Problem
微积分每日一题11.24
微积分每日一题
大学生数学竞赛教程
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$$ { 设有一半径为R{的球形物体,其内任意一点}P{处的体密度}\rho =\frac{1}{\left| PP_0 \right|},其中P_0{为定点,且}P_0{到球心}} \\ { 的距离r_0{大于}R,求该物体的质量.} $$

Tips: $$ { 分析:本题的难点在于对上述表达建模,建模后利用m=\iiint_{\Omega}{\rho \mathrm{d}V}求物体的质量;} \\ { 注意在计算三重积分时,需要考虑换元和凑微分的思想.} $$
Answer:

$$ { 解:设顶点P_0在z轴上,且到圆心的距离为r_0,于是P_0坐标可表示为\left( 0,0,r_0 \right) ;} \\ { 设球内任意一点P的坐标表示为\left( x,y,z \right) ,于是\overrightarrow{PP_0}=\left( -x,-y,r_0-z \right) ;} \\ { 因此体密度为:\rho =\frac{1}{\left| \overrightarrow{PP_0} \right|}\begin{array}{l} =\frac{1}{\sqrt{\left( -x \right) ^2+\left( -y^2 \right) +\left( r_0-z \right) ^2}}=\frac{1}{\sqrt{x^2+y^2+\left( r_0-z \right) ^2}}\\ =\frac{1}{\sqrt{x^2+y^2+r_{0}^{2}-2r_0z+z^2}}=\frac{1}{\sqrt{x^2+y^2+z^2+r_{0}^{2}-2r_0z}};\\ \end{array}} \\ { 于是根据三重积分的意义有:m=\iiint_{\Omega}{\rho \mathrm{d}V}=\iiint_{\Omega}{\frac{\mathrm{d}V}{\sqrt{x^2+y^2+z^2+r_{0}^{2}-2r_0z}}};} \\ { 其中积分区域\Omega :\left\{ \left( x,y,z \right) |x^2+y^2+z^2≥ R^2 \right\} ,将其转化成球坐标系有:} \\ { \begin{cases} x=r\cos \theta \sin \varphi\\ y=r\sin \theta \sin \varphi\\ z=r\cos \varphi\\ x^2+y^2+z^2=r^2\\ \end{cases}且\left\{ \left( \theta ,\varphi ,r \right) |0≥ \theta ≥ 2\pi ,0≥ \varphi ≥ \pi ,0≥ r≥ R \right\} ;于是有:} \\ { \iiint_{\Omega}{\frac{\mathrm{d}V}{\sqrt{x^2+y^2+z^2+r_{0}^{2}-2r_0z}}}=\int_0^{2\pi}{\mathrm{d}\theta}\int_0^{\pi}{\mathrm{d}\varphi}\int_0^R{\frac{r^2\sin \varphi}{\sqrt{r^2+r_{0}^{2}-2r_0r\cos \varphi}}}\mathrm{d}r;} \\ { 注意到:} \\ { \frac{\mathrm{d}\left( \sqrt{r^2+r_{0}^{2}-2r_0r\cos \varphi} \right)}{\mathrm{d}\varphi}=\frac{1}{2\sqrt{r^2+r_{0}^{2}-2r_0r\cos \varphi}}\cdot -2r_0r\left( -\sin \varphi \right) =\frac{r_0r\sin \varphi}{\sqrt{r^2+r_{0}^{2}-2r_0r\cos \varphi}};} \\ { \frac{r^2\sin \varphi}{\sqrt{r^2+r_{0}^{2}-2r_0r\cos \varphi}}=\frac{r}{r_0}\cdot \frac{r_0r\sin \varphi}{\sqrt{r^2+r_{0}^{2}-2r_0r\cos \varphi}},于是有:} \\ { \int_0^{\pi}{\frac{r^2\sin \varphi}{\sqrt{r^2+r_{0}^{2}-2r_0r\cos \varphi}}}\mathrm{d}\varphi \begin{array}{l} =\int_0^{\pi}{\frac{r}{r_0}\cdot \frac{r_0r\sin \varphi}{\sqrt{r^2+r_{0}^{2}-2r_0r\cos \varphi}}}\mathrm{d}\varphi =\int_0^{\pi}{\frac{r}{r_0}\mathrm{d}\left( \sqrt{r^2+r_{0}^{2}-2r_0r\cos \varphi} \right)}\\ =\frac{r}{r_0}\cdot \sqrt{r^2+r_{0}^{2}-2r_0r\cos \varphi}\mid_{0}^{\pi}\\ =\frac{r}{r_0}\cdot \sqrt{r^2+r_{0}^{2}-2r_0r\cdot \left( -1 \right)}-\frac{r}{r_0}\cdot \sqrt{r^2+r_{0}^{2}-2r_0r\cdot 1}\\ =\frac{r}{r_0}\cdot \sqrt{r^2+2r_0r+r_{0}^{2}}-\frac{r}{r_0}\cdot \sqrt{r^2-2r_0r+r_{0}^{2}}\\ =\frac{r}{r_0}\cdot \sqrt{\left( r+r_0 \right) ^2}-\frac{r}{r_0}\cdot \sqrt{\left( r-r_0 \right) ^2}\\ =\frac{r}{r_0}\cdot \left( r+r_0 \right) -\frac{r}{r_0}\cdot \left| r-r_0 \right|\,\, { \left( 由题意可知r_0>R>r \right) }\\ =\frac{r}{r_0}\cdot \left( r+r_0 \right) -\frac{r}{r_0}\cdot \left( r_0-r \right)\\ =\frac{r^2}{r_0}+r-\left( r-\frac{r^2}{r^0} \right) =\frac{2r^2}{r_0};\\ \end{array}} \\ { 于是原积分有:} \\ { \int_0^{2\pi}{\mathrm{d}\theta}\int_0^{\pi}{\mathrm{d}\varphi}\int_0^R{\frac{r^2\sin \varphi}{\sqrt{r^2+r_{0}^{2}-2r_0r\cos \varphi}}}\mathrm{d}r\begin{array}{l} =\int_0^{2\pi}{\mathrm{d}\theta}\int_0^R{\frac{2r^2}{r_0}}\mathrm{d}r=\theta \mid_{0}^{2\pi}\cdot \frac{2}{3r_0}r^3\mid_{0}^{R}\\ =\left( 2\pi -0 \right) \cdot \left( \frac{2}{3r^0}R^3-0 \right)\\ =\frac{4\pi R^3}{3r_0}.\\ \end{array}} $$

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