## 张祖锦2023年数学专业真题分类70天之第32天
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714、 2、 (20 分) 计算 $\displaystyle I=\iiint\_\varOmega z\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z$, 其中 $\displaystyle \varOmega$ 是两个球面 $\displaystyle x^2+y^2+z^2=2z$ 与 $\displaystyle x^2+y^2+z^2=z$ 之间的点集. (上海交通大学2023年数学分析考研试题) [重积分 ]
[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 $\displaystyle x=r\sin\phi\cos\theta, y=r\sin\phi \sin\theta, z=r\cos\phi$, 则
\begin\{aligned\} r\cos\phi\leq r^2\leq 2r\cos\phi\Leftrightarrow \cos\phi\leq r\leq 2\cos\phi. \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=\int\_0^\frac\{\pi\}\{2\}\mathrm\{ d\}\phi \int\_0^\{2\pi\}\mathrm\{ d\} \theta\int\_\{\cos\phi\}^\{2\cos\phi\}r\cos\phi\cdot r^2\sin\phi\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](
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715、 10、 设 $\displaystyle f$ 连续, $\displaystyle F(t)=\iiint\_\{\varOmega\_t\}\left\[x^2+f(x^2+y^2)\right\]\mathrm\{ d\} V$, 其中
\begin\{aligned\} \varOmega\_t=\left\\{(x,y,z);x^2+y^2\leq t^2, 0\leq z\leq \sqrt\{x^2+y^2\}\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(1)、 求 $\displaystyle \frac\{\mathrm\{ d\} F\}\{\mathrm\{ d\} t\}$; (2)、 求 $\displaystyle \lim\_\{t\to 0\}\frac\{F(t)\}\{t^3\}$. (首都师范大学2023年数学分析考研试题) [重积分 ]
[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)、
\begin\{aligned\} F(t)\xlongequal\{\tiny\mbox\{对称性\}\}&\int\_0^t \mathrm\{ d\} z\iint\_\{z^2\leq x^2+y^2\leq t^2\}\left\[\frac\{x^2+y^2\}\{2\}+f(x^2+y^2)\right\]\mathrm\{ d\} x\mathrm\{ d\} y\\\\ =&\int\_0^t \mathrm\{ d\} z\int\_z^t \left\[\frac\{r^2\}\{2\}+f(r^2)\right\]\cdot 2\pi r\mathrm\{ d\} r. \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\} F'(t)=\int\_0^t \left\[\frac\{t^2\}\{2\}+f(t)^2\right\]\cdot 2\pi t\mathrm\{ d\} r=2\pi\left\[\frac\{t^2\}\{2\}+f(t^2)\right\]t^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、
\begin\{aligned\} \mbox\{原式\}&\xlongequal\{\tiny\mbox\{L'Hospital\}\} \lim\_\{t\to 0\}\frac\{F'(t)\}\{3t^2\} =\frac\{1\}\{3\}\lim\_\{t\to 0\}2\pi\left\[\frac\{t^2\}\{2\}+f(t^2)\right\] =\frac\{2\pi f(0)\}\{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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716、 5、 (15 分) 求球体
\begin\{aligned\} (x-2)^2+(y-2)^2+(z-2)^2\leq 12 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
在平面 $\displaystyle x+y+z=3$ 上方的体积. (苏州大学2023年数学分析考研试题) [重积分 ]
[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) / 设球体在平面下方的部分为 $\displaystyle \varOmega$, 则所求
\begin\{aligned\} =\frac\{4\pi\}\{3\}(2\sqrt\{3\})^3-|\varOmega|. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设 $\displaystyle x-2=u, y-2=v, z-2=w$, 则
\begin\{aligned\} |\varOmega|=\iiint\_\{u^2+v^2+w^2\leq 12\atop u+v+w\leq -3\}\mathrm\{ d\} u\mathrm\{ d\} v\mathrm\{ d\} w. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
将 $\displaystyle \eta\_1=\frac\{(1,1,1)^\mathrm\{T\}\}\{\sqrt\{3\}\}$ 扩充为 $\displaystyle \mathbb\{R\}^3$ 的一组标准正交基 $\displaystyle P=(\eta\_1,\eta\_2,\eta\_3)$, 并作正交线性变换
\begin\{aligned\} \left(\begin\{array\}\{cccccccccccccccccccc\}X\\\\Y\\\\Z\end\{array\}\right)=\left(\begin\{array\}\{cccccccccccccccccccc\}\eta\_1\\\\\eta\_2\\\\\eta\_3\end\{array\}\right)\left(\begin\{array\}\{cccccccccccccccccccc\}u\\\\v\\\\w\end\{array\}\right)=P^\mathrm\{T\}\left(\begin\{array\}\{cccccccccccccccccccc\}u\\\\v\\\\w\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\}
则
\begin\{aligned\} |\varOmega|=&\iiint\_\{X^2+Y^2+Z^2\leq 12\atop X\leq -\sqrt\{3\}\}\mathrm\{ d\} X\mathrm\{ d\} Y\mathrm\{ d\} Z =\int\_\{-2\sqrt\{3\}\}^\{\sqrt\{3\}\}\pi(12-X^2)\mathrm\{ d\} X=5\sqrt\{3\}\pi. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是所求 $\displaystyle =32\sqrt\{3\}\pi-5\sqrt\{3\}\pi=27\sqrt\{3\}\pi$.跟锦数学微信公众号. [在线资料](
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717、 4、 设 $\displaystyle z=z(x,y)$ 是由 $\displaystyle x-az=\varphi(y-bz)$ 确定的隐函数, 其中 $\displaystyle \varphi$ 是连续可微函数, 且 $\displaystyle a-b\varphi'\neq 0$. 求重积分
\begin\{aligned\} \iint\_D (az\_x+bz\_y)\mathrm\{e\}^\frac\{x-y\}\{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 $\displaystyle D$ 是由 $\displaystyle x=0, y=0, x+y=1$ 所围成的区域. (太原理工大学2023年数学分析考研试题) [重积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle x-az=\varphi(y-bz)$ 知
\begin\{aligned\} 1-az\_x=\varphi'(y-bz)(-bz\_x)\Rightarrow& z\_x=\frac\{1\}\{a-b\varphi'(y-bz)\},\\\\ -az\_y=\varphi'(y-bz)(1-bz\_y)\Rightarrow& z\_y=\frac\{-\varphi'(y-bz)\}\{a-b\varphi'(y-bz)\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle az\_x+bz\_y=1$. 设
\begin\{aligned\} u=x-y, v=x+y, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则
\begin\{aligned\} \frac\{\partial(u,v)\}\{\partial(x,y)\}=2, x=\frac\{u+v\}\{2\}, y=\frac\{v-u\}\{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\} x\geq 0\Leftrightarrow u\geq -v, y\geq 0\Leftrightarrow u\leq v, x+y\leq 1\Leftrightarrow v\leq 1. \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\}=&\iint\_D \mathrm\{e\}^\frac\{x-y\}\{x+y\}\mathrm\{ d\} x\mathrm\{ d\} y =\int\_0^1 \mathrm\{ d\} v\int\_\{-v\}^v \mathrm\{e\}^\frac\{u\}\{v\}\cdot \frac\{1\}\{2\}\mathrm\{ d\} u\\\\ =&\frac\{1\}\{2\}\int\_0^1 v(\mathrm\{e\}-\mathrm\{e\}^\{-1\})\mathrm\{ d\} v =\frac\{1\}\{4\}\left(\mathrm\{e\}-\frac\{1\}\{\mathrm\{e\}\}\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](
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718、 (2)、 讨论 $\displaystyle \iint\_D \frac\{\sin^3x\}\{x\}\mathrm\{ d\} x\mathrm\{ d\} y$ 的敛散性, 其中 $\displaystyle D=(0,\infty)\times [0,1]$. (天津大学2023年数学分析考研试题) [重积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 注意: 反常二重积分的收敛就是它的绝对收敛性 (参考华东师大数学分析第 5 版下册第 251 页定理 21.19). 由
\begin\{aligned\} &\iint\_D \left|\frac\{\sin^3x\}\{x\}\right|\mathrm\{ d\} x\mathrm\{ d\} y \geq \iint\_D \frac\{\sin^4x\}\{x\}\mathrm\{ d\} x\mathrm\{ d\} y\\\\ =&\int\_0^1 \mathrm\{ d\} y\int\_0^\infty \left(\frac\{3\}\{8\}-\frac\{\cos 2x\}\{2\}+\frac\{\cos 4x\}\{8\}\right)\mathrm\{ d\} x\mathrm\{ d\} y=+\infty \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知原式发散!跟锦数学微信公众号. [在线资料](
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---
719、 (3)、 求二重积分 $\displaystyle \iint\_D \frac\{\mathrm\{ d\} \sigma\}\{\sqrt\{x+y+4\}\}$, 其中 $\displaystyle D: |x|+|y|\leq 1$. (武汉理工大学2023年数学分析考研试题) [重积分 ]
[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 $\displaystyle x+y=u, x-y=v$, 则 $\displaystyle \left|\frac\{\partial(x,y)\}\{\partial(u,v)\}\right|=2$, 而
\begin\{aligned\} \mbox\{原式\}=&\int\_\{-1\}^1 \mathrm\{ d\} u\int\_\{-1\}^1 \frac\{1\}\{\sqrt\{u+4\}\}\cdot \frac\{1\}\{2\}\mathrm\{ d\} v\\\\ =&\int\_\{-1\}^1 \frac\{1\}\{\sqrt\{u+4\}\}\mathrm\{ d\} u=2\left(\sqrt\{5\}-\sqrt\{3\}\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](
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720、 (4)、 求 $\displaystyle I=\int\_0^1 \mathrm\{ d\} x\int\_0^\{1-x\}\mathrm\{ d\} z\int\_0^\{1-x-z\}(1-y)\mathrm\{e\}^\{-(1-y-z)^2\}\mathrm\{ d\} y$. (武汉理工大学2023年数学分析考研试题) [重积分 ]
[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\_0^1 \mathrm\{ d\} y\int\_0^\{1-y\}\mathrm\{ d\} z\int\_0^\{1-y-z\}(1-y)\mathrm\{e\}^\{-(1-y-z)^2\}\mathrm\{ d\} x\\\\ =&\int\_0^1 (1-y)\mathrm\{ d\} y\int\_0^\{1-y\} (1-y-z)\mathrm\{e\}^\{-(1-y-z)^2\}\mathrm\{ d\} z\\\\ \stackrel\{1-y-z=t\}\{=\}&\int\_0^1 (1-y)\mathrm\{ d\} y\int\_0^\{1-y\} t\mathrm\{e\}^\{-t^2\}\mathrm\{ d\} t\\\\ =&\int\_0^1 (1-y)\cdot\frac\{1\}\{2\}\left\[1-\mathrm\{e\}^\{-(1-y)^2\}\right\]\mathrm\{ d\} y\\\\ \stackrel\{1-y=s\}\{=\}&\cdots\xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} \cdots=\frac\{1\}\{4\mathrm\{e\}\}. \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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---
721、 (9)、 $\displaystyle \iint\_\{x^2+y^2\leq 1\}x^2\sin y\mathrm\{ d\} x\mathrm\{ d\} y=\underline\{\ \ \ \ \ \ \ \ \ \ \}$. (西安交通大学2023年数学分析考研试题) [重积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / $\displaystyle \mbox\{原式\}\xlongequal\{\tiny\mbox\{对称性\}\} 0$.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](
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722、 5、 (20 分) 计算二重积分
\begin\{aligned\} \iint\_D r^2\sin\theta\sqrt\{1-r^2\cos2\theta\}\mathrm\{ d\} r\mathrm\{ d\} \theta, \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 D=\left\\{(\theta,r); 0\leq r\leq \sec\theta, \theta\in\left\[0,\frac\{\pi\}\{4\}\right\]\right\\}$. (西南财经大学2023年数学分析考研试题) [重积分 ]
[纸质资料](
https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](
https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/[pdf1](
https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/[pdf2](
https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 由 $\displaystyle r\cos \theta\leq 1, 0\leq\theta\leq\frac\{\pi\}\{4\}\Leftrightarrow x\leq 1, 0\leq y\leq x$ 知
\begin\{aligned\} \mbox\{原式\}=&\iint\_D r\sin\theta\sqrt\{1-r^2(\cos^2\theta-\sin^2\theta)\}\cdot r\mathrm\{ d\} r\mathrm\{ d\} \theta\\\\ =&\int\_0^1 \mathrm\{ d\} x\int\_0^x y\sqrt\{1-x^2+y^2\}\mathrm\{ d\} y\\\\ \stackrel\{y^2=s\}\{=\}&\frac\{1\}\{2\}\int\_0^1 \mathrm\{ d\} x\int\_0^\{x^2\} \sqrt\{1-x^2+s\}\mathrm\{ d\} s =\frac\{1\}\{3\}\int\_0^1 [1-(1-x^2)^\frac\{3\}\{2\}]\mathrm\{ d\} x\\\\ \stackrel\{x=\sin t\}\{=\}&\cdots=\frac\{1\}\{3\}-\frac\{\pi\}\{16\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
723、 (2)、 (10 分) 求 $\displaystyle \int\_0^1 \mathrm\{ d\} y\int\_y^1 \frac\{y\}\{\sqrt\{1+x^3\}\}\mathrm\{ d\} x$. (西南大学2023年数学分析考研试题) [重积分 ]
[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\_0^1\frac\{\mathrm\{ d\} x\}\{\sqrt\{1+x^3\}\}\int\_0^x y\mathrm\{ d\} y =\frac\{1\}\{2\}\int\_0^1 \frac\{x^2\}\{\sqrt\{1+x^3\}\}\mathrm\{ d\} x\\\\ \stackrel\{x^3=t\}\{=\}&\frac\{1\}\{6\}\int\_0^1 \frac\{\mathrm\{ d\} t\}\{\sqrt\{1+t\}\} =\left.\frac\{1\}\{3\}\sqrt\{1+t\}\right|\_0^1=\frac\{\sqrt\{2\}-1\}\{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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---
724、 (3)、 (10 分) 求
\begin\{aligned\} \iiint\_V (y-z)\arctan z\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 $\displaystyle V$ 是由曲面 $\displaystyle x^2+\frac\{(y-z)^2\}\{2\}=r^2, z=0$ 及 $\displaystyle z=h$ 所围成的立体. (西南大学2023年数学分析考研试题) [重积分 ]
[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^h \mathrm\{ d\} z\int\_\{-r\}^r \mathrm\{ d\} x\int\_\{z-\sqrt\{2(r^2-x^2)\}\}^\{z+\sqrt\{2(r^2-x^2)\}\}(y-z)\arctan z\mathrm\{ d\} y\\\\ \stackrel\{y-z=t\}\{=\}&\int\_0^h \mathrm\{ d\} z\int\_\{-r\}^r \mathrm\{ d\} x\int\_\{-\sqrt\{2(r^2-x^2)\}\}^\{\sqrt\{2(r^2-x^2)\}\}t\mathrm\{ d\} t \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\}
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)/
---
725、 10、 (10 分) 求 $\displaystyle \iiint\_V y\cos(x+z)\mathrm\{ d\} V$, 其中 $\displaystyle V$ 由 $\displaystyle y=\sqrt\{x\}, y=0, z=0$ 及 $\displaystyle x+z=\frac\{\pi\}\{2\}$ 所围成. (西南交通大学2023年数学分析考研试题) [重积分 ]
[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\}\mathrm\{ d\} x\int\_0^\{\sqrt\{x\}\}\mathrm\{ d\} y\int\_0^\{\frac\{\pi\}\{2\}-x\} y\cos (x+z)\mathrm\{ d\} z\\\\ =&\int\_0^\frac\{\pi\}\{2\}\mathrm\{ d\} x\int\_0^\{\sqrt\{x\}\} y(1-\sin x)\mathrm\{ d\} y=\cdots=\frac\{\pi^2-8\}\{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)/
---
726、 8、 (15 分) 计算 $\displaystyle \iiint\_V (x^2-x^2y+xy+y^2)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z$, 其中 $\displaystyle V: x^2+y^2+z^2\leq 1$. (云南大学2023年数学分析考研试题) [重积分 ]
[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\}\}&\iiint\_V (x^2+y^2)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z \xlongequal\{\tiny\mbox\{对称性\}\}\frac\{2\}\{3\}\iiint\_V (x^2+y^2+z^2)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z\\\\ =&\frac\{2\}\{3\}\int\_0^1 r^2\cdot 4\pi r^2\mathrm\{ d\} r=\frac\{8\pi\}\{15\}. \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)/
---
727、 (5)、 设函数 $\displaystyle f(x,y)$ 连续, 交换二重积分的次序 $\displaystyle \int\_0^4 \mathrm\{ d\} x\int\_x^\{2\sqrt\{x\}\} f(x,y)\mathrm\{ d\} y$ 得 $\displaystyle \underline\{\ \ \ \ \ \ \ \ \ \ \}$. (长安大学2023年数学分析考研试题) [重积分 ]
[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) / $\displaystyle \int\_0^4 \mathrm\{ d\} y\int\_\frac\{y^2\}\{4\}^y f(x,y)\mathrm\{ d\} x$.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)/
---
728、 (6)、 (10 分) 设 $\displaystyle f(x,y)$ 在 $\displaystyle D=\left\\{(x,y); x^2+y^2\leq R^2\right\\}$ 上有一阶连续偏导数, 在圆周 $\displaystyle x^2+y^2=R^2$ 上 $\displaystyle f(x,y)=0$, 且 $\displaystyle f(0,0)=2023$, 求
\begin\{aligned\} \lim\_\{\varepsilon\to 0^+\}\frac\{-1\}\{2\pi\}\iint\_\{\varepsilon^2\leq x^2+y^2\leq R^23\}\frac\{xf\_x'+yf\_y'\}\{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\}
(长安大学2023年数学分析考研试题) [重积分 ]
[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\} &\quad \lim\_\{\varepsilon\to 0^+\}\iint\_\{\varepsilon^2\leq x^2+y^2\leq 1\} \frac\{xf\_x'+yf\_y'\}\{x^2+y^2\}\mathrm\{ d\} x\mathrm\{ d\} y\\\\ &=\lim\_\{\varepsilon\to 0^+\} \int\_0^\{2\pi\} \int\_\varepsilon^1 \frac\{r\cos \theta f\_x'+r\sin \theta f\_y'\}\{r^2\}\cdot r\mathrm\{ d\} r \mathrm\{ d\} \theta\\\\ &\quad \left(x=r\cos\theta, y=r\sin \theta\right)\\\\ &=\lim\_\{\varepsilon\to 0^+\}\int\_0^\{2\pi\} \int\_\varepsilon^1 (\cos \theta f\_x'+\sin \theta f\_y')\mathrm\{ d\} r \mathrm\{ d\} \theta\\\\ &=\lim\_\{\varepsilon\to 0^+\}\int\_0^\{2\pi\} \int\_\varepsilon^1 \frac\{\partial f\}\{\partial r\}\mathrm\{ d\} r\mathrm\{ d\} \theta\\\\ &=\lim\_\{\varepsilon\to 0^+\}\int\_0^\{2\pi\} -f(\varepsilon\cos\theta,\varepsilon\sin \theta)\mathrm\{ d\} \theta\left(f|\_\{\partial D\}=0\right)\\\\ &=-2\pi\cdot \lim\_\{\varepsilon\to 0^+\} \frac\{1\}\{2\pi\}\int\_0^\{2\pi\} f(\varepsilon \cos\theta, \varepsilon\sin\theta)\mathrm\{ d\} \theta\\\\ &=-2\pi f(0) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知原式 $\displaystyle =f(0,0)=2023$.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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---
729、 3、 (10 分) 设 $\displaystyle f(x)$ 在 $\displaystyle \mathbb\{R\}$ 上是连续函数, 求二重积分
\begin\{aligned\} \iint\_D x[1-yf(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 $\displaystyle D$ 为 $\displaystyle y=x^3, y=1, x=-1$ 所围成的区域. (郑州大学2023年数学分析考研试题) [重积分 ]
[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 $\displaystyle F=xyf(x^2-y^2)$, 则 $\displaystyle F(-x,-y)=F(x,y)$. 而可将 $\displaystyle D$ 在 $\displaystyle x\geq 0$ 的部分移到它关于原点的对称部分.
\begin\{aligned\} \mbox\{原式\}\xlongequal\{\tiny\mbox\{对称性\}\}&\int\_D x\mathrm\{ d\} x\mathrm\{ d\} y+\int\_\{[-1,0]\times [-1,1]\}F(x,y)\mathrm\{ d\} x\mathrm\{ d\} y\\\\ \xlongequal\{\tiny\mbox\{对称性\}\}&\int\_\{-1\}^1 x\mathrm\{ d\} x\int\_\{x^3\}^\{1\}\mathrm\{ d\} y+0\left(\mbox\{$F$ 关于 $\displaystyle y$ 是奇函数\}\right)\\\\ \xlongequal\{\tiny\mbox\{对称性\}\}&-\frac\{2\}\{5\}+0=-\frac\{2\}\{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)/
---
730、 8、 (15 分) 设 $\displaystyle r > 0$, 计算球面 $\displaystyle x^2+y^2+z^2=r^2$ 与圆柱面 $\displaystyle x^2+y^2=rx\ (r > 0)$ 所围区域的体积. (中国人民大学2023年数学分析考研试题) [重积分 ]
[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 $\displaystyle x=\rho \cos\theta, y=\rho \sin\theta$, 则 $\displaystyle \rho ^2\leq r\rho \cos\theta\Leftrightarrow \rho \leq r\cos\theta$, 而
\begin\{aligned\} \mbox\{原式\}=&\int\_\{-\frac\{\pi\}\{2\}\}^\frac\{\pi\}\{2\}\mathrm\{ d\} \theta\int\_0^\{r\cos\theta\}\rho \mathrm\{ d\} \rho \int\_\{-\sqrt\{r^2-\rho ^2\}\}^\{\sqrt\{r^2-\rho ^2\}\}\mathrm\{ d\} z\\\\ =&2\int\_\{-\frac\{\pi\}\{2\}\}^\frac\{\pi\}\{2\}\mathrm\{ d\} \theta\int\_0^\{r\cos\theta\}\sqrt\{r^2-\rho ^2\}\rho \mathrm\{ d\} \rho \\\\ \stackrel\{\rho ^2=s\}\{=\}&4\int\_0^\frac\{\pi\}\{2\}\mathrm\{ d\} \theta\int\_0^\{r^2\cos^2\theta\} \sqrt\{r^2-s\}\frac\{\mathrm\{ d\} s\}\{2\}\\\\ =&2\int\_0^\frac\{\pi\}\{2\} \left.-\frac\{2\}\{3\}(r^2-s)^\frac\{3\}\{2\}\right|\_\{s=0\}^\{s=r^2\cos^2\theta\}\mathrm\{ d\} \theta\\\\ =&\frac\{4\}\{3\}\int\_0^\frac\{\pi\}\{2\} (r^3-r^3\sin^3\theta)\mathrm\{ d\} \theta =\frac\{2(3\pi-4)r^3\}\{9\}. \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)/
---
731、 6、 (15 分) 求重积分 $\displaystyle \iint\_\{|x|+|y|\leq 1\}\cos(x+y)\mathrm\{ d\} x\mathrm\{ d\} y$. (中山大学2023年数学分析考研试题) [重积分 ]
[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 $\displaystyle u=x+y, v=x-y$, 则 $\displaystyle \left|\frac\{\partial(u,v)\}\{\partial(x,y)\}\right|=2$, 而
\begin\{aligned\} \mbox\{原式\}=&\iint\_\{[0,1]^2\} \cos u\cdot \frac\{1\}\{2\}\mathrm\{ d\} u\mathrm\{ d\} v =\frac\{1\}\{2\}\int\_\{-1\}^1 \mathrm\{ d\} v\int\_\{-1\}^1 \cos u\mathrm\{ d\} u=2\sin 1. \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)/
---
732、 9、 (10 分) 设函数 $\displaystyle f(x)$ 在区间 $\displaystyle [0,1]$ 上连续且有 $\displaystyle \int\_0^1 f(x)\mathrm\{ d\} x=A$. 证明:
\begin\{aligned\} \int\_0^1 \mathrm\{ d\} x\int\_x^1 f(x)f(y)\mathrm\{ d\} y=\frac\{1\}\{2\}A^2. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(重庆大学2023年数学分析考研试题) [重积分 ]
[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) / 交换 $\displaystyle x,y$ 的位置, 有
\begin\{aligned\} \int\_0^1\mathrm\{ d\} x\int\_x^1 f(x)f(y)\mathrm\{ d\} y =\int\_0^1 \mathrm\{ d\} y\int\_y^1 f(y)f(x)\mathrm\{ d\} x. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
交换积分次序, 又有
\begin\{aligned\} \int\_0^1 \mathrm\{ d\} y\int\_y^1 f(y)f(x)\mathrm\{ d\} x =\int\_0^1 \mathrm\{ d\} x\int\_0^x f(x)f(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\}
于是
\begin\{aligned\} \int\_0^1\mathrm\{ d\} x\int\_x^1 f(x)f(y)\mathrm\{ d\} y =&\frac\{1\}\{2\}\left\[ \int\_0^1\mathrm\{ d\} x\int\_x^1 f(x)f(y)\mathrm\{ d\} y+\int\_0^1 \mathrm\{ d\} x\int\_0^x f(x)f(y)\mathrm\{ d\} y\right\]\\\\ =&\frac\{1\}\{2\}\int\_0^1\mathrm\{ d\} x\int\_0^1 f(x)f(y)\mathrm\{ d\} y\\\\ =&\frac\{1\}\{2\}\int\_0^1f(x)\mathrm\{ d\} x\int\_0^1f(y)\mathrm\{ d\} y =\frac\{A^2\}\{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)/
---
733、 (6)、 求 $\displaystyle \iiint\_V xyz\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z$, 其中 $\displaystyle V: x^2+y^2+z^2\leq a^2, x\geq 0, y\geq 0, z\geq 0$. (重庆师范大学2023年数学分析考研试题) [重积分 ]
[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^a \mathrm\{ d\} r\int\_0^\frac\{\pi\}\{2\}\mathrm\{ d\} \phi \int\_0^\frac\{\pi\}\{2\} r\sin\phi \cos\theta\cdot r\sin\phi \sin\theta\cdot r\cos \phi\cdot r^2\sin\phi\mathrm\{ d\} \theta\\\\ &=\int\_0^a r^5\mathrm\{ d\} r\int\_0^\frac\{\pi\}\{2\}\sin^3\phi \cos\phi\mathrm\{ d\} \phi\int\_0^\frac\{\pi\}\{2\}\sin\theta\cos\theta\mathrm\{ d\} \theta\\\\ &\stackrel\{\sin\phi=t, \sin\theta=s\}\{=\}\cdots=\frac\{a^6\}\{6\}\cdot \frac\{1\}\{4\}\cdot \frac\{1\}\{2\} =\frac\{a^6\}\{36\}. \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)/
---
734、 9、 (10 分) 求积分
\begin\{aligned\} I=\int\_L \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 $\displaystyle L$ 为曲线 $\displaystyle y=\sin x$ 从 $\displaystyle (0,0)$ 到 $\displaystyle (\pi,0)$ 的一段. (安徽大学2023年高等代数考研试题) [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\}=&-\left(-\mbox\{原式\}+\int\_L\cdots\right)\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\{\mathrm\{e\}^\pi-1\}\{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)/
---
735、 9、 求第二型曲面积分
\begin\{aligned\} \iint\_S (xz^2+y^2)\mathrm\{ d\} y\mathrm\{ d\} z+(x^2y+z^4)\mathrm\{ d\} z\mathrm\{ d\} x+(3x^2+4y^2+y^2z)\mathrm\{ d\} x\mathrm\{ d\} y, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle S$ 为椭球面 $\displaystyle x^2+2xy+2y^2+z^2=1$, 方向取其外侧. (北京工业大学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) /
\begin\{aligned\} \mbox\{原式\}&\xlongequal\{\tiny\mbox\{Gauss\}\}\iiint\_\{x^2+2xy+2y^2+z^2\leq 1\}(z^2+x^2+y^2)\mathrm\{ d\} V. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
作变换 $\displaystyle x+y=u, y=v, z=w\Rightarrow \left|\frac\{\partial(u,v,w)\}\{\partial(x,y,z)\}\right|=1$, 则
\begin\{aligned\} \mbox\{原式\}=&\iiint\_\{u^2+v^2+w^2\leq 1\}[w^2+(u-v)^2+v^2]\mathrm\{ d\} u\mathrm\{ d\} v\mathrm\{ d\} w\\\\ \xlongequal\{\tiny\mbox\{对称性\}\}&\iiint\_\{u^2+v^2+w^2\leq 1\}(u^2+2v^2+w^2)\mathrm\{ d\} u\mathrm\{ d\} v\mathrm\{ d\} w\\\\ \xlongequal\{\tiny\mbox\{对称性\}\}&\frac\{4\}\{3\}\iiint\_\{u^2+v^2+w^2\leq 1\}(u^2+v^2+w^2)\mathrm\{ d\} u\mathrm\{ d\} v\mathrm\{ d\} w =\frac\{4\}\{3\}\int\_0^1 r^2\cdot 4\pi r^2\mathrm\{ d\} r=\frac\{16\pi\}\{15\}. \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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736、 8、 (15 分) 设 $\displaystyle a,b$ 为正常数, 函数 $\displaystyle f(x)$ 连续可微, $\displaystyle S$ 是曲面 $\displaystyle z=x^2+y^2$ 与 $\displaystyle z=8-x^2-y^2$ 所围成, 方向取外侧, 计算
\begin\{aligned\} I=\iint\_S \frac\{2\}\{b+y\}f\left(\frac\{a+x\}\{(b+y)^2\}\right)\mathrm\{ d\} y\mathrm\{ d\} z +\frac\{1\}\{a+x\}f\left(\frac\{a+x\}\{(b+y)^2\}\right)\mathrm\{ d\} z\mathrm\{ d\} x+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\}
(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
\begin\{aligned\} P=\frac\{2\}\{b+y\}f\left(\frac\{a+x\}\{(b+y)^2\}\right), Q=\frac\{1\}\{a+x\}f\left(\frac\{a+x\}\{(b+y)^2\}\right), R=z, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
则 $\displaystyle P\_x+Q\_y+R\_z=1$. 再设 $\displaystyle S$ 包围的区域为 $\displaystyle V$, 则
\begin\{aligned\} I&\xlongequal\{\tiny\mbox\{Gauss\}\} \iiint\_V \mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z =\int\_0^4\pi z\mathrm\{ d\} z+\int\_4^8 \pi(8-z)\mathrm\{ d\} z=16\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)/
发表于 2023-3-5 09:12:24
