## 张祖锦2023年数学专业真题分类70天之第33天
---
737、 9、 (10 分) 设 $\displaystyle F(x), G(x)$ 连续可微, 且 $\displaystyle F(1)=F(4), G(1)=G(4)$, 区域 $\displaystyle D$ 由 $\displaystyle y=x, y=4x, xy=1, xy=4$ 围成, 其边界 $\displaystyle \partial D$ 取逆时针方向. 计算
\begin\{aligned\} I=\int\_\{\partial D\}\frac\{F(xy)\}\{x\}\mathrm\{ d\} x+\frac\{G(xy)\}\{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\}
(北京科技大学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 $\displaystyle F'(x)=f(x), G'(x)=g(x)$, 则
\begin\{aligned\} I&\xlongequal\{\tiny\mbox\{Green\}\} \iint\_D \left\\{\frac\{\partial\}\{\partial x\}\left\[\frac\{G(xy)\}\{y\}\right\] -\frac\{\partial\}\{\partial y\}\left\[\frac\{F(xy)\}\{x\}\right\]\right\\}\mathrm\{ d\} x\mathrm\{ d\} y\\\\ &=\iint\_D[g(xy)-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 $\displaystyle xy=u, \frac\{y\}\{x\}=v$, 则 $\displaystyle \frac\{\partial(u,v)\}\{\partial(x,y)\}=2v$, 而
\begin\{aligned\} I&=\iint\_\{1\leq u,v\leq 4\}[g(u)-f(u)]\frac\{1\}\{2v\}\mathrm\{ d\} u\mathrm\{ d\} v\\\\ &=\int\_1^4 [g(u)-f(u)]\mathrm\{ d\} u\cdot\frac\{1\}\{2\}\int\_1^4 \frac\{\mathrm\{ d\} v\}\{v\}\\\\ &=\left\\{[G(4)-G(1)]-[F(4)-F(1)]\right\\}\ln 2=0. \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)/
---
738、 8、 (15 分) 计算第二型曲面积分
\begin\{aligned\} \iint\_S \frac\{xz\}\{a^2\}\mathrm\{ d\} y\mathrm\{ d\} z+\frac\{yz\}\{b^2\}\mathrm\{ d\} z\mathrm\{ d\} x+\frac\{z^2\}\{c^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 $\displaystyle S=\left\\{(x,y,z; \frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}+\frac\{z^2\}\{c^2\}=1, z\geq 0\right\\}$ (上半椭球面). (北京师范大学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 $\displaystyle \varSigma: \frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}\leq 1, z=0$, 取上侧, 则 $\displaystyle \iint\_\varSigma\cdots=0$, 而
\begin\{aligned\} \mbox\{原式\}=&\mbox\{原式\}-\iint\_\varSigma\cdots\\\\ \xlongequal\{\tiny\mbox\{Gauss\}\}&\iiint\_\{\frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}+\frac\{z^2\}\{c^2\}\leq 1\atop z\geq 0\} \left(\frac\{z\}\{a^2\}+\frac\{z\}\{b^2\}+\frac\{2z\}\{c^2\}\right)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z\\\\ =&\int\_0^c\frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\} z\mathrm\{ d\} z\iint\_\{\frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}\leq 1-\frac\{z^2\}\{c^2\}\}\mathrm\{ d\} x\mathrm\{ d\} y\\\\ =&\frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}\int\_0^c z\cdot \pi \left(1-\frac\{z^2\}\{c^2\}\right) ab\mathrm\{ d\} z =\frac\{\pi abc^2\}\{4\}\left(\frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}\right). \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)/
---
739、 6、 求曲线积分
\begin\{aligned\} \oint\_L y\cos x\mathrm\{ d\} x+(xy^2+\sin 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 $\displaystyle (x-1)^2+(y-1)^2=1$, 取顺时针方向. (北京邮电大学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\{Green\}\} -\iint\_\{(x-1)^2+(y-1)^2\leq 1\}[y^2+\cos x-\cos x]\mathrm\{ d\} x\mathrm\{ d\} y\\\\ &\stackrel\{x-1=u, y-1=v\}\{=\}-\iint\_\{u^2+v^2\leq 1\}(v+1)^2\mathrm\{ d\} u\mathrm\{ d\} v\\\\ &\xlongequal\{\tiny\mbox\{对称性\}\} -\iint\_\{u^2+v^2\leq 1\}\left(\frac\{u^2+v^2\}\{2\}+1\right)\mathrm\{ d\} u\mathrm\{ d\} v =-\int\_0^1 \left(\frac\{r^2\}\{2\}+1\right)\cdot 2\pi r\mathrm\{ d\} r=-\frac\{5\pi\}\{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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
740、 7、 计算曲面积分
\begin\{aligned\} I=\iint\_\varSigma (x^3+Rx)\mathrm\{ d\} y\mathrm\{ d\} z +y^3\mathrm\{ d\} z\mathrm\{ d\} x+(1-3z)(x^2+y^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 $\displaystyle \varSigma: z=\sqrt\{R^2-x^2-y^2\}$, 方向取下侧, $\displaystyle R > 0$. (北京邮电大学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 $\displaystyle S: x^2+y^2\leq R^2, z=0$, 取上侧, 则
\begin\{aligned\} \iint\_S\cdots=\iint\_\{x^2+y^2\leq R^2\}(x^2+y^2)\mathrm\{ d\} x\mathrm\{ d\} y=\int\_0^R r^2\cdot 2\pi r\mathrm\{ d\} r=\frac\{\pi R^4\}\{2\}. \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=&\left(-I-\iint\_S\cdots\right)-\iint\_S\cdots\\\\ \xlongequal\{\tiny\mbox\{Gauss\}\}&-\iiint\_\{x^2+y^2+z^2\leq R^2\atop z\geq 0\}R\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z-\frac\{\pi R^4\}\{2\}\\\\ =&-R\cdot\frac\{2\pi R^3\}\{3\}-\frac\{\pi R^4\}\{2\}=-\frac\{7\pi R^4\}\{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)/
---
741、 (4)、 设 $\displaystyle \varGamma$ 是以 $\displaystyle (1,0)$ 为心, $\displaystyle r$ 为半径的圆周 ($r\neq 1$), 逆时针方向为正向, 求第二型曲线积分
\begin\{aligned\} I=\oint\_\varGamma \frac\{-y\mathrm\{ d\} x+x\mathrm\{ d\} y\}\{4x^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\}
(电子科技大学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 $\displaystyle (P,Q)=\frac\{(-y,x)\}\{4x^2+y^2\}$, 则 $\displaystyle Q\_x=P\_y$. (4-1)、 若 $\displaystyle 0 < r < 1$, 则 $\displaystyle I\xlongequal\{\tiny\mbox\{Green\}\} 0$. (4-2)、 若 $\displaystyle r > 1$, 则
\begin\{aligned\} I&\xlongequal\{\tiny\mbox\{Green\}\} \oint\_\{4x^2+y^2=\varepsilon^2\}P\mathrm\{ d\} x+Q\mathrm\{ d\} y =\frac\{1\}\{\varepsilon^2\}\oint\_\{4x^2+y^2=\varepsilon^2\}-y\mathrm\{ d\} x+x\mathrm\{ d\} y\\\\ \xlongequal\{\tiny\mbox\{Green\}\}&\frac\{1\}\{\varepsilon^2\}\iint\_\{4x^2+y^2\leq\varepsilon^2\} 2\mathrm\{ d\} x\mathrm\{ d\} y \stackrel\{2x=u, y=v\}\{=\} \frac\{2\}\{\varepsilon^2\}\iint\_\{u^2+v^2\leq \varepsilon^2\} \frac\{\mathrm\{ d\} u\mathrm\{ d\} v\}\{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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
742、 (4)、 计算 $\displaystyle \int\_L (x^2+y^2+z^2)\mathrm\{ d\} s$, 其中
\begin\{aligned\} L: x=a\cos t, y=a\sin t, z=bt, t\in [0,2\pi]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(东北师范大学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) /
\begin\{aligned\} \mbox\{原式\}=\int\_0^\{2\pi\}(a^2+b^2t^2) \sqrt\{a^2+b^2\}\mathrm\{ d\} t =\frac\{2\pi(3a^2+4\pi^2b^2)\}\{3\}\sqrt\{a^2+b^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)/
---
743、 3、 计算
\begin\{aligned\} \int\_\varGamma \mathrm\{e\}^x(1-\cos y)\mathrm\{ d\} x-\mathrm\{e\}^x(y-\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 $\displaystyle \varGamma: y=\sin x, x\in [0,\pi]$, 方向从 $\displaystyle (\pi,0)$ 到 $\displaystyle (0,0)$. (东北师范大学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 $\displaystyle P=\mathrm\{e\}^x(1-\cos y), Q=-\mathrm\{e\}^x(y-\sin y)$, 则 $\displaystyle Q\_x-P\_y=-y\mathrm\{e\}^x$. 再设 $\displaystyle L: (0,0)\xrightarrow\{y=0\}(\pi,0)$, 则 $\displaystyle \int\_L \cdots =0$, 而
\begin\{aligned\} \mbox\{原式\}=&\mbox\{原式\}+\int\_L\cdots\xlongequal\{\tiny\mbox\{Green\}\} \iint\_\{0\leq y\leq \sin x\atop 0\leq x\leq \pi\}(-y\mathrm\{e\}^x)\mathrm\{ d\} x\mathrm\{ d\} y\\\\ =&-\int\_0^\pi \mathrm\{e\}^x \mathrm\{ d\} x\int\_0^\{\sin x\}y\mathrm\{ d\} y=\frac\{1-\mathrm\{e\}^\pi\}\{5\}. \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)/
---
744、 16、 设 $\displaystyle P(x,y)$ 和 $\displaystyle Q(x,y)$ 在 $\displaystyle \mathbb\{R\}^2$ 上具有连续的偏导数, 且对任意 $\displaystyle (x\_0,y\_0)\in\mathbb\{R\}^2$, 以及任意 $\displaystyle r > 0$, 总有
\begin\{aligned\} \int\_L P\mathrm\{ d\} x+Q\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\}
其中 $\displaystyle L$ 是以 $\displaystyle (x\_0,y\_0)$ 为心, $\displaystyle r > 0$ 为半径的上半圆周, 方向是逆时针. 证明: 在 $\displaystyle \mathbb\{R\}^2$ 上, $\displaystyle P(x,y)\equiv 0, \frac\{\partial Q\}\{\partial x\}\equiv 0$. (东南大学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
\begin\{aligned\} C: x\_0-r\leq x\leq x\_0+r, y=y\_0, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
取逆时针, 则
\begin\{aligned\} \int\_L P\mathrm\{ d\} x+Q\mathrm\{ d\} y-\int\_C P\mathrm\{ d\} x+Q\mathrm\{ d\} y =\iint\_\{(x-x\_0)^2+(y-y\_0)^2\leq r^2\atop y\geq y\_0\} (Q\_x-P\_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\}
此即
\begin\{aligned\} \int\_\{x\_0-r\}^\{x\_0+r\} P(x,y\_0)\mathrm\{ d\} x=\iint\_\{(x-x\_0)^2+(y-y\_0)^2\leq r^2\atop y\geq y\_0\} (Q\_x-P\_y)\mathrm\{ d\} x\mathrm\{ d\} y.\qquad(\star) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
两边除以 $\displaystyle 2r$ 并令 $\displaystyle r\to 0^+$ 得
\begin\{aligned\} P(x\_0,y\_0)&=\lim\_\{r\to 0\} \frac\{1\}\{2r\}\int\_\{x\_0-r\}^\{x\_0+r\} P(x,y\_0)\mathrm\{ d\} x\\\\ &=\lim\_\{r\to 0\}\frac\{\pi r\}\{4\}\cdot \frac\{1\}\{\frac\{\pi r^2\}\{2\}\}\iint\_\{(x-x\_0)^2+(y-y\_0)^2\leq r^2\atop y\geq y\_0\} (Q\_x-P\_y)\mathrm\{ d\} x\mathrm\{ d\} y\\\\ &=0\cdot \left\[Q\_x(x\_0,y\_0)-P\_y(x\_0,y\_0)\right\]=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 $\displaystyle (x\_0,y\_0)$ 的任意性知 $\displaystyle P\equiv 0$. 代入 $\displaystyle (\star)$ 得
\begin\{aligned\} 0=\iint\_\{(x-x\_0)^2+(y-y\_0)^2\leq r^2\atop y\geq y\_0\} Q\_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\}
两边除以 $\displaystyle \frac\{\pi r^2\}\{2\}$ 后令 $\displaystyle r\to 0^+$ 得 $\displaystyle Q\_x(x\_0,y\_0)=0$. 由 $\displaystyle (x\_0,y\_0)$ 的任意性知 $\displaystyle Q\_x\equiv 0$.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)/
---
745、 6、 (20 分) 求平面 $\displaystyle \alpha x+\beta y+\gamma z=0$ $\displaystyle (\gamma\neq 0)$ 与圆柱面 $\displaystyle \frac\{x^2\}\{A^2\}+\frac\{y^2\}\{B^2\}=1$ 截出的椭圆面积. (福州大学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) / 面积
\begin\{aligned\} =&\iint\_\{\frac\{x^2\}\{A^2\}+\frac\{y^2\}\{B^2\}\leq 1\atop \alpha x+\beta y+\gamma z=0\}\mathrm\{ d\} S =\iint\_\{\frac\{x^2\}\{A^2\}+\frac\{y^2\}\{B^2\}\leq 1\} \frac\{\sqrt\{\alpha^2+\beta^2+\gamma^2\}\}\{|\gamma|\}\mathrm\{ d\} x\mathrm\{ d\} y =\pi AB\frac\{\sqrt\{\alpha^2+\beta^2+\gamma^2\}\}\{|\gamma|\}. \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)/
---
746、 7、 (15 分) 求曲面积分
\begin\{aligned\} \iint\_S (x^2\cos^2\alpha+y^2\cos^2\beta+z^2\cos^2\gamma)\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 $\displaystyle S: x^2+y^2=z^2, 0\leq z\leq 1, \cos\alpha,\cos\beta,\cos\gamma$ 为外法线方向余弦. (福州大学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) / 设 $\displaystyle F=x^2+y^2-z^2$, 则
\begin\{aligned\} \frac\{1\}\{2\}\mathrm\{ grad\} F=\left\\{x,y,-z\right\\}\Rightarrow (\cos^2\alpha,\cos^2\beta,\cos^2\gamma)=\frac\{(x^2,y^2,z^2)\}\{x^2+y^2+z^2\}, \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\}=&\iint\_S \frac\{x^4+y^4+z^4\}\{x^2+y^2+z^2\}\mathrm\{ d\} S =\iint\_\{x^2+y^2\leq 1\} \frac\{x^4+y^4+(x^2+y^2)^2\}\{2(x^2+y^2)\}\cdot \sqrt\{2\}\mathrm\{ d\} x\mathrm\{ d\} y\\\\ =&\frac\{1\}\{\sqrt\{2\}\}\int\_0^\{2\pi\}\mathrm\{ d\} \theta\int\_0^1 \frac\{r^4\cos^4\theta+r^4\sin^4\theta+r^4\}\{r^2\}\cdot r\mathrm\{ d\} r =\frac\{7\pi\}\{8\sqrt\{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)/
---
747、 (3)、 设 $\displaystyle S$ 是单位球面 $\displaystyle x^2+y^2+z^2=1$ 被锥 $\displaystyle z > \sqrt\{x^2+y^2\}$ 所截部分曲面, 定向取外侧为正向, 则对于
\begin\{aligned\} \vec\{F\}=(xy+\cos z)\vec\{i\} +(-xy+x^2)\vec\{j\} +(x+2z^2)\vec\{k\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
曲面积分 $\displaystyle \iint\_S \vec\{F\}\cdot \mathrm\{ d\} \vec\{S\}=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (复旦大学2023年分析(第6,7,8,9,10题没做)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 $\displaystyle \varSigma: x^2+y^2\leq \frac\{1\}\{2\}, z=\frac\{1\}\{\sqrt\{2\}\}$, 取上侧, 则
\begin\{aligned\} \iint\_\varSigma\cdots=\iint\_\{x^2+y^2\leq\frac\{1\}\{2\}\} (x+1)\mathrm\{ d\} x\mathrm\{ d\} y \xlongequal\{\tiny\mbox\{对称性\}\} \iint\_\{x^2+y^2\leq \frac\{1\}\{2\}\} \mathrm\{ d\} x\mathrm\{ d\} y=\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\}
故
\begin\{aligned\} \mbox\{原式\}=&\left\[\mbox\{原式\}-\iint\_\varSigma\cdots\right\]+\iint\_\varSigma\cdots\\\\ \xlongequal\{\tiny\mbox\{Gauss\}\}&\iiint\_\{x^2+y^2+z^2\leq 1\atop \frac\{1\}\{2\}\leq z\leq 1\} [y+(-x)+4z]\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z+\frac\{\pi\}\{2\}\\\\ \xlongequal\{\tiny\mbox\{对称性\}\}&4\iiint\_\{x^2+y^2+z^2\leq 1\atop \frac\{1\}\{2\}\leq z\leq 1\} z\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z+\frac\{\pi\}\{2\}\\\\ =&4\int\_\frac\{1\}\{2\}^1 z\cdot \pi(1-z^2)\mathrm\{ d\} z+\frac\{\pi\}\{2\}=\frac\{17\pi\}\{16\}. \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)/
---
748、 5、 设 $\displaystyle P(x,y,z), Q(x,y,z), R(x,y,z)$ 是 $\displaystyle \mathbb\{R\}^3$ 上的二阶连续可微函数, 满足
\begin\{aligned\} \frac\{\partial R\}\{\partial y\}-\frac\{\partial Q\}\{\partial z\}=a, \frac\{\partial P\}\{\partial z\}-\frac\{\partial R\}\{\partial x\}=b, \frac\{\partial Q\}\{\partial x\}-\frac\{\partial P\}\{\partial y\}=c, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle a,b,c$ 为常数. 证明: 存在 $\displaystyle \mathbb\{R\}^3$ 上的线性函数 $\displaystyle f,g,h$ 使得
\begin\{aligned\} (P-f)\mathrm\{ d\} x+(Q-g)\mathrm\{ d\} y+(R-h)\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\}
是全微分. (复旦大学2023年分析(第6,7,8,9,10题没做)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) / 只要取 $\displaystyle f=bz, g=cx, h=ay$, 就有
\begin\{aligned\} \frac\{\partial (R-h)\}\{\partial y\}-\frac\{\partial (Q-g)\}\{\partial z\}&=0,\\\\ \frac\{\partial (P-f)\}\{\partial z\}-\frac\{\partial (R-h)\}\{\partial x\}&=0,\\\\ \frac\{\partial (Q-g)\}\{\partial x\}-\frac\{\partial (P-f)\}\{\partial y\}&=0. \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\} (P-f)\mathrm\{ d\} x+(Q-g)\mathrm\{ d\} y+(R-h)\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\}
是全微分.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)/
---
749、 (6)、 求第一型曲线积分 $\displaystyle \int\_\varGamma \sqrt\{1-x^2-y^2\}\mathrm\{ d\} s$, 其中 $\displaystyle \varGamma$ 是曲线 $\displaystyle x^2+y^2=x$. (广西大学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 $\displaystyle x=\frac\{1+\cos\theta\}\{2\}, y=\frac\{\sin\theta\}\{2\}$, 则
\begin\{aligned\} \mbox\{原式\}=\int\_\varGamma \sqrt\{1-x\}\mathrm\{ d\} s=\int\_0^\{2\pi\} \sqrt\{\frac\{1-\cos \theta\}\{2\}\} \cdot\frac\{1\}\{2\}\mathrm\{ d\} \theta =\frac\{1\}\{2\}\int\_0^\{2\pi\}\sin\frac\{\theta\}\{2\}\mathrm\{ d\}\theta=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)/
---
750、 (9)、 求曲面积分 $\displaystyle I=\iint\_\{x^2+y^2+z^2=R^2\}\frac\{\mathrm\{ d\} S\}\{\sqrt\{x^2+y^2+(z-h)^2\}\}$, 其中 $\displaystyle h\neq R$. (广西大学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\}=\iint\_\{x^2+y^2+z^2=R^2\}\frac\{\mathrm\{ d\} S\}\{\sqrt\{R^2+h^2-2hz\}\}\\\\ =&\iint\_\{x^2+y^2\leq R^2\} \frac\{1\}\{\sqrt\{R^2+h^2-2h\sqrt\{R^2-x^2-y^2\}\}\} \cdot\frac\{R\}\{\sqrt\{R^2-x^2-y^2\}\}\mathrm\{ d\} x\mathrm\{ d\} y\\\\ =&2\pi R\int\_0^R \frac\{r\}\{\sqrt\{R^2+h^2-2h\sqrt\{R^2-r^2\}\}\}\frac\{\mathrm\{ d\} r\}\{\sqrt\{R^2-r^2\}\}\\\\ \stackrel\{\sqrt\{R^2-r^2\}=s\}\{=\}&-2\pi R\int\_R^0 \frac\{\mathrm\{ d\} s\}\{\sqrt\{R^2+h^2-2hs\}\} =\frac\{2\pi R\left(\sqrt\{h^2+R^2\}-|h-R|\right)\}\{h\}. \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)/
---
751、 (6)、 证明 Green 定理: 设闭区域 $\displaystyle D$ 由分段光滑的简单曲线 $\displaystyle L$ 围成, 函数 $\displaystyle P(x,y)$ 及 $\displaystyle Q(x,y)$ 在 $\displaystyle D$ 上有连续的一阶偏导数, 则有
\begin\{aligned\} \iint\_D \left(\frac\{\partial Q\}\{\partial x\}-\frac\{\partial P\}\{\partial y\}\right)\mathrm\{ d\} x\mathrm\{ d\} y=\oint\_L P\mathrm\{ d\} x+Q\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 $\displaystyle D$ 的取正向的边界曲线. (广西大学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) / (6-1)、 若 $\displaystyle D$ 既是 $\displaystyle x$ 型区域, 也是 $\displaystyle y$ 型区域, 则可设
\begin\{aligned\} D=&\left\\{(x,y); a\leq x\leq b, \varphi\_1(x)\leq \varphi\_2(x)\right\\}\\\\ =&\left\\{(x,y); \psi\_1(y)\leq x\leq \psi\_2(y), \alpha\leq y\leq \beta\right\\}. \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\} &\iint\_D \frac\{\partial Q\}\{\partial x\}\mathrm\{ d\} x\mathrm\{ d\} y =\int\_\alpha^\beta \mathrm\{ d\} y\int\_\{\psi\_1(y)\}^\{\psi\_2(y)\}\frac\{\partial Q\}\{\partial x\}\mathrm\{ d\} x\\\\ =&\int\_\alpha^\beta Q\left(\psi\_2(y),y\right)\mathrm\{ d\} y -\int\_\alpha^\beta Q\left(\psi\_1(y),y\right)\mathrm\{ d\} y =\oint\_L Q(x,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\}
同理, $\displaystyle -\iint\_D \frac\{\partial P\}\{\partial y\}\mathrm\{ d\} x\mathrm\{ d\} y=\oint\_L P(x,y)\mathrm\{ d\} x$. 上面两个式子相加即得结论. (6-2)、 对一般的 $\displaystyle D$, 将 $\displaystyle D$ 分成有限个既是 $\displaystyle x$ 型又是 $\displaystyle y$ 型的子区域, 然后逐块利用第 1 步的结果后, 相加即得结论.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)/
---
752、 4、 设 $\displaystyle S$ 是光滑封闭曲面, 点 $\displaystyle (x,y,z)$ 是一不在 $\displaystyle S$ 上的定点, 点 $\displaystyle (x,y,z)$ 到 $\displaystyle S$ 上任意点 $\displaystyle (\xi,\eta,\zeta)$ 的向径为 $\displaystyle \vec\{r\}$, 且
\begin\{aligned\} r=\sqrt\{(x-\xi)^2+(y-\eta)^2+(z-\zeta)^2\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
求 $\displaystyle \iint\_S\frac\{\cos\angle(\vec\{n\},\vec\{r\})\}\{r^2\}\mathrm\{ d\} S$, 其中 $\displaystyle \cos\angle(\vec\{n\},\vec\{r\})$ 是曲面 $\displaystyle S$ 上点的法向量与向径 $\displaystyle \vec\{r\}$ 的夹角的余弦, $\displaystyle S$ 取正侧. (哈尔滨工程大学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\}=\iint\_S \frac\{\vec\{r\}\cdot \vec\{n\}\}\{r^3\}\mathrm\{ d\} S =\iint\_S \frac\{(\xi-x)\mathrm\{ d\} \eta\mathrm\{ d\}\zeta +(\eta-y)\mathrm\{ d\} \zeta\mathrm\{ d\} \xi+(\zeta-z)\mathrm\{ d\} \xi\mathrm\{ d\} \eta\}\{\left\[ (\xi-x)^2+(\eta-y)^2+(\zeta-z)^2\right\]^\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\}
令
\begin\{aligned\} (P,Q,R)=\frac\{(\xi-x,\eta-y,\zeta-z)\}\{\left\[ (\xi-x)^2+(\eta-y)^2+(\zeta-z)^2\right\]^\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\}
则 $\displaystyle P\_\xi+Q\_\eta+R\_\zeta=0$. 令 $\displaystyle V$ 是 $\displaystyle S$ 所围例题, 则 (1)、 当曲面 $\displaystyle S$ 不包围点 $\displaystyle (x,y,z)$ 时,
\begin\{aligned\} \mbox\{原式\}\xlongequal\{\tiny\mbox\{Gauss\}\} \iiint\_V \left\[P\_\xi+Q\_\eta+R\_\zeta\right\]\mathrm\{ d\} \xi\mathrm\{ d\} \eta\mathrm\{ d\} \zeta=0. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 当曲面 $\displaystyle S$ 包围点 $\displaystyle (x,y,z)$ 时,
\begin\{aligned\} &\mbox\{原式\}-\iint\_\{(\xi-x)^2+(\eta-y)^2+(\zeta-z)^2=\varepsilon^2\ll 1\} P\mathrm\{ d\} \eta\mathrm\{ d\} \zeta +Q\mathrm\{ d\} \zeta\mathrm\{ d\} \xi+R\mathrm\{ d\} \xi\mathrm\{ d\} \eta\\\\ \xlongequal\{\tiny\mbox\{Gauss\}\}& \iiint\_V \left\[P\_\xi+Q\_\eta+R\_\zeta\right\]\mathrm\{ d\} \xi\mathrm\{ d\} \eta\mathrm\{ d\} \zeta=0. \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\}&=\frac\{1\}\{\varepsilon^3\} \iint\_\{(\xi-x)^2+(\eta-y)^2+(\zeta-z)^2=\varepsilon^2\} \left\[\begin\{array\}\{c\}(\xi-x)\mathrm\{ d\} \eta\mathrm\{ d\} \zeta +(\eta-y)\mathrm\{ d\} \zeta\mathrm\{ d\} \xi\\\\ +(\zeta-z)\mathrm\{ d\} \xi\mathrm\{ d\} \eta\end\{array\}\right\]\\\\ &\xlongequal\{\tiny\mbox\{Gauss\}\} \frac\{1\}\{\varepsilon^3\}\iiint\_\{(\xi-x)^2+(\eta-y)^2+(\zeta-z)^2 < \varepsilon^2\} 3\mathrm\{ d\} \xi\mathrm\{ d\} \eta\mathrm\{ d\} \zeta =\frac\{3\}\{\varepsilon^3\}\cdot \frac\{4\pi \varepsilon^3\}\{3\}=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)/
---
753、 10、 计算曲面积分
\begin\{aligned\} \iint\_S (xy^2+z^3)\mathrm\{ d\} y\mathrm\{ d\} z+(yz^2+x^3)\mathrm\{ d\} z\mathrm\{ d\} x+(zx^2+a^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 $\displaystyle S$ 为上半球面 $\displaystyle z=\sqrt\{a^2-x^2-y^2\}\ (a > 0)$, 方向向上. (哈尔滨工业大学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 $\displaystyle \varSigma: x^2+y^2\leq a^2, z=0$, 取上侧, 则
\begin\{aligned\} \iint\_\varSigma \cdots=\iint\_\{x^2+y^2\leq a^2\}a^3\mathrm\{ d\} x\mathrm\{ d\} y=\pi a^5. \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\}=&\left(\mbox\{原式\}-\iint\_\varSigma\cdots\right)+\iint\_\varSigma \cdots\\\\ \xlongequal\{\tiny\mbox\{Gauss\}\}&\iiint\_\{x^2+y^2+z^2\leq a^2\atop z\geq 0\} (y^2+z^2+x^2)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z+\pi a^5\\\\ =&\int\_0^a r^2\frac\{4\pi r^2\}\{2\}\mathrm\{ d\} r+\pi a^5 =\frac\{7\pi a^5\}\{5\}. \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)/
---
754、 10、 (15 分) 计算
\begin\{aligned\} I=\iint\_\varSigma \frac\{x^3\}\{3\}\mathrm\{ d\} y\mathrm\{ d\} z+\frac\{y^3\}\{3\}\mathrm\{ d\} z\mathrm\{ d\} x+(z+1)\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 $\displaystyle \varSigma: z=1-x^2-y^2\ (z\geq 0)$, 取上侧. (合肥工业大学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 $\displaystyle \varSigma: x^2+y^2\leq 1, z=0$, 取上侧, 则
\begin\{aligned\} \iint\_\varSigma\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\} I=&\left(I-\iint\_\varSigma\cdots\right)+\iint\_\varSigma\cdots\\\\ \xlongequal\{\tiny\mbox\{Gauss\}\}&\iiint\_\{x^2+y^2\leq 1-z\atop 0\leq z\leq 1\}(x^2+y^2+1)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z+\pi\\\\ =&\int\_0^1 \mathrm\{ d\} z\int\_0^\{\sqrt\{1-z\}\} (r^2+1)\cdot 2\pi r\mathrm\{ d\} r+\pi=\frac\{5\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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
755、 11、 (15 分) 计算 $\displaystyle I=\iint\_\{x^2+y^2+z^2=4\}f(x,y,z)\mathrm\{ d\} S$, 其中
\begin\{aligned\} f(x,y,z)=\left\\{\begin\{array\}\{llllllllllll\}x^2+y^2,&z\geq \sqrt\{x^2+y^2\},\\\\ 0,&z < \sqrt\{x^2+y^2\}.\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\}
(合肥工业大学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\} z=\sqrt\{x^2+y^2\}\geq 0, x^2+y^2+z^2=4\Rightarrow 2z^2=4, z\geq 0\Rightarrow z=\sqrt\{2\} \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\} I\stackrel\{z=\sqrt\{4-x^2-y^2\}\}\{=\}&\iint\_\{x^2+y^2\leq 2\}(x^2+y^2)\sqrt\{1+z\_x^2+z\_y^2\}\mathrm\{ d\} x\mathrm\{ d\} y\\\\ =&\iint\_\{x^2+y^2\leq 2\}(x^2+y^2)\frac\{2\}\{\sqrt\{4-x^2-y^2\}\}\mathrm\{ d\} x\mathrm\{ d\} y\\\\ =&\int\_0^\{\sqrt\{2\}\} \frac\{2r^2\}\{\sqrt\{4-r^2\}\}\cdot 2\pi r\mathrm\{ d\} r \stackrel\{\sqrt\{4-r^2\}=\rho\}\{=\}\cdots =\frac\{8\left(8-5\sqrt\{2\}\right)\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](
https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
756、 (4)、 叙述 Green 公式的条件及结论. (河海大学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) / (Green 公式) 设 (4-1)、 $\displaystyle D$ 是闭区域, 它的边界 $\displaystyle \partial D$ 由一条或几条光滑曲线所组成, (4-2)、 $\displaystyle P(x,y), Q(x,y)$ 在 $\displaystyle D$ 上连续, 且有连续的一阶偏导数, 则
\begin\{aligned\} \iint\_D \left(Q\_x-P\_y\right)\mathrm\{ d\} x\mathrm\{ d\} y=\oint\_\{\partial D\}P\mathrm\{ d\} x+Q\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\}
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)/
---
757、 (5)、 计算曲面积分
\begin\{aligned\} \iint\_\varSigma \frac\{x\mathrm\{ d\} y\mathrm\{ d\} z+y\mathrm\{ d\} z\mathrm\{ d\} x+z\mathrm\{ d\} x\mathrm\{ d\} y\}\{(x^2+y^2+z^2)^\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\}
其中 $\displaystyle \varSigma$ 为椭球面 $\displaystyle x^2+4y^2+9z^2=1$, 取外侧. (河海大学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 $\displaystyle (P,Q,R)=\frac\{(x,y,z)\}\{(x^2+y^2+z^2)^\frac\{3\}\{2\}\}$, 则 $\displaystyle P\_x+Q\_y+R\_z=0$. 故
\begin\{aligned\} I&\xlongequal\{\tiny\mbox\{Gauss\}\} \iint\_\{x^2+y^2+z^2=\varepsilon^2\}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\_\{x^2+y^2+z^2=\varepsilon^2\}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\_\{x^2+y^2+z^2\leq \varepsilon^2\}3\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z =\frac\{3\}\{\varepsilon^3\}\cdot \frac\{4\pi\varepsilon^3\}\{3\}=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)/
---
758、 (5)、 求 $\displaystyle \iint\_\varSigma |xyz|\mathrm\{ d\} S$, 其中 $\displaystyle \varSigma: |x|+|y|+|z|=1$. (河南大学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\{对称性\}\}&8\iint\_\{x+y+z=1\atop x,y,z\geq 0\} xyz\mathrm\{ d\} S =8\iint\_\{x+y\leq 1\atop x,y\geq 0\} xy(1-x-y)\cdot \sqrt\{3\}\mathrm\{ d\} x\mathrm\{ d\} y\\\\ =&8\sqrt\{3\}\int\_0^1 \mathrm\{ d\} x\int\_0^\{1-x\}xy(1-x-y)\mathrm\{ d\} y =\frac\{1\}\{5\sqrt\{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)/
---
759、 5、 (15 分) 求 $\displaystyle \int\_L x\mathrm\{ d\} y+y\mathrm\{ d\} z+z\mathrm\{ d\} x$, 其中 $\displaystyle L$ 是 $\displaystyle \frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}+\frac\{z^2\}\{c^2\}=1$ 与 $\displaystyle \frac\{x\}\{a\}+\frac\{y\}\{b\}=1$ 的交线上从 $\displaystyle (a,0,0)$ 到 $\displaystyle (0,0,c)$ 的劣弧. [题目有问题, 跟锦数学微信公众号没法做哦.] (河南大学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) / [题目有问题, 跟锦数学微信公众号没法做哦.]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 09:12:46
