## 张祖锦2023年数学专业真题分类70天之第34天
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760、 7、 设 $\displaystyle L: x^2+y^2=1$, 求 $\displaystyle \oint\_L(|x|+|y|)\mathrm\{ d\} s$. (黑龙江大学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\}\}&2\oint\_L |x|\mathrm\{ d\} s =2\int\_0^\{2\pi\}|\cos \theta|\mathrm\{ d\} \theta =2\cdot 4\int\_0^\frac\{\pi\}\{2\}\cos\theta\mathrm\{ d\}\theta=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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---
761、 9、 计算积分
\begin\{aligned\} \iint\_S (y-z)\mathrm\{ d\} y\mathrm\{ d\} z+(z-x)\mathrm\{ d\} z\mathrm\{ d\} x+(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 $\displaystyle S$ 为圆锥面 $\displaystyle z=\sqrt\{x^2+y^2\}, z\leq h\ (h > 0)$, 并取曲面外侧. (湖南大学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 $\displaystyle \varSigma: x^2+y^2\leq h^2, z=h$, 取上侧, 则在 $\displaystyle \varSigma$ 上, $\displaystyle z=h, \mathrm\{ d\} z=0$, 而
\begin\{aligned\} \iint\_\varSigma\cdots=\iint\_\{x^2+y^2\leq h^2\}(x-y)\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+y^2\leq z\atop 0\leq z\leq h\}0\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z=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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---
762、 8、 设 $\displaystyle \varGamma$ 是平面上过原点的简单光滑闭曲线, $\displaystyle L\_\varepsilon$ 是以原点为圆心, 半径为 $\displaystyle \varepsilon$ 的圆周, $\displaystyle \varGamma\_\varepsilon$ 表示 $\displaystyle \varGamma$ 上截取含在 $\displaystyle L\_\varepsilon$ 中的曲线段得到的曲线, 取正向. 求极限 $\displaystyle \lim\_\{\varepsilon\to 0^+\}\int\_\{\varGamma\_\varepsilon\} \frac\{x\mathrm\{ d\} y-y\mathrm\{ d\} x\}\{x^2+y^2\}$. (华东理工大学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 $\displaystyle \varGamma\_\{\varepsilon\delta\}$ 表示 $\displaystyle \varGamma\_\varepsilon$ 在 $\displaystyle x^2+y^2=\delta^2$ 外的部分, 其中 $\displaystyle 0 < \delta < \varepsilon$. 再设 $\displaystyle L\_\{\varepsilon\varepsilon\}$ 表示 $\displaystyle L\_\varepsilon$ 含在 $\displaystyle \varGamma$ 中的部分: $\displaystyle x=\varepsilon\cos\theta, y=\varepsilon\sin\theta, \theta: \alpha\_\varepsilon\to \beta\_\varepsilon$, 则
\begin\{aligned\} &\int\_\{\varGamma\_\varepsilon\} \frac\{x\mathrm\{ d\} y-y\mathrm\{ d\} x\}\{x^2+y^2\} =\lim\_\{\delta\to 0^+\}\int\_\{\varGamma\_\{\varepsilon\delta\}\}\frac\{x\mathrm\{ d\} y-y\mathrm\{ d\} x\}\{x^2+y^2\} \xlongequal\{\tiny\mbox\{Green\}\}\lim\_\{\delta\to 0^+\}\int\_\{L\_\{\varepsilon\varepsilon\}\}\frac\{x\mathrm\{ d\} y-y\mathrm\{ d\} x\}\{x^2+y^2\}\\\\ =&\int\_\{L\_\{\varepsilon\varepsilon\}\}\frac\{x\mathrm\{ d\} y-y\mathrm\{ d\} x\}\{x^2+y^2\} =\frac\{1\}\{\varepsilon^2\}\int\_\{\alpha\_\varepsilon\}^\{\beta\_\varepsilon\}\varepsilon^2\mathrm\{ d\} \theta =(\beta\_\varepsilon-\alpha\_\varepsilon)\xrightarrow\{\varepsilon\to 0^+\}\pi. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
最后一个极限是因为 $\displaystyle \varGamma$ 是光滑曲线, 而在原点附近的曲线段趋于 $\displaystyle \varGamma$ 在原点的切线, 所夹的角 $\displaystyle =\pi$.跟锦数学微信公众号. [在线资料](
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763、 4、 求曲面积分 $\displaystyle \iint\_\varSigma (x+y+z)\mathrm\{ d\} S$, 其中 $\displaystyle \varSigma$ 为上半球面 $\displaystyle x^2+y^2+z^2=1\ (z\geq 0)$. (华南理工大学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) / 由
\begin\{aligned\} z=\sqrt\{1-x^2-y^2\}\Rightarrow \sqrt\{1+z\_x^2+z\_y^2\}=\frac\{1\}\{z\} \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\}\xlongequal\{\tiny\mbox\{对称性\}\} \iint\_S z\mathrm\{ d\} S =\iint\_\{x^2+y^2\leq 1\} 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\}
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---
764、 8、 (10 分) 求曲面积分 $\displaystyle \iint\_S x\mathrm\{ d\} y\mathrm\{ d\} z$, 其中 $\displaystyle S$ 为 $\displaystyle z=\sqrt\{1-x^2-y^2\}$, 方向取下侧. (华南师范大学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 $\displaystyle \varSigma: x^2+y^2\leq 1, z=0$, 取上侧, 则 $\displaystyle \iint\_\varSigma\cdots=0$. 而
\begin\{aligned\} \mbox\{原式\}=&-\left(-\mbox\{原式\}-\iint\_\varSigma\cdots\right)\\\\ \xlongequal\{\tiny\mbox\{Gauss\}\}&-\iiint\_\{x^2+y^2+z^2\leq 1\atop z\geq 0\}\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z=-\frac\{2\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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---
765、 4、 (15 分) 求极限 $\displaystyle \lim\_\{x\_0\to 0\atop x\_1\to+\infty\} \iint\_S \frac\{\mathrm\{e\}^\{-x^9\}\}\{\sqrt\{y^2+z^2\}\}\mathrm\{ d\} y\mathrm\{ d\} z$, 其中 $\displaystyle S$ 由 $\displaystyle x=y^2+z^2$ 和 $\displaystyle x=x\_0, x=x\_1$ ($a > 0, x\_0 < x\_1$) 所围成, 方向取取外侧. (华中科技大学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 $\displaystyle \varSigma\_i: x=x\_i, y^2+z^2\leq x\_i$, 朝 $\displaystyle x$ 轴正向, 则
\begin\{aligned\} \iint\_\{\varSigma\_i\}\cdots=&\iint\_\{y^2+z^2\leq x\_i\} \frac\{\mathrm\{e\}^\{-x\_i^9\}\}\{\sqrt\{y^2+z^2\}\}\mathrm\{ d\} y\mathrm\{ d\} z\\\\ =&\mathrm\{e\}^\{-x\_i^9\}\int\_0^\{\sqrt\{x\_i\}\} \frac\{1\}\{r\}\cdot 2\pi r\mathrm\{ d\} r =2\pi \sqrt\{x\_i\}\mathrm\{e\}^\{-x\_i^9\}\to 0, x\_i\to 0\mbox\{或\} +\infty. \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\}=&\lim\_\{x\_0\to 0\atop x\_1\to+\infty\}\left\[\iint\_S\cdots +\iint\_\{\varSigma\_1\}\cdots -\iint\_\{\varSigma\_0\}\cdots\right\]\\\\ \xlongequal\{\tiny\mbox\{Gauss\}\}&\iiint\_\{y^2+z^2\leq x\} \frac\{-9x^8\mathrm\{e\}^\{-x^9\}\}\{\sqrt\{y^2+z^2\}\}\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z\\\\ =&\int\_0^\infty -9x^8\mathrm\{e\}^\{-x^9\}\mathrm\{ d\} x\iint\_\{y^2+z^2\leq x\} \frac\{\mathrm\{ d\} y\mathrm\{ d\} z\}\{\sqrt\{y^2+z^2\}\}\\\\ =&\int\_0^\infty -9x^8\mathrm\{e\}^\{-x^9\}\mathrm\{ d\} x\int\_0^\{\sqrt\{x\}\} \frac\{1\}\{r\}\cdot 2\pi r\mathrm\{ d\} r\\\\ =&-18\pi \int\_0^\infty x^\frac\{17\}\{2\} \mathrm\{e\}^\{-x^9\}\mathrm\{ d\} x\\\\ \stackrel\{x^9=s\}\{=\}&-2\pi \int\_0^\infty s^\frac\{1\}\{18\} \mathrm\{e\}^\{-s\}\mathrm\{ d\} s =-2\pi \varGamma\left(\frac\{19\}\{18\}\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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---
766、 (4)、 计算积分 $\displaystyle \int\_\varGamma (x^\frac\{4\}\{3\}+y^\frac\{4\}\{3\})\mathrm\{ d\} s$, 其中 $\displaystyle \varGamma$ 为方程 $\displaystyle x^\frac\{2\}\{3\}+y^\frac\{2\}\{3\}=a^\frac\{2\}\{3\}\ (a > 0)$ 所确定的曲线. (华中师范大学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 $\displaystyle x=a\cos^3\theta, y=a\sin^3\theta$, 则
\begin\{aligned\} 0\leq \theta\leq \frac\{\pi\}\{2\}\Rightarrow \sqrt\{x'^2+y'^2\}=3a\sin\theta\cos\theta. \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\}\xlongequal\{\tiny\mbox\{对称性\}\}&4\int\_\{\varGamma\cap \left\\{x\geq 0, y\geq 0\right\\}\} (x^\frac\{4\}\{3\}+y^\frac\{4\}\{3\})\mathrm\{ d\} s\\\\ =&4\int\_0^\frac\{\pi\}\{2\} a^\frac\{4\}\{3\}(\cos^4\theta+\sin^4\theta)\cdot 3a\sin\theta\cos\theta\mathrm\{ d\} \theta\\\\ =&4\int\_0^\frac\{\pi\}\{2\} a^\frac\{4\}\{3\}\left(1-2\sin^2\theta\cos^2\theta\right)\cdot 3a\sin\theta\cos\theta\mathrm\{ d\} \theta =4a^\frac\{7\}\{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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---
767、 (5)、 计算积分 $\displaystyle \iint\_S x^3\mathrm\{ d\} y\mathrm\{ d\} z$, 其中 $\displaystyle S$ 为方程 $\displaystyle \frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}+\frac\{z^2\}\{c^2\}=1$ $\displaystyle (a > 0, b > 0, c > 0)$ 所确定的曲面的上半部分 (即 $\displaystyle z\geq 0$ 部分) 的上侧. (华中师范大学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) / 令 $\displaystyle S\_1: \frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}\leq 1, z=0$, 取上侧, 则 $\displaystyle \iint\_\{S\_1\} x^3\mathrm\{ d\} y\mathrm\{ d\} z\stackrel\{\mathrm\{ d\} z=0\}\{=\}0$. 故
\begin\{aligned\} \mbox\{原式\}=&\mbox\{原式\}-\iint\_\{S\_1\} x^3\mathrm\{ d\} y\mathrm\{ d\} z \xlongequal\{\tiny\mbox\{Gauss\}\} \iint\_\{\frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}+\frac\{z^2\}\{c^2\}\leq 1\atop z\geq 0\} 3x^2\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z\\\\ \stackrel\{x=au, y=bv, zcw\}\{=\}&3a^3bc\iiint\_\{u^2+v^2+w^2\leq 1\atop w\geq 0\} u^2\mathrm\{ d\} u\mathrm\{ d\} v\mathrm\{ d\} w\\\\ \xlongequal\{\tiny\mbox\{对称性\}\}&a^3bc\iiint\_\{u^2+v^2+w^2\leq 1\atop w\geq 0\} (u^2+v^2+w^2)\mathrm\{ d\} u\mathrm\{ d\} v\mathrm\{ d\} w\\\\ =&a^3bc\int\_0^1 r^2\cdot \frac\{4\pi r^2\}\{2\}\mathrm\{ d\} r =\frac\{2\pi a^3bc\}\{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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---
768、 (6)、 计算第二型曲线积分 $\displaystyle \int\_L xy\mathrm\{ d\} x$, 其中 $\displaystyle L$ 是圆周
\begin\{aligned\} (x-a)^2+y^2=a^2\left(a > 0\right) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
及 $\displaystyle x$ 轴围成的在第一象限内的区域的边界, 取逆时针方向. (吉林大学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-a)^2+y^2\leq a^2\atop y\geq 0\}(-x)\mathrm\{ d\} x\mathrm\{ d\} y\\\\ &\stackrel\{x=a+r\cos\theta, y=r\sin \theta\}\{=\}\int\_0^\pi \mathrm\{ d\} \theta \int\_0^a -(a+r\cos\theta)\cdot r\mathrm\{ d\} r=-\frac\{\pi a^3\}\{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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---
769、 (0-18)、 计算
\begin\{aligned\} \oint\_L (x^2-2y+3)\mathrm\{ d\} x+(y^2+x+1)\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 $\displaystyle L$ 为正向圆周 $\displaystyle x^2+y^2=4$. (吉林师范大学2023年(学科数学)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^2+y^2\leq 4\}[1-(-2)]\mathrm\{ d\} x\mathrm\{ d\} y =3\cdot \pi 4=12\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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770、 (0-25)、 设 $\displaystyle L$ 为曲线 $\displaystyle |2x|+|2y|=1$, 取逆时针方向, 计算
\begin\{aligned\} \oint\_L \frac\{(x-y)\mathrm\{ d\} x+(x+4y)\mathrm\{ d\} y\}\{x^2+4y^2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(吉林师范大学2023年(学科数学)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 $\displaystyle (P,Q)=\frac\{(x-y,x+4y)\}\{x^2+4y^2\}$, 则 $\displaystyle Q\_x=P\_y$, 而
\begin\{aligned\} \mbox\{原式\}&\xlongequal\{\tiny\mbox\{Green\}\} \oint\_\{x^2+4y^2=\varepsilon^2 \ll 1\}P\mathrm\{ d\} x+Q\mathrm\{ d\} y\\\\ &=\frac\{1\}\{\varepsilon^2\}\oint\_\{x^2+4y^2=\varepsilon^2\}(x-y)\mathrm\{ d\} x+(x+4y)\mathrm\{ d\} y\\\\ &\xlongequal\{\tiny\mbox\{Green\}\}\frac\{1\}\{\varepsilon^2\}\iint\_\{x^2+4y^2\leq \varepsilon^2\}2\mathrm\{ d\} x\mathrm\{ d\} y \stackrel\{2y=v\}\{=\}\frac\{1\}\{\varepsilon^2\}\iint\_\{x^2+v^2\leq \varepsilon^2\}\mathrm\{ d\} x\mathrm\{ d\} v =\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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---
771、 (3)、 求曲线积分 $\displaystyle \int\_L xy\mathrm\{ d\} s$, 其中 $\displaystyle L$ 为 $\displaystyle \frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}=1\ (a,b > 0)$ 在第一象限的部分. (暨南大学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\}=&\int\_0^\frac\{\pi\}\{2\}a\cos\theta \cdot b\sin\theta\sqrt\{b^2\cos^2\theta+a^2\sin^2\theta\}\mathrm\{ d\} \theta\\\\ =&\frac\{ab\}\{2\}\int\_0^\frac\{\pi\}\{2\} \sin2\theta \sqrt\{b^2+(a^2-b^2)\sin^2\theta\}\mathrm\{ d\} \theta\\\\ \stackrel\{\sin^2\theta=t\}\{=\}&\frac\{ab\}\{2\}\int\_0^1 \sqrt\{b^2+(a^2-b^2)t\}\mathrm\{ d\} t\\\\ =&\left.\frac\{ab\}\{3(a^2+b^2)\}[b^2+(a^2-b^2)t]^\frac\{3\}\{2\}\right|\_\{t=0\}^\{t=1\} =\frac\{ab(a^2+ab+b^2)\}\{3(a+b)\}. \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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---
772、 (2)、 求曲面积分
\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, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle S$ 为曲面 $\displaystyle (x-a)^2+(y-b)^2+(z-c)^2=R^2\ (R > 0)$ 的外侧. (暨南大学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\}\} \iiint\_\{(x-a)^2+(y-b)^2+(z-c)^2\leq R^2\} (2x+2y+2z)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z\\\\ &\stackrel\{x-a=u, y-b=v, z-c=w\}\{=\}\iiint\_\{u^2+v^2+w^2\leq R^2\} (a+b+c+u+v+w)\mathrm\{ d\} u\mathrm\{ d\} v\mathrm\{ d\} w\\\\ &\xlongequal\{\tiny\mbox\{对称性\}\} 2(a+b+c)\cdot \frac\{4\pi R^3\}\{3\} =\frac\{8\pi R^3(a+b+c)\}\{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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---
773、 3、 求
\begin\{aligned\} I=\iint\_\varSigma (x^3+z^2)\mathrm\{ d\} y\mathrm\{ d\} z+(y^3+ax^2)\mathrm\{ d\} z\mathrm\{ d\} x+(z^3+ay^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 $\displaystyle \varSigma$ 为上半球面 $\displaystyle z=\sqrt\{a^2-x^2-y^2\}$ ($a > 0$), 取上侧. (南昌大学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 $\displaystyle S: z=0, x^2+y^2\leq a^2$, 取上侧, 则
\begin\{aligned\} \iint\_S \cdots&=\iint\_\{x^2+y^2\leq a^2\} ay^2\mathrm\{ d\} x\mathrm\{ d\} y =\frac\{a\}\{2\}\iint\_\{x^2+y^2\leq a^2\} (x^2+y^2)\mathrm\{ d\} x\mathrm\{ d\} y\\\\ &=\frac\{a\}\{2\}\int\_0^a r^2\cdot 2\pi r\mathrm\{ d\} r=\frac\{\pi a^5\}\{4\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由 Gauss 公式,
\begin\{aligned\} I&=\left\[\iint\_\{\varSigma\}-\iint\_S\cdots\right\]+\iint\_S \cdots\\\\ &=\iiint\_\{x^2+y^2+z^2\leq a^2\atop z\geq 0\} (3x^2+3y^2+3z^2)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z+\frac\{\pi a^5\}\{4\}\\\\ &=3\int\_0^a r^2\cdot 2\pi r^2\mathrm\{ d\} r+\frac\{\pi a^5\}\{4\} =\frac\{6\pi a^5\}\{5\}+\frac\{\pi a^5\}\{4\} =\frac\{29\pi\}\{20\}. \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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---
774、 2、 (20 分) (1)、 (10 分) 设 $\displaystyle C$ 为三维空间中环绕 $\displaystyle z$ 轴一周的光滑简单闭曲线 (与 $\displaystyle z$ 轴无交点), 其定向与 $\displaystyle z$ 轴成右手系, 记
\begin\{aligned\} (P,Q,R)=\left(\frac\{-y\}\{x^2+y^2\},\frac\{x\}\{x^2+y^2\},0\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
求 $\displaystyle \int\_C P\mathrm\{ d\} x+Q\mathrm\{ d\} y+R\mathrm\{ d\} z$. (南京大学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) / $\displaystyle \left|\begin\{array\}\{cccccccccc\}\vec\{i\}&\vec\{j\}&\vec\{k\}\\\\ \frac\{\partial\}\{\partial x\}&\frac\{\partial\}\{\partial y\}&\frac\{\partial\}\{\partial z\}\\\\ P&Q&R\end\{array\}\right|=\vec\{0\}$. 令 $\displaystyle C\_\varepsilon: x^2+y^2=\varepsilon^2\ll 1, z=0$, 逆时针为正向; 并取以 $\displaystyle C\_\varepsilon,C$ 为边界的曲面 $\displaystyle S$, 则
\begin\{aligned\} \mbox\{原式\}=&\oint\_\{C\_\varepsilon\}\cdots+\left\[-\oint\_\{C\_\varepsilon\}\cdots+\oint\_C \cdots\right\]\\\\ \xlongequal\{\tiny\mbox\{Stokes\}\}&\frac\{1\}\{\varepsilon^2\} (-y\mathrm\{ d\} x+x\mathrm\{ d\} y)+\iint\_S (Q\_x-P\_y)\mathrm\{ d\} x\mathrm\{ d\} y\\\\ \xlongequal\{\tiny\mbox\{Green\}\}&\frac\{1\}\{\varepsilon^2\}\iint\_\{x^2+y^2\leq\varepsilon^2\}2\mathrm\{ d\} x\mathrm\{ d\} y+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\}
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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---
775、 (2)、 (10 分) 设 $\displaystyle S$ 为三维空间中半径为 $\displaystyle R$ 的球面, $\displaystyle \mathbb\{R\}^3\backslash\left\\{0\right\\}$ 中光滑向量场
\begin\{aligned\} \vec\{F\}(x,y,z)=f\left(\left\Vert \vec\{r\}\right\Vert \right)\vec\{r\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle \vec\{r\}=(x,y,z)$, $\displaystyle f$ 只依赖于 $\displaystyle \left\Vert \vec\{r\}\right\Vert $, $\displaystyle \vec\{n\}$ 为 $\displaystyle S$ 上单位外法向量. 若 $\displaystyle \int\_S \vec\{F\}\cdot\vec\{n\}\mathrm\{ d\} \sigma$ 不依赖于 $\displaystyle R$, 求证: 存在常数 $\displaystyle C$ 使得
\begin\{aligned\} f\left(\left\Vert \vec\{r\}\right\Vert \right)=C\left\Vert \vec\{r\}\right\Vert ^\{-3\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(南京大学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 r=\left\Vert \vec\{r\}\right\Vert $, 则
\begin\{aligned\} &\vec\{F\}=f(r)\vec\{r\}=\left\\{f(r)x,f(r)y,f(r)z\right\\}\\\\ \Rightarrow&\mathrm\{ div\} \vec\{F\}=\frac\{\partial\}\{\partial x\}[f(r)x] +\frac\{\partial\}\{\partial y\}[f(r)y] +\frac\{\partial\}\{\partial z\}[f(r)z]=3f(r)+rf'(r). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由题设,
\begin\{aligned\} 0&=\frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} R\}\int\_\{\partial B\_R\}\vec\{F\}\cdot \vec\{n\}\mathrm\{ d\} S \xlongequal\{\tiny\mbox\{Gauss\}\} \frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} R\}\int\_\{B\_R\} \mathrm\{ div\}\vec\{F\}\mathrm\{ d\} V\\\\ &=\frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} R\}\int\_0^R [3f(r)+rf'(r)]\cdot 4\pi r^2\mathrm\{ d\} r\\\\ &=[3f(R)+Rf'(R)]\cdot 4\pi R^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\} &3f(R)+Rf'(R)=0\\\\ \Rightarrow& 0=3R^2f(R)+R^3f(R)=\frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} R\}[R^3f(R)]\\\\ \Rightarrow&R^3f(R)=C\Rightarrow f(r)=\frac\{C\}\{r^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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---
776、 10、 计算曲面积分
\begin\{aligned\} \iint\_S \left(\frac\{x^3\}\{a^2\}+yz\right)\mathrm\{ d\} y\mathrm\{ d\} z +\left(\frac\{y^3\}\{b^2\}+z^2x^2\right)\mathrm\{ d\} z\mathrm\{ d\} x +\left(\frac\{z^3\}\{c^2\}+x^3y^3\right)\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 $\displaystyle \frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}+\frac\{z^2\}\{c^2\}=1\ (x\geq 0)$, 取后侧, $\displaystyle a,b,c > 0$. (南京航空航天大学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 $\displaystyle \varSigma: \frac\{y^2\}\{b^2\}+\frac\{z^2\}\{c^2\}\leq 1, x=0$, 取后侧, 则
\begin\{aligned\} \iint\_\varSigma \cdots=-\iint\_\{\frac\{y^2\}\{b^2\}+\frac\{z^2\}\{c^2\}\leq 1\} yz\mathrm\{ d\} y\mathrm\{ d\} z\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\}=&-\left\[-\mbox\{原式\}+\iint\_\varSigma\cdots\right\] \xlongequal\{\tiny\mbox\{Gauss\}\}-3\iiint\_\{\frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}+\frac\{z^2\}\{c^2\}\leq 1\atop x\geq 0\} \left(\frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}+\frac\{z^2\}\{c^2\}\right)\mathrm\{ d\} x \mathrm\{ d\} y\mathrm\{ d\} z\\\\ =&-3abc\iiint\_\{u^2+v^2+w^2\leq 1\atop u\geq 0\} (u^2+v^2+w^2)\mathrm\{ d\} u\mathrm\{ d\} v\mathrm\{ d\} w\\\\ =&-3abc\int\_0^1 r^2\mathrm\{ d\} r\int\_0^\pi \mathrm\{ d\} \phi \int\_\{-\frac\{\pi\}\{2\}\}^\frac\{\pi\}\{2\} r^2\sin\phi \mathrm\{ d\} \theta =-\frac\{6\pi abc\}\{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](
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---
777、 12、 求均匀曲线 $\displaystyle x^2+y^2+z^2=a^2, xyz=0, x\geq 0, y\geq 0, z\geq 0$ 的重心. (南京航空航天大学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) / 由对称性知题中曲线 $\displaystyle L$ 的重心为 $\displaystyle (b,b,b)$, 其中 $\displaystyle b=\frac\{\displaystyle\int\_L x\mathrm\{ d\} s\}\{\displaystyle\int\_L \mathrm\{ d\} x\}$. 由
\begin\{aligned\} \int\_L \mathrm\{ d\} s=3\cdot \frac\{2\pi\}\{4\}=\frac\{3\pi\}\{2\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
\begin\{aligned\} \int\_L x\mathrm\{ d\} s=&\int\_\{x^2+z^2=a^2\atop x,z\geq 0\}x\mathrm\{ d\} s+\int\_\{x^2+y^2=a^2\atop x,y\geq 0\}x\mathrm\{ d\} s =2\int\_0^\frac\{\pi\}\{2\}a\cos\theta\cdot a\mathrm\{ d\}\theta=2a^2 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知 $\displaystyle b=\frac\{4a^2\}\{3\pi\}$. 故重心为 $\displaystyle \left(\frac\{4a^2\}\{3\pi\},\frac\{4a^2\}\{3\pi\},\frac\{4a^2\}\{3\pi\}\right)$.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)/
---
778、 3、 (25 分)计算曲面积分
\begin\{aligned\} \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 $\displaystyle S$ 为 $\displaystyle z=4, z=4-(x^2+y^2)$ 与 $\displaystyle x^2+y^2=4$ 所围外侧曲面. (南开大学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\}=&3\iiint\_\{0\leq 4-z\leq x^2+y^2\leq 4\}(x^2+y^2+z^2)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z\\\\ =&3\int\_0^4 \mathrm\{ d\} z\iint\_\{4-z\leq x^2+y^2\leq 4\}(x^2+y^2+z^2)\mathrm\{ d\} x\mathrm\{ d\} y\\\\ =&3\int\_0^4 \mathrm\{ d\} z\int\_\{\sqrt\{4-z\}\}^2 (r^2+z^2)\cdot 2\pi r\mathrm\{ d\} r=256\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)/
---
779、 8、 计算
\begin\{aligned\} \iint\_\varSigma (x+y^2+z^2)\mathrm\{ d\} y\mathrm\{ d\} z-z\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 $\displaystyle \varSigma$ 为旋转抛物面 $\displaystyle z=x^2+y^2$ 被 $\displaystyle z=2$ 所截的部分的下侧. (厦门大学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 $\displaystyle S: x^2+y^2\leq 2, z=2$, 取上侧, 则
\begin\{aligned\} \iint\_S \cdots=-2\iint\_\{x^2+y^2\leq 2\}\mathrm\{ d\} x\mathrm\{ d\} y=-2\cdot \pi 2=-4\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\}=&\left\[\iint\_\varSigma\cdots+\iint\_S\cdots\right\]-\iint\_S\cdots\\\\ \xlongequal\{\tiny\mbox\{Gauss\}\}&\iiint\_\{x^2+y^2\leq 2\}(1-1)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z-(-4\pi)=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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780、 6、 (15 分) 计算 $\displaystyle \oint\_L \frac\{x\mathrm\{ d\} y-y\mathrm\{ d\} x\}\{4x^2+y^2\}$, 其中 $\displaystyle L$ 是以 $\displaystyle (1,0)$ 为圆心, 以 $\displaystyle R$ 为半径的圆周 $\displaystyle (R\neq 1)$. (陕西师范大学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 $\displaystyle (P,Q)=\frac\{(-y,x)\}\{4x^2+y^2\}$, 则 $\displaystyle Q\_x=P\_y$. (1)、 若 $\displaystyle R < 1$, 则 $\displaystyle P,Q$ 在 $\displaystyle L$ 的内部连续可微, 而由 Green 公式知原式 $\displaystyle =0$. (2)、 若 $\displaystyle R > 1$, 则 $\displaystyle L$ 的内部有 $\displaystyle P,Q$ 的奇点 $\displaystyle (0,0)$, 而
\begin\{aligned\} \mbox\{原式\}&\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\} x\mathrm\{ d\} y-y\mathrm\{ d\} x\\\\ &\xlongequal\{\tiny\mbox\{Green\}\} \frac\{1\}\{\varepsilon^2\}\iint\_\{4x^2+y^2 < \varepsilon^2\} 2\mathrm\{ d\} x\mathrm\{ d\} y \stackrel\{2x=u, y=v\}\{=\}\frac\{1\}\{\varepsilon^2\}\iint\_\{u^2+v^2 < \varepsilon^2\}\mathrm\{ d\} u\mathrm\{ d\} v=\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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781、 10、 (15 分) 求曲面积分 $\displaystyle \iint\_\varSigma \frac\{\mathrm\{ d\} S\}\{z\}$, 其中 $\displaystyle \varSigma$ 为 $\displaystyle x^2+y^2+z^2=a^2$ 被平面 $\displaystyle z=h\ (0 < h < a)$ 所截的顶部. (陕西师范大学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\} I&=\iint\_\{x^2+y^2\leq a^2-h^2\} \frac\{1\}\{z\}\sqrt\{1+z\_x^2+z\_y^2\}\mathrm\{ d\} x\mathrm\{ d\} y =\iint\_\{x^2+y^2\leq a^2-h^2\} \frac\{a\}\{a^2-(x^2+y^2)\}\mathrm\{ d\} x\mathrm\{ d\} y\\\\ &=\int\_0^\{\sqrt\{a^2-h^2\}\} \frac\{a\}\{a^2-r^2\}\cdot 2\pi r\mathrm\{ d\} r =2\pi a\ln \frac\{a\}\{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)/
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782、 2、 求曲线曲面积分. (1)、 求
\begin\{aligned\} \int\_L(y+z)\mathrm\{ d\} x-(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\}
其中 $\displaystyle L$ 是 $\displaystyle x^2+y^2=1$ 被 $\displaystyle x-y+z=1$ 所截曲线, 从 $\displaystyle z$ 轴正向看为顺时针. (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) /
\begin\{aligned\} \mbox\{原式\}&\xlongequal\{\tiny\mbox\{Stokes\}\} -\iint\_\{x^2+y^2\leq 1\atop x-y+z=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&-z-x&x-y \end\{array\}\right|\mathrm\{ d\} S\\\\ &=-\frac\{1\}\{\sqrt\{3\}\}\iint\_\{x^2+y^2\leq 1\atop x-y+z=1\} (-2)\mathrm\{ d\} S =2\iint\_\{x^2+y^2\leq 1\} \mathrm\{ d\} x\mathrm\{ d\} y=2\pi. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
当然也可用参数方程来做: $\displaystyle x=\cos\theta, y=\sin\theta, z=1-\cos\theta+\sin\theta, \theta: 2\pi\to 0$.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)/
发表于 2023-3-5 09:13:08
