## 张祖锦2023年数学专业真题分类70天之第31天
---
691、 7、 (15 分) 计算由球面 $\displaystyle x^2+y^2+z^2=4a^2$ 与柱面 $\displaystyle x^2+y^2=2ax$ 所围成的立体图形的体积. (北京师范大学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) / 设 $\displaystyle x=r\cos\theta, y=r\sin\theta$, 则 $\displaystyle r^2\leq 2ar\cos\theta\Leftrightarrow r\leq 2a\cos\theta$, 而
\begin\{aligned\} \mbox\{原式\}=&\int\_\{-\frac\{\pi\}\{2\}\}^\frac\{\pi\}\{2\}\mathrm\{ d\} \theta\int\_0^\{2a\cos\theta\}r\mathrm\{ d\} r \int\_\{-\sqrt\{4a^2-r^2\}\}^\{\sqrt\{4a^2-r^2\}\}\mathrm\{ d\} z\\\\ =&2\int\_\{-\frac\{\pi\}\{2\}\}^\frac\{\pi\}\{2\}\mathrm\{ d\} \theta\int\_0^\{2a\cos\theta\}\sqrt\{4a^2-r^2\}r\mathrm\{ d\} r\\\\ \stackrel\{r^2=s\}\{=\}&4\int\_0^\frac\{\pi\}\{2\}\mathrm\{ d\} \theta\int\_0^\{4a^2\cos^2\theta\} \sqrt\{4a^2-s\}\frac\{\mathrm\{ d\} s\}\{2\}\\\\ =&2\int\_0^\frac\{\pi\}\{2\} \left.-\frac\{2\}\{3\}(4a^2-s)^\frac\{3\}\{2\}\right|\_\{s=0\}^\{s=4a^2\cos^2\theta\}\mathrm\{ d\} \theta\\\\ =&\frac\{4\}\{3\}\int\_0^\frac\{\pi\}\{2\} (8a^3-8a^3\sin^3\theta)\mathrm\{ d\} \theta =\frac\{16(3\pi-4)a^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)/
---
692、 2、 计算题. (1)、 设 $\displaystyle D=\left\\{(x,y); x^2+y^2\leq 1\right\\}$, 实数 $\displaystyle \alpha,\beta$ 满足 $\displaystyle \alpha^2+\beta^2=1$. 计算二重积分
\begin\{aligned\} \iint\_D \frac\{\mathrm\{ d\} x\mathrm\{ d\} y\}\{\sqrt\{(1-\alpha x+\beta y)^2+(\beta x+\alpha y)^2\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(大连理工大学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 $\displaystyle u=\alpha x-\beta y, v=\beta x+\alpha y$, 则 $\displaystyle \frac\{\partial(u,v)\}\{\partial(x,y)\}=\alpha^2+\beta^2=1$,
\begin\{aligned\} x=\alpha u+\beta v, y=-\beta u+\alpha v, x^2+y^2=u^2+v^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\} \mbox\{原式\}=\iint\_\{u^2+v^2\leq 1\} \frac\{1\}\{\sqrt\{(1-u)^2+v^2\}\}\mathrm\{ d\} u\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\}
设 $\displaystyle u=1+r\cos\theta, v=r\sin\theta$, 则
\begin\{aligned\} &1\geq u^2+v^2=1+r^2+2r\cos\theta\Leftrightarrow r\leq -2\cos \theta\\\\ \Rightarrow& \cos\theta\leq 0 \Leftrightarrow \frac\{\pi\}\{2\}\leq \theta \leq \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\}
故
\begin\{aligned\} \mbox\{原式\}=\int\_\frac\{\pi\}\{2\}^\frac\{3\pi\}\{2\}\mathrm\{ d\} \theta\int\_0^\{-2\cos\theta\}\frac\{1\}\{r\}\cdot r\mathrm\{ d\} r=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)/
---
693、 (3)、 求三重积分
\begin\{aligned\} I=\iiint\_\varOmega x^2\sqrt\{x^2+y^2\}\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 $\displaystyle \varOmega$ 是由锥面 $\displaystyle z=\sqrt\{x^2+y^2\}$ 与旋转抛物面 $\displaystyle z=x^2+y^2$ 所围成的有界区域. (电子科技大学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\}\}\frac\{1\}\{2\}\iiint\_\varOmega (x^2+y^2)\sqrt\{x^2+y^2\}\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z\\\\ &=\frac\{1\}\{2\}\int\_0^1 \mathrm\{ d\} z \iint\_\{z^2\leq x^2+y^2\leq z\}(x^2+y^2)\sqrt\{x^2+y^2\}\mathrm\{ d\} x\mathrm\{ d\} y\\\\ &=\frac\{1\}\{2\}\int\_0^1 \mathrm\{ d\} z \int\_z^\{\sqrt\{z\}\} r^3\cdot 2\pi r\mathrm\{ d\} r =\frac\{\pi\}\{42\}. \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)/
---
694、 3、 计算积分
\begin\{aligned\} \iiint\_\varOmega (x^2+y^2+z^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\}
其中$\varOmega$ 是由 $\displaystyle 1\leq x^2+y^2+z^2\leq 4$ 与 $\displaystyle z\geq \sqrt\{x^2+y^2\}$ 围成的区域. (东北大学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\} 1\leq r\leq 2, r\cos\phi\geq r\sin\phi\Leftrightarrow \cos \phi\geq \sin\phi\Leftrightarrow 0\leq \phi\leq\frac\{\pi\}\{4\}, \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\}=\int\_0^\frac\{\pi\}\{4\}\mathrm\{ d\} \phi \int\_0^\{2\pi\}\mathrm\{ d\} \theta \int\_1^2 r^2\cdot r^2\sin\phi\mathrm\{ d\} \phi=\frac\{31\}\{5\}(2-\sqrt\{2\})\pi. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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)/
---
695、 4、 求 $\displaystyle \int\_0^1 \mathrm\{ d\} x\int\_x^\{\sqrt\{x\}\} \frac\{\sin y\}\{y\}\mathrm\{ d\} y$. (东南大学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\{\sin y\}\{y\}\mathrm\{ d\} y\int\_\{y^2\}^y \mathrm\{ d\} x =\int\_0^1 \sin y(1-y)\mathrm\{ d\} y\xlongequal[\tiny\mbox\{积分\}]\{\tiny\mbox\{分部\}\} 1-\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)/
---
696、 5、 求 $\displaystyle \iiint\_\varOmega (x^2+y^2)\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z$, 其中
\begin\{aligned\} \varOmega=\left\\{(x,y,z); x^2+y^2+(z-1)^2\geq 1, x^2+y^2+z^2\leq 4, z\geq 0\right\\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(东南大学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^\frac\{\pi\}\{2\}\mathrm\{ d\} \phi \int\_0^\{2\pi\}\mathrm\{ d\} \theta\int\_\{2\cos\phi\}^2 r^2\sin^2\phi\cdot r^2\sin\phi\mathrm\{ d\} \phi=\cdots=8\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)/
---
697、 (7)、 求三重积分 $\displaystyle \iiint\_V \frac\{\mathrm\{ d\} x\mathrm\{ d\} y\mathrm\{ d\} z\}\{(1+x+y+z)^3\}$, 其中 $\displaystyle V$ 是由平面 $\displaystyle x+y+z=1$ 及三个坐标平面围成的立体图形. (广西大学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\} x\int\_0^\{1-x\}\mathrm\{ d\} y\int\_0^\{1-x-y\}\frac\{\mathrm\{ d\} z\}\{(1+x+y+z)^3\}\\\\ =&\int\_0^1 \mathrm\{ d\} \int\_0^\{1-x\}\left\[\frac\{1\}\{2(1+x+y)^2\}-\frac\{1\}\{8\}\right\]\mathrm\{ d\} y\\\\ =&\frac\{1\}\{8\}\int\_0^1 \left\[x-1+4\left(\frac\{1\}\{1+x\}-\frac\{1\}\{2\}\right)\right\]\mathrm\{ d\} x=\frac\{8\ln 2-5\}\{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)/
---
698、 8、 求二重积分
\begin\{aligned\} \iint\_D \left(\sqrt\{x\}+\sqrt\{y\}\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 $\displaystyle D$ 是由曲线 $\displaystyle \sqrt\{x\}+\sqrt\{y\}=1, x=0$ 及 $\displaystyle y=0$ 围成的闭区域. (哈尔滨工程大学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\}&\stackrel\{\sqrt\{x\}=u, \sqrt\{y\}=v\}\{=\}\iint\_\{0\leq u,v\leq 1\atop u+v\leq 1\} (u+v)\cdot 2u\cdot 2v\mathrm\{ d\} u\mathrm\{ d\} v\\\\ &=4\int\_0^1 \mathrm\{ d\} u\int\_0^\{1-u\} (u+v)uv\mathrm\{ d\} v =\frac\{2\}\{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)/
---
699、 9、 计算极限
\begin\{aligned\} \lim\_\{t\to 0^+\}\frac\{1\}\{t^4\}\iiint\_\{x^2+y^2+z^2\leq t^2\}f\left(\sqrt\{x^2+y^2+z^2\}\right)\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 $\displaystyle f(0)=0, f'(0)=2$. (哈尔滨工业大学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\}=&\lim\_\{t\to 0^+\}\frac\{1\}\{t^4\}\int\_0^t f(t)\cdot4\pi r^2\mathrm\{ d\} r \xlongequal\{\tiny\mbox\{L'Hospital\}\} \lim\_\{t\to 0^+\}\frac\{f(t)\cdot 4\pi t^2\}\{4t^3\}\\\\ =&\pi \lim\_\{t\to 0^+\}\frac\{f(t)-f(0)\}\{t\} =\pi f'(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](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
700、 2、 已知 $\displaystyle f(x,y)$ 为二元连续函数, 交换累次积分 $\displaystyle \int\_0^\pi \mathrm\{ d\} x\int\_\{-\sin\frac\{x\}\{2\}\}^\{\sin x\}f(x,y)\mathrm\{ d\} y$ 的顺序, 使其值保持不变. (华东理工大学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\_\{-1\}^0\mathrm\{ d\} y\int\_\{-2\arcsin y\}^\pi f(x,y)\mathrm\{ d\} x +\int\_0^1 \mathrm\{ d\} y\int\_\{\arcsin y\}^\{\pi-\arcsin y\}f(x,y)\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\}
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)/
---
701、 4、 设 $\displaystyle D$ 是由 $\displaystyle y=x^3, y=c^3, x=-c\ (c\neq 0)$ 围成的积分区域, 且 $\displaystyle f(x)$ 是 $\displaystyle \mathbb\{R\}$ 上的连续函数, 求二重积分
\begin\{aligned\} \iint\_D x\left\[1+yf(1+|\sin x|+\cos y)\right\]\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\}
(华东师范大学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) / 不妨设 $\displaystyle c > 0$. 设 $\displaystyle F=xyf(1+|\sin x|+\cos y)$, 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\_\{[-c,0]\times [-c^3,c^3]\}F(x,y)\mathrm\{ d\} x\mathrm\{ d\} y\\\\ \xlongequal\{\tiny\mbox\{对称性\}\}&\int\_\{-c\}^c x\mathrm\{ d\} x\int\_\{x^3\}^\{c^3\}\mathrm\{ d\} y+0\left(\mbox\{$F$ 关于 $\displaystyle y$ 是奇函数\}\right)\\\\ \xlongequal\{\tiny\mbox\{对称性\}\}&-\frac\{2c^5\}\{5\}+0=-\frac\{2c^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)/
---
702、 5、 设立体区域 $\displaystyle \varOmega$ 是由 $\displaystyle yOz$ 平面上的曲线 $\displaystyle y^2+z^4-4z^2=0, z\geq 0$ 绕 $\displaystyle z$ 轴旋转一周所成的曲面, 点 $\displaystyle (x,y,z)\in \varOmega$ 处的体密度为 $\displaystyle u(x,y,z)=z$, 求 $\displaystyle \varOmega$ 的重心坐标. (华东师范大学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) / $\displaystyle \varOmega$ 由 $\displaystyle x^2+y^2+z^4-4z^2=0, z\geq 0$ 围成. 由对称性知 $\displaystyle \varOmega$ 的重心坐标为 $\displaystyle (0,0,z\_0)$. 由
\begin\{aligned\} \iiint\_\varOmega u(x,y,z)\mathrm\{ d\} v=&\int\_0^2 z\mathrm\{ d\} z\iint\_\{x^2+y^2\leq 4z^2-z^4\}\mathrm\{ d\} x\mathrm\{ d\} y\\\\ =&\int\_0^2 z\cdot\pi (4z^2-z^4)\mathrm\{ d\} z=\frac\{16\pi\}\{3\},\\\\ \iiint\_\varOmega zu(x,y,z)\mathrm\{ d\} v=&\int\_0^2 z^2\mathrm\{ d\} z\iint\_\{x^2+y^2\leq 4z^2-z^4\}\mathrm\{ d\} x\mathrm\{ d\} y\\\\ =&\int\_0^2 z^2\cdot\pi (4z^2-z^4)\mathrm\{ d\} z=\frac\{256\pi\}\{35\} \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\} z\_0=\frac\{1\}\{\displaystyle\iiint\_\varOmega u(x,y,z) \mathrm\{ d\} v\}\iiint\_\varOmega z u(x,y,z)\mathrm\{ d\} v=\frac\{48\}\{35\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \varOmega$ 的重心为 $\displaystyle \left(0,0,\frac\{48\}\{35\}\right)$.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)/
---
703、 (3)、 求二重积分 $\displaystyle \iint\_D \mathrm\{e\}^\frac\{y-x\}\{x+y\}\mathrm\{ d\} x\mathrm\{ d\} y$, 其中 $\displaystyle D$ 为 $\displaystyle x=0, y=0$ 与 $\displaystyle x+y=2$ 围成的区域. (华南师范大学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) / 设 $\displaystyle y-x=v, x+y=u$, 则 $\displaystyle \frac\{\partial(u,v)\}\{\partial(x,y)\}=\left|\begin\{array\}\{cccccccccc\}1&1\\\\ -1&1\end\{array\}\right|=2$, 而
\begin\{aligned\} \mbox\{原式\}=&\int\_0^2 \mathrm\{ d\} u\int\_\{-u\}^u \mathrm\{e\}^\frac\{v\}\{u\}\frac\{1\}\{2\}\mathrm\{ d\} v =\frac\{1\}\{2\}\int\_0^2 u \left.\mathrm\{e\}^\frac\{v\}\{u\}\right|\_\{v=-u\}^\{v=u\}\mathrm\{ d\} u\\\\ =&\frac\{1\}\{2\}\int\_0^2 u(\mathrm\{e\}-\mathrm\{e\}^\{-1\})\mathrm\{ d\} u=\mathrm\{e\}-\mathrm\{e\}^\{-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)/
---
704、 (3)、 计算积分
\begin\{aligned\} \iint\_D (x+y)\sin (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 $\displaystyle D: \left\\{(x,y); 0\leq x+y\leq \pi, 0\leq x-y\leq \pi\right\\}$. (华中师范大学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) / 设
\begin\{aligned\} \left.\begin\{array\}\{rrrr\} x+y=u\\\\ x-y=v\end\{array\}\right\\}\Leftrightarrow \left\\{\begin\{array\}\{llllllllllll\}x=\frac\{u+v\}\{2\}\\\\ y=\frac\{u-v\}\{2\}\end\{array\}\right., \frac\{\partial(x,y)\}\{\partial(u,v)\}=-\frac\{1\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} \mbox\{原式\}&=\int\_0^\pi \mathrm\{ d\} u\int\_0^\pi u\sin v\frac\{1\}\{2\}\mathrm\{ d\} v\\\\ &=\frac\{1\}\{2\}\int\_0^\pi u\mathrm\{ d\} u\int\_0^\pi \sin v\mathrm\{ d\} v=\frac\{\pi^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)/
---
705、 (0-23)、 计算二重积分 $\displaystyle \iint\_D(x+y)\mathrm\{ d\} x\mathrm\{ d\} y$, 其中 $\displaystyle D$ 为 $\displaystyle y=x^2, x=1$ 与 $\displaystyle x$ 轴所围成的区域. (吉林师范大学2023年(学科数学)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\} x\int\_0^\{x^2\}(x+y)\mathrm\{ d\} y=\frac\{7\}\{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](
https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
706、 2、 将积分
\begin\{aligned\} \int\_0^1 \mathrm\{ d\} x\int\_0^\{1-x\}\mathrm\{ d\} y\int\_0^\{x+y\}f(x,y,z)\mathrm\{ d\} z \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
交换积分次序, 变为先对 $\displaystyle x$ 再对 $\displaystyle y$ 最后对 $\displaystyle z$ 的累次积分. (南昌大学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) / 题中的四棱锥在 $\displaystyle z\in [0,1]$ 处的截面是一个等腰梯形, 其边由 $\displaystyle x+y=z, x+y=1, x=0, y=0$ 围成. 故
\begin\{aligned\} \mbox\{原式\}=&\int\_0^1 \mathrm\{ d\} z\int\_0^z \mathrm\{ d\} y\int\_\{z-y\}^\{1-y\}f(x,y,z)\mathrm\{ d\} x +\int\_0^1 \mathrm\{ d\} z\int\_z^1 \mathrm\{ d\} y\int\_0^\{1-y\}f(x,y,z)\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\}
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)/
---
707、 4、 (15 分) 设 $\displaystyle B\_1$ 为单位圆, $\displaystyle a,b\in\mathbb\{R\}$ 且 $\displaystyle a^2+b^2 > 1$. 求证:
\begin\{aligned\} f(a,b)=\iint\_\{B\_1\} \ln \left\[(x\_1-a)^2+(x\_2-b)^2\right\]\mathrm\{ d\} x\_1\mathrm\{ d\} x\_2 \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
只依赖于 $\displaystyle \sqrt\{a^2+b^2\}$, 并求其值. (南京大学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)、 先给出调和函数的平均值公式. 若在 $\displaystyle D$ 内, $\displaystyle \Delta u=\frac\{\partial^2u\}\{\partial x\}+\frac\{\partial^2u\}\{\partial y^2\}=0$, $\displaystyle C\_R$ 是 $\displaystyle D$ 内以 $\displaystyle (x\_0,y\_0)$ 为圆心, $\displaystyle R$ 为半径的圆周, 则
\begin\{aligned\} u(x\_0,y\_0)=&\frac\{1\}\{2\pi R\}\oint\_\{C\_R\}u(x,y)\mathrm\{ d\} s\\\\ =&\frac\{1\}\{\pi R^2\}\iint\_\{B\_R\} u\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\}
事实上, 由
\begin\{aligned\} 0=\Delta u=(u\_x)\_x-(-u\_y)\_y \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\} 0=&\oint\_\{B\_r\}[(u\_x)\_x-(-u\_y)\_y]\mathrm\{ d\} x\mathrm\{ d\} y\\\\ &\quad \left(B\_r=\left\\{(x,y);(x-x\_0)^2+(y-y\_0)^2 < r^2\right\\}\right)\\\\ =&\oint\_\{C\_r\} \left\[-u\_y(-\cos(n,y))+u\_x\cos(n,x)\right\]\mathrm\{ d\} s\\\\ =&\oint\_\{C\_r\} \frac\{\partial u\}\{\partial n\}\mathrm\{ d\} s =\oint\_\{C\_r\} \frac\{\partial u\}\{\partial r\}\mathrm\{ d\} s =\int\_0^\{2\pi\} \frac\{\partial u\}\{\partial r\}(r\cos \theta,r\sin\theta)\cdot r\mathrm\{ d\} \theta\\\\ =&r\frac\{\partial\}\{\partial r\}\int\_0^\{2\pi\} u(r\cos \theta,r\sin\theta)\mathrm\{ d\} \theta =r\frac\{\partial \}\{\partial r\}\left\[\frac\{1\}\{r\}\oint\_\{C\_r\}u\mathrm\{ d\} s\right\]. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} \frac\{1\}\{R\}\oint\_\{C\_R\} u\mathrm\{ d\} s =\lim\_\{r\to 0^+\}\frac\{1\}\{r\}\oint\_\{C\_r\}u\mathrm\{ d\} s =2\pi u(x\_0,y\_0). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
进一步,
\begin\{aligned\} 2\pi r u(x\_0,y\_0)=\oint\_\{x^2+y^2=r^2\}u\mathrm\{ d\} s =\int\_0^\{2\pi\}u(x\_0+r\cos \theta, y\_0+r\sin\theta)\cdot 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\}
关于 $\displaystyle r$ 在 $\displaystyle [0,R]$ 上积分即得
\begin\{aligned\} \pi R^2 u(x\_0,y\_0)=\iint\_\{x^2+y^2\leq R^2\}u\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\}
(2)、 设 $\displaystyle u(x,y)=\ln(x^2+y^2)$, 则 $\displaystyle \Delta u=0$, 而
\begin\{aligned\} f(a,b)&\stackrel\{x\_1-a=u, x\_2-b=v\}\{=\}\iint\_\{(u-a)^2+(v-b)^2\leq 1\} \ln (u^2+v^2)\mathrm\{ d\} u\mathrm\{ d\} v\\\\ &=\pi 1^2\cdot \ln (a^2+b^2)=\pi \ln (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)/
---
708、 11、 求三重积分
\begin\{aligned\} \iiint\_\varOmega \sqrt\{1-\frac\{x^2\}\{a^2\}-\frac\{y^2\}\{b^2\}-\frac\{z^2\}\{c^2\}\}\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 $\displaystyle \varOmega$ 为 $\displaystyle \left\\{(x,y,z); \frac\{x^2\}\{a^2\}+\frac\{y^2\}\{b^2\}+\frac\{z^2\}\{c^2\}\leq 1\right\\}, a,b,c > 0$. (南京航空航天大学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\}=&abc\iiint\_\{u^2+v^2+w^2\leq 1\}\sqrt\{1-u^2-v^2-w^2\}\mathrm\{ d\} u\mathrm\{ d\} v\mathrm\{ d\} w\\\\ =&abc\int\_0^1 \sqrt\{1-r^2\}\cdot 4\pi r^2\mathrm\{ d\} r \stackrel\{r=\sin\theta\}\{=\}\cdots=\frac\{\pi^2abc\}\{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)/
---
709、 7、 讨论 $\displaystyle \iint\_\{(x,y)\in\mathbb\{R\}^2; x\geq 1, y\geq 1\} \frac\{x^2-y^2\}\{(x^2+y^2)^2\}\mathrm\{ d\} x\mathrm\{ d\} y$ 的敛散性. (厦门大学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)、 二重反常积分的定义. 设 $\displaystyle f(x,y)$ 为定义在无界区域 $\displaystyle D$ 上的二元函数. 若对于平面上任一包围原点的光滑封闭曲线 $\displaystyle \gamma$, $\displaystyle f(x,y)$ 在曲线 $\displaystyle \gamma$ 所围的有界区域 $\displaystyle E\_\gamma$ 与 $\displaystyle D$ 的交集是$E\_\gamma\cap D=D\_\gamma$ 上恒可积. 令
\begin\{aligned\} d\_\gamma=\inf\left\\{\sqrt\{x^2+y^2\}; (x,y)\in\gamma\right\\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
若极限 $\displaystyle \lim\_\{d\_\gamma\to+\infty\}\iint\_\{D\_\gamma\}f(x,y)\mathrm\{ d\} \sigma$ 存在且有限, 且与 $\displaystyle \gamma$ 的取法无关, 则称 $\displaystyle f(x,y)$ 在 $\displaystyle D$ 上的反常二重积分收敛, 并记
\begin\{aligned\} \iint\_D f(x,y)\mathrm\{ d\} \sigma=\lim\_\{d\_\gamma\to+\infty\}\iint\_\{D\_\gamma\}f(x,y)\mathrm\{ d\} \sigma. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
(2)、 回到题目. 设 $\displaystyle D: x\geq 1, y\geq 1$, 取 $\displaystyle \gamma\_r: x^2+y^2=r^2$, 则
\begin\{aligned\} \iint\_\{D\_\gamma\}\frac\{x^2-y^2\}\{(x^2+y^2)^2\}\mathrm\{ d\}\sigma\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\}
取 $\displaystyle \varGamma\_r$:
\begin\{aligned\} &\left\\{(R,\theta); \frac\{\pi\}\{4\}\leq \theta\leq 2\pi\right\\}\cup \left\\{\left(r,\frac\{\pi\}\{4\}\right); R\leq r\leq 2R\right\\}\\\\ \cup&\left\\{(r,0); R\leq r\leq 2R\right\\}\cup \left\\{(2R,\theta); 0\leq \theta\leq\frac\{\pi\}\{4\}\right\\}, \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
令 $\displaystyle \tilde\{D\}\_r$ 是 $\displaystyle \varGamma\_r$ 围成的区域与 $\displaystyle x\geq 1, y\geq 1$ 的交, 则
\begin\{aligned\} &\iint\_\{\tilde\{D\}\_r\}\frac\{x^2-y^2\}\{(x^2+y^2)^2\}\mathrm\{ d\} \sigma \xlongequal\{\tiny\mbox\{对称性\}\} \iint\_\{R^2\leq x^2+y^2\leq 4R^2\atop 0\leq\theta\leq\frac\{\pi\}\{4\}\}\frac\{x^2-y^2\}\{(x^2+y^2)^2\}\mathrm\{ d\} \sigma\\\\ =&\int\_0^\frac\{\pi\}\{4\}\mathrm\{ d\} \theta\int\_R^\{2R\}\frac\{r^2\cos2\theta\}\{r^4\}\cdot r\mathrm\{ d\} r =\int\_0^\frac\{\pi\}\{4\} \cos2\theta\mathrm\{ d\}\theta\int\_R^\{2R\} \frac\{\mathrm\{ d\} r\}\{r\}=\frac\{\ln 2\}\{2\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故 $\displaystyle \lim\_\{d\_\gamma\to+\infty\}\iint\_\{D\_\gamma\}\frac\{x^2-y^2\}\{(x^2+y^2)^2\}\mathrm\{ d\} \sigma$ 与 $\displaystyle \gamma$ 的选取有关, 而原反常二重积分不存在.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)/
---
710、 (2)、 (20 分) 求积分 $\displaystyle \iint\_A \mathrm\{e\}^\{-x^2-y^2\}\mathrm\{ d\} x\mathrm\{ d\} y$, 其中 $\displaystyle A: 1\leq x^2+y^2\leq 3$. (山东大学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\_1^\{\sqrt\{3\}\}\mathrm\{e\}^\{-r^2\}\cdot 2\pi r\mathrm\{ d\} r=\pi(\mathrm\{e\}^\{-1\}-\mathrm\{e\}^\{-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)/
---
711、 9、 (15 分) 求 $\displaystyle f(x,y,z)=x^2+y^2+z^2$ 在区域
\begin\{aligned\} D=\left\\{(x,y,z); x^2+y^2+z^2 < x+y+z\right\\} \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
上的积分平均值. (山西大学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) / $\displaystyle D: \left(x-\frac\{1\}\{2\}\right)^2+\left(y-\frac\{1\}\{2\}\right)^2+\left(z-\frac\{1\}\{2\}\right)^2\leq\frac\{3\}\{4\}$. 故
\begin\{aligned\} A=\iiint\_D \mathrm\{ d\} v=\frac\{4\}\{3\}\pi \left(\frac\{\sqrt\{3\}\}\{2\}\right)^3=\frac\{\sqrt\{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\} B=&\iiint\_D (x^2+y^2+z^2)\mathrm\{ d\} v\\\\ =&\iiint\_\{u^2+v^2+w^2\leq \frac\{3\}\{4\}\}\left\[\left(\frac\{1\}\{2\}+u\right)^2+\left(\frac\{1\}\{2\}+v\right)^2+\left(\frac\{1\}\{2\}+w\right)^2\right\]\mathrm\{ d\} u\mathrm\{ d\} v\mathrm\{ d\} w\\\\ \xlongequal\{\tiny\mbox\{对称性\}\}&\iiint\_\{u^2+v^2+w^2\leq \frac\{3\}\{4\}\}\left\[\frac\{3\}\{4\}+(u^2+v^2+w^2)\right\]\mathrm\{ d\} u\mathrm\{ d\} v\mathrm\{ d\} w\\\\ =&\int\_0^\frac\{\sqrt\{3\}\}\{2\} \left(\frac\{3\}\{4\}+r^2\right)\cdot 4\pi r^2\mathrm\{ d\} r =\frac\{3\sqrt\{3\}\pi\}\{5\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故所求 $\displaystyle =\frac\{B\}\{A\}=\frac\{6\}\{5\}$.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)/
---
712、 (4)、 求重积分 $\displaystyle \iint\_D \frac\{\sin x\}\{x\}\mathrm\{ d\} x\mathrm\{ d\} y$, $\displaystyle D$ 为 $\displaystyle y=x, y=\frac\{x\}\{2\}, x=2$ 所围成的区域. (陕西师范大学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\_0^2 \frac\{\sin x\}\{x\}\mathrm\{ d\} x\int\_\frac\{x\}\{2\}^x \mathrm\{ d\} y =\frac\{1\}\{2\}\int\_0^2 \sin x\mathrm\{ d\} x =\frac\{1-\cos 2\}\{2\}=\sin^21. \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)/
---
713、 8、 解答如下问题: (1)、 设 $\displaystyle f(x)=\int\_\{\sqrt\{\frac\{\pi\}\{2\}\}\}^\{\sqrt\{x\}\} \cos(t^2)\mathrm\{ d\} t$, 求 $\displaystyle I=\int\_0^\frac\{\pi\}\{2\}\frac\{f(x)\}\{\sqrt\{x\}\}\mathrm\{ d\} x$. (上海大学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 $\displaystyle f(x)\stackrel\{t^2=\tau\}\{=\}\int\_\frac\{\pi\}\{2\}^x \frac\{\cos\tau\}\{2\sqrt\{\tau\}\}\mathrm\{ d\} \tau$ 知
\begin\{aligned\} I=&-\int\_0^\frac\{\pi\}\{2\} \frac\{\mathrm\{ d\} x\}\{\sqrt\{x\}\} \int\_x^\frac\{\pi\}\{2\} \frac\{\cos\tau\}\{2\sqrt\{\tau\}\}\mathrm\{ d\} \tau\\\\ =&-\int\_0^\frac\{\pi\}\{2\} \frac\{\cos \tau\}\{2\sqrt\{\tau\}\}\mathrm\{ d\} \tau\int\_0^\tau \frac\{\mathrm\{ d\} x\}\{\sqrt\{x\}\} =-\int\_0^\frac\{\pi\}\{2\} \frac\{\cos \tau\}\{\sqrt\{\tau\}\}\cdot \sqrt\{\tau\}\mathrm\{ d\} \tau=-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)/
发表于 2023-3-5 09:12:01
