高等数学（第七版）同济大学习题10-4 （后7题）个人解答

高等数学（第七版）同济大学习题10-4（后7题）

函数作图软件：Mathematica

$\begin{aligned}&8. \ 设球占有闭区域\Omega=\{(x, \ y, \ z)\ |\ x^2+y^2+z^2 \le 2Rz\}，它在内部各点处的密度的大小等于该点到坐标\\\\&\ \ \ \ 原点的距离的平方，试求这球的质心.&\end{aligned}$

解：

$\begin{aligned} &\ \ 在球面坐标系中，\Omega表示为0 \le r \le 2Rcos\ \varphi，0 \le \varphi \le \frac{\pi}{2}，0 \le \theta \le 2\pi，球内任意一点(x, \ y, \ z)处的密度大小为\\\\ &\ \ \rho=x^2+y^2+z^2=r^2，因为球体的形状和质量分布关于z轴对称，可知质心位于z轴上，因此\overline{x}=\overline{y}=0，\\\\ &\ \ M=\iiint_{\Omega}\rho dv=\int_{0}^{2\pi}d\theta \int_{0}^{\frac{\pi}{2}}d\varphi \int_{0}^{2Rcos\ \varphi}r^2\cdot r^2sin\ \varphi dr=2\pi \int_{0}^{\frac{\pi}{2}}\frac{32}{5}R^5cos^5\ \varphi sin\ \varphi d\varphi=\frac{32}{15}\pi R^5，\\\\ &\ \ \overline{z}=\frac{1}{M}\iiint_{\Omega}\rho zdv=\frac{1}{M}\int_{0}^{2\pi}d\theta \int_{0}^{\frac{\pi}{2}}d\varphi \int_{0}^{2Rcos\ \varphi}r^2\cdot rcos\ \varphi \cdot r^2sin\ \varphi dr=\frac{2\pi}{M}\int_{0}^{\frac{\pi}{2}}\frac{64}{6}R^6cos^7\ \varphi sin\ \varphi d\varphi=\frac{5}{4}R，\\\\ &\ \ 所求质心为\left(0, \ 0, \ \frac{5}{4}R\right). & \end{aligned}$

$\begin{aligned}&9. \ 设均匀薄片（面密度为常数1）所占闭区域D如下，求指定的转动惯量：&\end{aligned}$

$\begin{aligned} &\ \ (1)\ \ D=\left\{(x, \ y)\ |\ \frac{x^2}{a^2}+\frac{y^2}{b^2} \le 1\right\}，求I_y；\\\\ &\ \ (2)\ \ D由抛物线y^2=\frac{9}{2}x与直线x=2所围成，求I_x和I_y；\\\\ &\ \ (3)\ \ D为矩形闭区域\{(x, \ y)\ |\ 0 \le x \le a，0 \le y \le b\}，求I_x和I_y. & \end{aligned}$

解：

$\begin{aligned} &\ \ (1)\ I_y=\iint_{D}x^2dxdy=\int_{-a}^{a}x^2dx\int_{-\frac{b}{a}\sqrt{a^2-x^2}}^{\frac{b}{a}\sqrt{a^2-x^2}}dy=\frac{2b}{a}\int_{-a}^{a}x^2\sqrt{a^2-x^2}dx=\frac{4b}{a}\int_{0}^{a}x^2\sqrt{a^2-x^2}dx，令x=asin\ t，\\\\ &\ \ \ \ \ \ \ \ 则上式=\frac{4b}{a}\int_{0}^{\frac{\pi}{2}}a^3sin^2\ tcos\ t\cdot acos\ tdt=4a^3b\left[\int_{0}^{\frac{\pi}{2}}sin^2\ tdt-\int_{0}^{\frac{\pi}{2}}sin^4\ tdt\right]=4a^3b\left(\frac{1}{2}\cdot \frac{\pi}{2}-\frac{3}{4}\cdot \frac{1}{2}\cdot \frac{\pi}{2}\right)=\\\\ &\ \ \ \ \ \ \ \ \frac{1}{4}\pi a^3b.\\\\ &\ \ (2)\ D=\left\{(x, \ y)\ \bigg|\ -3\sqrt{\frac{x}{2}} \le y \le 3\sqrt{\frac{x}{2}}，0 \le x \le 2\right\}，I_x=\iint_{D}y^2dxdy \underline{\underline{对称性}}2\int_{0}^{2}dx\int_{0}^{3\sqrt{\frac{x}{2}}}y^2dy=\\\\ &\ \ \ \ \ \ \ \ \frac{2}{3}\int_{0}^{2}\frac{27}{2\sqrt{2}}x^{\frac{3}{2}}dx=\frac{72}{5}，I_y=\iint_{D}x^2dxdy\underline{\underline{对称性}}2\int_{0}^{2}x^2dx\int_{0}^{3\sqrt{\frac{x}{2}}}dy=2\int_{0}^{2}\frac{3}{\sqrt{2}}x^{\frac{5}{2}}dx=\frac{96}{7}.\\\\ & \end{aligned}$
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$\begin{aligned} &\ \ (3)\ I_x=\iint_{D}y^2dxdy=\int_{0}^{a}dx\int_{0}^{b}y^2dy=\frac{ab^3}{3}，I_y=\iint_{D}x^2dxdy=\int_{0}^{a}x^2dx\int_{0}^{b}dy=\frac{a^3b}{3}. & \end{aligned}$

$\begin{aligned}&10. \ 已知均匀矩形板（面密度为常量\mu）的长和宽分别为b和h，计算此矩形板对于通过其形心且分别与\\\\&\ \ \ \ \ \ 一边平行的两轴的转动惯量.&\end{aligned}$

解：

$\begin{aligned} &\ \ 建立直角坐标系，原点O为矩形板的形心，x轴和y轴分别平行于矩形的两边，所求转动惯量为\\\\ &\ \ I_x=\iint_{D}y^2\mu dxdy=\mu\int_{-\frac{b}{2}}^{\frac{b}{2}}dx\int_{-\frac{h}{2}}^{\frac{h}{2}}y^2dy=\frac{1}{12}\mu bh^3，I_y=\iint_{D}x^2\mu dxdy=\mu\int_{-\frac{b}{2}}^{\frac{b}{2}}x^2dx\int_{-\frac{h}{2}}^{\frac{h}{2}}dy=\frac{1}{12}\mu hb^3. & \end{aligned}$
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$\begin{aligned}&11. \ 一均匀物体（密度\rho为常量）占有的闭区域\Omega由曲面z=x^2+y^2和平面z=0，|x|=a，|y|=a所围成，&\end{aligned}$

$\begin{aligned} &\ \ (1)\ \ 求物体的体积；\\\\ &\ \ (2)\ \ 求物体的质心；\\\\ &\ \ (3)\ \ 求物体关于z轴的转动惯量. & \end{aligned}$

解：

$\begin{aligned} &\ \ (1)\ 由\Omega对称性可知，V=4\int_{0}^{a}dx\int_{0}^{a}dy\int_{0}^{x^2+y^2}dz=4\int_{0}^{a}dx\int_{0}^{a}(x^2+y^2)dy=4\int_{0}^{a}\left(ax^2+\frac{a^3}{3}\right)dx=\frac{8}{3}a^4.\\\\ &\ \ (2)\ 由对称性可知，质心位于z轴上，所以\overline{x}=\overline{y}=0，\overline{z}=\frac{1}{M}\iiint_{\Omega}\rho zdv\underline{\underline{对称性}}\frac{4}{V}\int_{0}^{a}dx\int_{0}^{a}dy\int_{0}^{x^2+y^2}zdz=\\\\ &\ \ \ \ \ \ \ \ \frac{4}{V}\int_{0}^{a}dx\int_{0}^{a}\frac{1}{2}(x^4+2x^2y^2+y^4)dy=\frac{2}{V}\int_{0}^{a}\left(ax^4+\frac{2}{3}a^3x^2+\frac{1}{5}a^5\right)dx=\frac{7}{15}a^2.\\\\ &\ \ (3)\ I_z=\iiint_{\Omega}\rho(x^2+y^2)dv\underline{\underline{对称性}}4\rho \int_{0}^{a}dx\int_{0}^{a}dy\int_{0}^{x^2+y^2}(x^2+y^2)dz=4\rho \int_{0}^{a}dx\int_{0}^{a}(x^4+2x^2y^2+y^4)dy=\frac{112}{45}\rho a^6. & \end{aligned}$
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$\begin{aligned}&12. \ 求半径为a、高为h的均匀圆柱体对于过中心而平行于母线的轴的转动惯量（设密度\rho=1）.&\end{aligned}$

解：

$\begin{aligned} &\ \ 建立空间直角坐标系，使原点位于圆柱体的中心，z轴平行于母线，圆柱体所占的空间闭区域\\\\ &\ \ \Omega=\left\{(x, \ y, \ z)\ \bigg|\ x^2+y^2 \le a^2，-\frac{h}{2} \le z \le \frac{h}{2}\right\}\underline{\underline{柱面坐标}}\left\{(\rho, \ \theta, \ z)\ \bigg|\ 0 \le 0 \le 2\pi，0 \le \rho \le a，-\frac{h}{2} \le z \le \frac{h}{2}\right\}，\\\\ &\ \ 所求转动惯量为I_z=\iiint_{\Omega}(x^2+y^2)dv=\iiint_{\Omega}\rho^2 \cdot \rho d\rho d\theta dz=\int_{0}^{2\pi}d\theta \int_{0}^{a}\rho^3 d\rho \int_{-\frac{h}{2}}^{\frac{h}{2}}dz=2\pi \cdot \frac{a^4}{4}\cdot h=\frac{1}{2}\pi ha^4. & \end{aligned}$

$\begin{aligned}&13. \ 设面密度为常量\mu的质量均匀的半圆环形薄片占有区域D=\{(x, \ y, \ 0)\ |\ R_1 \le \sqrt{x^2+y^2} \le R_2，x \ge 0\}，\\\\&\ \ \ \ \ \ 求它对位于z轴上点M_0(0, \ 0, \ a)（a \gt 0）处单位质量的质点的引力F.&\end{aligned}$

解：

$\begin{aligned} &\ \ 引力元素dF沿x轴和z轴的分量分别为dF_x=G\frac{\mu x}{(x^2+y^2+a^2)^{\frac{3}{2}}}d\sigma和dF_z=G\frac{\mu(-a)}{(x^2+y^2+a^2)^{\frac{3}{2}}}d\sigma，则\\\\ &\ \ F_x=G\mu\iint_{D}\frac{x}{(x^2+y^2+a^2)^{\frac{3}{2}}}d\sigma\underline{\underline{极坐标}}G\mu\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d\theta \int_{R_1}^{R_2}\frac{\rho cos\ \theta}{(\rho^2+a^2)^{\frac{3}{2}}}\cdot \rho d\rho=G\mu \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}cos\ \theta d\theta \int_{R_1}^{R_2}\frac{\rho^2}{(\rho^2+a^2)^{\frac{3}{2}}}d\rho=\\\\ &\ \ 2G\mu \int_{R_1}^{R_2}\frac{\rho^2}{(\rho^2+a^2)^{\frac{3}{2}}}d\rho，令\rho=atan\ t，则上式=2G\mu \int_{arctan\ \frac{R_1}{a}}^{arctan\ \frac{R_2}{a}}\frac{a^2tan^2\ t}{a^3sec^3\ t}\cdot asec^2\ tdt=\\\\ &\ \ 2G\mu \int_{arctan\ \frac{R_1}{a}}^{arctan\ \frac{R_2}{a}}(sec\ t-cos\ t)dt=2G\mu \left[ln(sec\ t+tan\ t)-sin\ t\right]_{arctan\ \frac{R_1}{a}}^{arctan\ \frac{R_2}{a}}=\\\\ &\ \ 2G\mu \left(ln\frac{\sqrt{R_2^2+a^2}+R_2}{\sqrt{R_1^2+a^2}+R_1}-\frac{R_2}{\sqrt{R_2^2+a^2}}+\frac{R_1}{\sqrt{R_1^2+a^2}}\right)，\\\\ &\ \ F_z=-Ga\mu \iint_{D}\frac{d\sigma}{(x^2+y^2+a^2)^{\frac{3}{2}}}\underline{\underline{极坐标}}-Ga\mu \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d\theta \int_{R_1}^{R_2}\frac{\rho}{(\rho^2+a^2)^{\frac{3}{2}}}d\rho=\pi Ga\mu \left[\frac{1}{\sqrt{\rho^2+a^2}}\right]_{R_1}^{R_2}=\\\\ &\ \ \pi Ga\mu\left(\frac{1}{\sqrt{R_2^2+a^2}}-\frac{1}{\sqrt{R_1^2+a^2}}\right)，因为D关于x轴对称，并且质量均匀分布，所以F_y=0，\\\\ &\ \ 因此引力F=\left(2G\mu \left(ln\frac{\sqrt{R_2^2+a^2}+R_2}{\sqrt{R_1^2+a^2}+R_1}-\frac{R_2}{\sqrt{R_2^2+a^2}}+\frac{R_1}{\sqrt{R_1^2+a^2}}\right), \ 0, \ \pi Ga\mu\left(\frac{1}{\sqrt{R_2^2+a^2}}-\frac{1}{\sqrt{R_1^2+a^2}}\right)\right). & \end{aligned}$

$\begin{aligned}&14. \ 设均匀柱体密度为\rho，占有闭区域\Omega=\{(x, \ y, \ z)\ |\ x^2+y^2 \le R^2，0 \le z \le h\}，求它对于位于\\\\&\ \ \ \ \ \ 点M_0(0, \ 0, \ a)（a \gt h）处的单位质量的质点的引力.&\end{aligned}$

解：

$\begin{aligned} &\ \ 根据柱体的对称性和质量分布的均匀性可知F_x=F_y=0，引力沿z轴的分量F_z=\iiint_{\Omega}G\rho \frac{z-a}{[x^2+y^2+(z-a)^2]^{\frac{3}{2}}}dv=\\\\ &\ \ G\rho \int_{0}^{h}(z-a)dz\iint_{x^2+y^2 \le R^2}\frac{dxdy}{[x^2+y^2+(z-a)^2]^{\frac{3}{2}}}\underline{\underline{柱面坐标}}G\rho \int_{0}^{h}(z-a)dz\int_{0}^{2\pi}d\theta \int_{0}^{R}\frac{rdr}{[r^2+(z-a)^2]^{\frac{3}{2}}}=\\\\ &\ \ 2\pi G\rho \int_{0}^{h}(z-a)\left[\frac{1}{a-z}-\frac{1}{\sqrt{R^2+(z-a)^2}}\right]dz=2\pi G\rho \int_{0}^{h}\left[-1-\frac{z-a}{\sqrt{R^2+(z-a)^2}}\right]dz=\\\\ &\ \ -2\pi G\rho[h+\sqrt{R^2+(h-a)^2}-\sqrt{R^2+a^2}]. & \end{aligned}$