq 1\}\mathrm\{ d\} x\mathrm\{ d\} y=4\cdot\frac\{\pi\}\{\sqrt\{2\}\}=2\sqrt\{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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---
813、 1、 计算题 (每题 10 分, 共 40 分). (1)、 求第一型曲面积分
\begin\{aligned\} \iint\_S \sqrt\{1+x^2+y^2\}\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\}
其中 $\displaystyle S$ 为螺旋面
\begin\{aligned\} x=u\cos v, y=u\sin v, z=v, \quad 0\leq u\leq 1, 0\leq v\leq \pi. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(中国海洋大学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) / 算出
\begin\{aligned\} E=&x\_u^2+y\_u^2+z\_u^2=\cos^2v+\sin^2v=1,\\\\ F=&x\_ux\_v+y\_uy\_v+z\_uz\_v=-u\sin v\cos v+u\sin v\cos v=0,\\\\ G=&x\_v^2+y\_v^2+z\_w^2=u^2\sin^2v+u^2\cos^2v+1=1+u^2, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
我们知 $\displaystyle \sqrt\{EG-F^2\}=\sqrt\{1+u^2\}$,
\begin\{aligned\} \mbox\{原式\}=&\iint\_\{0\leq u\leq 1\atop 0\leq v\leq \pi\} \sqrt\{1+u^2\}\cdot\sqrt\{1+u^2\}\mathrm\{ d\} u\mathrm\{ d\} v\\\\ =&\pi \int\_0^1 (1+u^2)\mathrm\{ d\} u=\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\}
跟锦数学微信公众号. [在线资料](
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---
814、 (4)、 设实轴上函数 $\displaystyle f(x)$ 有连续一阶导函数, $\displaystyle f(0)=-\frac\{1\}\{2\}$, 且曲线积分
\begin\{aligned\} \int\_C [\mathrm\{e\}^x-f(x)]y\mathrm\{ d\} x+f(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\}
与路径无关, 求 $\displaystyle f(x)$. (中国海洋大学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) / 由题设,
\begin\{aligned\} &f'(x)=\mathrm\{e\}^x-f(x)\Rightarrow [\mathrm\{e\}^x f(x)]'=\mathrm\{e\}^\{2x\}\\\\ \Rightarrow&\mathrm\{e\}^xf(x)-\mathrm\{e\}^0f(0)=\frac\{\mathrm\{e\}^\{2x\}-1\}\{2\}\\\\ \Rightarrow& f(x)=\mathrm\{e\}^\{-x\}\frac\{\mathrm\{e\}^\{2x\}-2\}\{2\}=\frac\{\mathrm\{e\}^x-2\mathrm\{e\}^\{-x\}\}\{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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---
815、 4、 (15 分) 求
\begin\{aligned\} I=\oint\_L y\mathrm\{ d\} x+z\mathrm\{ d\} y+x\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 $\displaystyle x^2+y^2+z^2=1$ 与 $\displaystyle x+y+z=1$ 的交线, 从 $\displaystyle x$ 轴正向看去为逆时针方向. (中国海洋大学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\} I&\xlongequal\{\tiny\mbox\{Stokes\}\} \iint\_\{x+y+z=1\atop x^2+y^2+z^2\leq 1\} \left|\begin\{array\}\{cccccccccc\}\frac\{1\}\{\sqrt\{3\}\}&\frac\{1\}\{\sqrt\{3\}\}&\frac\{1\}\{\sqrt\{3\}\}\\\\ \frac\{\partial\}\{\partial x\}&\frac\{\partial\}\{\partial y\}&\frac\{\partial\}\{\partial z\}\\\\ y&z&x\end\{array\}\right|\mathrm\{ d\} S\\\\ &=\iint\_\{x+y+z=1\atop x^2+y^2+z^2\leq 1\} \frac\{1\}\{\sqrt\{3\}\}(-1-1-1)\mathrm\{ d\} S =-\sqrt\{3\}\iint\_\{x+y+z=1\atop x^2+y^2+z^2\leq 1\} \mathrm\{ d\} S\\\\ &=-\sqrt\{3\}\cdot \pi \left\[1^2-\left(\frac\{1\}\{\sqrt\{3\}\}\right)^2\right\] =-\frac\{2\sqrt\{3\}\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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---
816、 7、 (15 分)求第二型曲面积分
\begin\{aligned\} I=\iint\_\varSigma \frac\{3x\mathrm\{ d\} y\mathrm\{ d\} z+(z+3)^2\mathrm\{ d\} x\mathrm\{ d\} y\}\{(x^2+y^2+z^2)^\frac\{1\}\{2\}\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle \varSigma$ 为下半球面 $\displaystyle z=-\sqrt\{9-x^2-y^2\}$, 方向为上侧. (中国海洋大学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\} I=\frac\{1\}\{3\}\iint\_\varSigma 3x\mathrm\{ d\} y\mathrm\{ d\} z+(z+3)^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\}
设 $\displaystyle S: x^2+y^2\leq 9, z=0$, 取上侧, 则
\begin\{aligned\} \iint\_S3x\mathrm\{ d\} y\mathrm\{ d\} z+(z+3)^2\mathrm\{ d\} x\mathrm\{ d\} y =9\iint\_\{x^2+y^2\leq 9\}\mathrm\{ d\} x\mathrm\{ d\} y =9\cdot \pi 9=81\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\} I=&-\frac\{1\}\{3\}\left\[\begin\{array\}\{c\}\iint\_\{\varSigma\mbox\{下侧\}\}3x\mathrm\{ d\} y\mathrm\{ d\} z+(z+3)^2\mathrm\{ d\} x\mathrm\{ d\} y\\\\+\iint\_S3x\mathrm\{ d\} y\mathrm\{ d\} z+(z+3)^2\mathrm\{ d\} x\mathrm\{ d\} y\end\{array\}\right\]\\\\ &+\frac\{1\}\{3\}\iint\_S 3x\mathrm\{ d\} y\mathrm\{ d\} z+(z+3)^2\mathrm\{ d\} x\mathrm\{ d\} y\\\\ \xlongequal\{\tiny\mbox\{Gauss\}\}&-\frac\{1\}\{3\}\iiint\_\{x^2+y^2+z^2\leq 9\atop -3\leq z\leq 0\} [3+2(z+3)]\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z+27\pi\\\\ =&-\frac\{1\}\{3\}\int\_\{-3\}^0 (2z+9)\cdot \pi(9-z^2)\mathrm\{ d\} z+27\pi=-\frac\{27\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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---
817、 5、 (15 分) 设 $\displaystyle \mathbb\{R\}^3$ 中有满足右手法则的坐标系 $\displaystyle (x,y,z)$. 又设 $\displaystyle \varGamma$ 是平面 $\displaystyle x+y=2$ 和球面
\begin\{aligned\} (x-1)^2+(y-1)^2+z^2=2 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
相交所得到的圆周. 从原点看过去, 以顺时针方向为 $\displaystyle \varGamma$ 的正定向. 计算
\begin\{aligned\} \int\_\varGamma y\mathrm\{ d\} x+z\mathrm\{ d\} y+x\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\}
(中国科学技术大学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) / 从 $\displaystyle y$ 轴正向看去, $\displaystyle \varGamma$ 为逆时针, 而
\begin\{aligned\} \mbox\{原式\}\xlongequal\{\tiny\mbox\{Stokes\}\}& \iint\_\{(x-1)^2+(y-1)^2+z^2\leq 2\atop x+y=2\} \left|\begin\{array\}\{cccccccccc\}\frac\{1\}\{\sqrt\{2\}\}&\frac\{1\}\{\sqrt\{2\}\}&0\\\\ \frac\{\partial\}\{\partial x\}&\frac\{\partial \}\{\partial y\}&\frac\{\partial\}\{\partial z\}\\\\ y&z&x \end\{array\}\right|\mathrm\{ d\} S\\\\ =&-\sqrt\{2\}\iint\_\{(x-1)^2+(y-1)^2+z^2\leq 2\atop x+y=2\} \mathrm\{ d\} S =-\sqrt\{2\}\cdot \pi (\sqrt\{2\})^2 =-2\sqrt\{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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---
818、 8、 计算
\begin\{aligned\} \int\_L \frac\{\left(x-\frac\{1\}\{2\}-y\right)\mathrm\{ d\} x+\left(x-\frac\{1\}\{2\}+y\right)\mathrm\{ d\} y\}\{\left(x-\frac\{1\}\{2\}\right)^2+y^2\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle L$ 是连接 $\displaystyle (0,-1), (0,1)$ 的曲线且位于 $\displaystyle \left(\frac\{1\}\{2\},0\right)$ 的右侧. (中国科学院大学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 $\displaystyle (P,Q)=\frac\{\left(x-\frac\{1\}\{2\}-y,x-\frac\{1\}\{2\}+y\right)\}\{\left(x-\frac\{1\}\{2\}\right)^2+y^2\}$, 则 $\displaystyle Q\_x=P\_y$. 设 $\displaystyle \ell: (1,0)\xrightarrow\{x=0\}(-1,0)$, 则在 $\displaystyle \ell$ 上, $\displaystyle x=0, \mathrm\{ d\} x=0$,
\begin\{aligned\} \iint\_\ell \cdots=&\int\_1^\{-1\}\frac\{\left(y-\frac\{1\}\{2\}\right)\mathrm\{ d\} y\}\{\frac\{1\}\{4\}+y^2\}\\\\ \xlongequal\{\tiny\mbox\{对称性\}\}&\int\_0^1 \frac\{\mathrm\{ d\} y\}\{\frac\{1\}\{4\}+y^2\}\stackrel\{y=\frac\{1\}\{2\}\tan\theta\}\{=\}\cdots =2\arctan 2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
再取 $\displaystyle C\_\varepsilon: \left(x-\frac\{1\}\{2\}\right)^2+y^2=\varepsilon^2$, 方向为逆时针, 由
\begin\{aligned\} \mbox\{原式\}+\int\_\ell \cdots-\int\_\{C\_\varepsilon\} \cdots\xlongequal\{\tiny\mbox\{Green\}\} 0 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} \mbox\{原式\}=&-\int\_\{\ell\}+\int\_\{C\_\varepsilon\}\cdots\\\\ =&-2\arctan 2+\frac\{1\}\{\varepsilon^2\}\oint\_\{C\_\varepsilon\} \left(x-\frac\{1\}\{2\}-y\right)\mathrm\{ d\} x+\left(x-\frac\{1\}\{2\}+y\right)\mathrm\{ d\} y\\\\ \xlongequal\{\tiny\mbox\{Green\}\}&-2\arctan 2+\frac\{1\}\{\varepsilon^2\} \iint\_\{\left(x-\frac\{1\}\{2\}\right)^2+y^2\leq \varepsilon^2\} 2\mathrm\{ d\} x\mathrm\{ d\} y\\\\ =&2\pi-2\arctan 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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---
819、 9、 证明:
\begin\{aligned\} \frac\{1\}\{2\}\iint\_\varSigma \cos(\vec\{r\},\vec\{n\})\mathrm\{ d\} S=\iiint\_V \frac\{1\}\{r\}\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 $\displaystyle \varSigma$ 是包围 $\displaystyle V$ 的曲面, 原点在 $\displaystyle V$ 的外面,
\begin\{aligned\} r=\sqrt\{x^2+y^2+z^2\}, \vec\{r\}=(x,y,z), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
$\displaystyle \vec\{n\}$ 是 $\displaystyle \varSigma$ 上的单位外法向量. (中国科学院大学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\} &\quad \frac\{1\}\{2\}\iint\_\{\partial \varOmega\} \cos (\vec\{r\},\vec\{n\})\mathrm\{ d\} S =\frac\{1\}\{2\}\iint\_\{\partial\varOmega\} \frac\{\vec\{r\}\}\{r\}\cdot\vec\{n\}\mathrm\{ d\} S\\\\ &\xlongequal\{\tiny\mbox\{Gauss\}\}\frac\{1\}\{2\}\iiint\_\varOmega\left\[\frac\{\partial\}\{\partial x\}\frac\{x\}\{r\} +\frac\{\partial\}\{\partial y\}\frac\{y\}\{r\} +\frac\{\partial\}\{\partial z\}\frac\{z\}\{r\}\right\]\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z\\\\ &=\frac\{1\}\{2\}\iiint\_\varOmega\left\[\frac\{3\}\{r\}-\frac\{x^2+y^2+z^2\}\{r^3\}\right\]\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z =\iiint\_\varOmega\frac\{\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z\}\{r\}. \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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---
820、 3、 (10 分) 求曲线积分 $\displaystyle I=\oint\_L\frac\{x\mathrm\{ d\} y-y\mathrm\{ d\} x\}\{4x^2+9y^2\}$, 其中 $\displaystyle L$ 是单位圆周 $\displaystyle x^2+y^2=1$, 方向为逆时针. (中国矿业大学(北京)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 $\displaystyle a=2, b=3$, 则 $\displaystyle (P,Q)=\frac\{(-y,x)\}\{a^2x^2+b^2y^2\}$, 算出 $\displaystyle Q\_x=P\_y$ 即知
\begin\{aligned\} \mbox\{原式\}\xlongequal\{\tiny\mbox\{Green\}\}&\int\_\{a^2x^2+b^2y^2=\varepsilon^2\ll 1\}P\mathrm\{ d\} x+Q\mathrm\{ d\} y =\frac\{1\}\{\varepsilon^2\}\int\_\{a^2x^2+b^2y^2=\varepsilon^2\}-y\mathrm\{ d\} x+x\mathrm\{ d\} y\\\\ \xlongequal\{\tiny\mbox\{Green\}\}&\frac\{1\}\{\varepsilon^2\}\iint\_\{a^2x^2+b^2y^2\leq \varepsilon^2\}2\mathrm\{ d\} x\mathrm\{ d\} y \stackrel\{ax=u, by=v\}\{=\}\frac\{2\}\{ab\varepsilon^2\}\iint\_\{u^2+v^2\leq \varepsilon^2\}\mathrm\{ d\} u\mathrm\{ d\} v =\frac\{2\pi\}\{ab\}=\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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---
821、 10、 (15 分) 求曲面积分
\begin\{aligned\} \iint\_\varSigma x^3\mathrm\{ d\} y\mathrm\{ d\} z+y^3\mathrm\{ d\} z\mathrm\{ d\} x+(z^3+x^3+y^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 $\displaystyle \varSigma$ 为上半球面 $\displaystyle z=\sqrt\{1-x^2-y^2\}$, 取上侧. (中国矿业大学(北京)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
\begin\{aligned\} \iint\_S\cdots=-\iint\_\{x^2+y^2\leq 1\}(x^3+y^3)\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\}
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\} (3x^2+3y^2+3z^2)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z\\\\ &=3\int\_0^1 r^2\cdot\frac\{4\pi r^2\}\{2\}\mathrm\{ d\} r=\frac\{6\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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https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
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---
822、 10、 证明: 若 $\displaystyle S$ 为封闭曲面, $\displaystyle \vec\{l\}$ 为任意固定的单位向量, 则 $\displaystyle \iint\_S \cos\angle(\vec\{n\},\vec\{l\})\mathrm\{ d\} S=0$, 其中 $\displaystyle \vec\{n\}$ 是 $\displaystyle S$ 的单位外法向量, $\displaystyle \cos\angle(\vec\{n\},\vec\{l\})$ 是 $\displaystyle \vec\{n\}$ 和 $\displaystyle \vec\{l\}$ 的夹角余弦. (中国矿业大学(徐州)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 $\displaystyle \vec\{l\}=\left\\{a,b,c\right\\}$, 则
\begin\{aligned\} \iint\_S \cos\angle(\vec\{n\},\vec\{l\})\mathrm\{ d\} S=\iint\_S a\mathrm\{ d\} y\mathrm\{ d\} z+b\mathrm\{ d\} z\mathrm\{ d\} x+c\mathrm\{ d\} x\mathrm\{ d\} y\xlongequal\{\tiny\mbox\{Gauss\}\} 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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---
823、 9、 (15 分) 设定义在 $\displaystyle \mathbb\{R\}\backslash\left\\{0\right\\}$ 的函数 $\displaystyle f(t)$ 在 $\displaystyle (-\infty, 0)$ 及 $\displaystyle (0,+\infty)$ 有连续导数, $\displaystyle f(1)=0$. 试确定 $\displaystyle f$, 使得对任意与直线 $\displaystyle y=x$ 及 $\displaystyle y=-x$ 不相交的光滑闭曲线 $\displaystyle L$, 都有
\begin\{aligned\} \int\_L [2-f(x^2-y^2)]y\mathrm\{ d\} x+f(x^2-y^2)x\mathrm\{ d\} y=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(中国人民大学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 Green 公式知
\begin\{aligned\} &f'(x^2-y^2)\cdot 2x\cdot x+f(x^2-y^2)\\\\ &=-f'(x^2-y^2)\cdot(-2y)\cdot y+[2-f(x^2-y^2)],\\\\ &f'(x^2-y^2)(x^2-y^2)+f(x^2-y^2)=1,\\\\ &tf'(t)+f(t)=1.\qquad(I) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由
\begin\{aligned\} &ty'+y=0\Rightarrow \frac\{\mathrm\{ d\} y\}\{y\}=-\frac\{\mathrm\{ d\} t\}\{t\} \Rightarrow \ln |y|=-\ln |t|\Rightarrow y=\frac\{C\}\{t\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及常数变易法知可设 $\displaystyle f(t)=\frac\{C(t)\}\{t\}$. 代入 $\displaystyle (I)$ 得
\begin\{aligned\} &C'(t)=1\Rightarrow C(t)=t+C\\\\ \Rightarrow& f(t)=\frac\{t+C\}\{t\}=1+\frac\{C\}\{t\} \stackrel\{f(1)=0\}\{\Rightarrow\}f(t)=1-\frac\{1\}\{t\}. \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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---
824、 (3)、 计算积分
\begin\{aligned\} \oint\_L (\mathrm\{e\}^x-y^2)\mathrm\{ d\} x+(x^2+\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\}
其中 $\displaystyle L: \frac\{(x-1)^2\}\{2\}+\frac\{(y-2)^2\}\{8\}=1$, 方向为逆时针. (中南大学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) /
\begin\{aligned\} \mbox\{原式\}\xlongequal\{\tiny\mbox\{Gauss\}\}&2\iint\_\{\frac\{(x-1)^2\}\{2\}+\frac\{(y-2)^2\}\{8\}\leq 1\} (x+y)\mathrm\{ d\} x\mathrm\{ d\} y\\\\ \stackrel\{\boxed\{\begin\{array\}\{c\}x=1+\sqrt\{2\}r\cos\theta\\\\y=2+2\sqrt\{2\}r\sin \theta\end\{array\}\}\}\{=\}&2\int\_0^\{2\pi\}\mathrm\{ d\} \theta \int\_0^1 (3+\sqrt\{2\}r\cos\theta+2\sqrt\{2\}r\sin\theta)\cdot 4r\mathrm\{ d\} r\\\\ =&24\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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---
825、 7、 (15 分) 设平面 $\displaystyle 3x+2z=3$ 与柱面 $\displaystyle x^2+y^2=1$ 的交线为 $\displaystyle L$. (1)、 在交线 $\displaystyle L$ 上求一点 $\displaystyle P$, 使其到平面 $\displaystyle xOy$ 平面的距离最短; (2)、 求交线 $\displaystyle L$ 上上述点 $\displaystyle P$ 的切线方程; (3)、 若规定 $\displaystyle L$ 的方向为: 从 $\displaystyle x$ 轴的正向看去为逆时针方向, 试计算积分
\begin\{aligned\} \int\_L (z-x)\mathrm\{ d\} y+(x-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\}
(中南大学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) / (1)、 即求 $\displaystyle z$ 在约束条件 $\displaystyle 3x+2z=3, x^2+y^2=1$ 的极值. 这不用 Lagrange 乘数法. 为啥? 因为 $\displaystyle z=\frac\{3(1-x)\}\{2\}, x\in [-1,1]\Rightarrow \min z=z(1)=0, \max z=z(-1)=3$. 故距离最短为 $\displaystyle 3$. 此时, $\displaystyle x=-1, y=0, z=3$. (2)、 $\displaystyle 3x+2z=3$ 与柱面 $\displaystyle x^2+y^2=1$ 在 $\displaystyle P(-1,0,3)$ 处的切向量为
\begin\{aligned\} \left.\frac\{1\}\{-2\}\left\\{3,0,2\right\\}\times\left\\{x,y,0\right\\}\right|\_P=\left\\{0,1,0\right\\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
而切线方程为 $\displaystyle \frac\{x+1\}\{0\}=\frac\{y\}\{1\}=\frac\{z-3\}\{0\}$. (3)、
\begin\{aligned\} \mbox\{原式\}&\xlongequal\{\tiny\mbox\{Stokes\}\} \iint\_\{3x+2z=3\atop x^2+y^2\leq 1\} \left|\begin\{array\}\{cccccccccc\}\mathrm\{ d\} y\mathrm\{ d\} z&\mathrm\{ d\} z\mathrm\{ d\} x&\mathrm\{ d\} x\mathrm\{ d\} y\\\\ \frac\{\partial\}\{\partial x\}&\frac\{\partial \}\{\partial y\}&\frac\{\partial\}\{\partial z\}\\\\ 0&z-x&x-y \end\{array\}\right|\\\\ &=\iint\_\{3x+2z=3\atop x^2+y^2\leq 1\} -2\mathrm\{ d\} y\mathrm\{ d\} z+\mathrm\{ d\} z\mathrm\{ d\} x-\mathrm\{ d\} x\mathrm\{ d\} y\\\\ &=\iint\_\{3x+2z=3\atop x^2+y^2\leq 1\} \frac\{-2\cdot 3+1\cdot 0-1\cdot 2\}\{\sqrt\{13\}\}\mathrm\{ d\} S =\iint\_\{x^2+y^2\leq 1\}\frac\{-8\}\{\sqrt\{13\}\}\cdot\frac\{\sqrt\{13\}\}\{2\}\mathrm\{ d\} x\mathrm\{ d\} y\\\\ &=-4\iint\_\{x^2+y^2\leq 1\}\mathrm\{ d\} x\mathrm\{ d\} y=-4\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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
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826、 8、 (15 分) 求曲线积分
\begin\{aligned\} \int\_L \mathrm\{e\}^y\cos x\mathrm\{ d\} x+(\mathrm\{e\}^y\sin x-x^2)\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 $\displaystyle L$ 为右半圆 $\displaystyle x^2+y^2=ay\ (y\geq 0)$, 方向从 $\displaystyle (0,0)$ 到 $\displaystyle (0,a)$. (中山大学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 $\displaystyle \ell: (0,a)\xrightarrow\{x=0\}(0,0)$, 则 $\displaystyle \int\_\ell\cdots=0$, 而
\begin\{aligned\} \mbox\{原式\}&=\mbox\{原式\}+\int\_\ell\cdots\xlongequal\{\tiny\mbox\{Green\}\} \iint\_\{x^2+y^2\leq ay\atop 0\leq x\leq \frac\{a\}\{2\}\} -2x\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\sin \theta\} (-2r\cos\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](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
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827、 9、 (15 分) 求 $\displaystyle \iint\_\varSigma x^2\mathrm\{ d\} y\mathrm\{ d\} z$, 其中 $\displaystyle \varSigma$ 为 $\displaystyle (x-1)^2+(y-2)^2+(z-3)^2=1$, 方向取外侧. (中山大学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 $\displaystyle V$ 为 $\displaystyle \varSigma$ 的内部, 则
\begin\{aligned\} \mbox\{原式\}\xlongequal\{\tiny\mbox\{Gauss\}\}&\iiint\_V 2x\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 < 1\}2(u+1)\mathrm\{ d\} u\mathrm\{ d\} v\mathrm\{ d\} w\\\\ \xlongequal\{\tiny\mbox\{对称性\}\}&2\iiint\_\{u^2+v^2+w^2 < 1\}\mathrm\{ d\} u\mathrm\{ d\} v\mathrm\{ d\} w =2\cdot \frac\{4\}\{3\}=\frac\{8\}\{3\}. \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](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
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828、 (5)、 计算曲面积分
\begin\{aligned\} I=\iint\_S x^3\mathrm\{ d\} y\mathrm\{ d\} z+y^3\mathrm\{ d\} z\mathrm\{ d\} x+z^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 $\displaystyle S$ 是球面 $\displaystyle (x-a)^2+(y-b)^2+(z-c)^2=R^2$ 的外侧, $\displaystyle a,b,c$ 为常数, $\displaystyle R > 0$ 为半径. (重庆大学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\} I&\xlongequal\{\tiny\mbox\{Gauss\}\} \iiint\_\{(x-a)^2+(y-b)^2+(z-c)^2\leq R^2\}(3x^2+3y^2+3z^2)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z\\\\ &\stackrel\{x-a=u,\cdots\}\{=\}3\iiint\_\{u^2+v^2+w^2\leq R^2\} [(a+u)^2+(b+v)^2+(c+w)^2]\mathrm\{ d\} u\mathrm\{ d\} v\mathrm\{ d\} w\\\\ &\xlongequal\{\tiny\mbox\{对称性\}\} 3\iiint\_\{u^2+v^2+w^2\leq R^2\} (a^2+b^2+c^2+u^2+v^2+w^2)\mathrm\{ d\} u\mathrm\{ d\} v\mathrm\{ d\} w\\\\ &=3\int\_0^R (a^2+b^2+c^2+r^2)\cdot 4\pi r^2\mathrm\{ d\} r =\frac\{4\pi R^5\}\{5\}(5a^2+5b^2+5c^2+3R^2). \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](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
发表于 2023-3-5 09:14:01
