高数强化NO21|质心和形心|液体的静压力

质心和形心

$$\begin{array}{rl}& \text{质心和形心问题，关键是记住质心公式}\bar{x}=\frac{\int xdm}{\int dm}，\bar{y}=\frac{\int ydm}{\int dm}，\text{其中}dm\text{为质量元。}\\ & \\ & \text{质量元}dm\text{，具体分为下面几种情况：}\\ & \\ & 1）\text{若为曲线型物体：}dm=\rho ds\text{，其中}\rho \text{为线密度，}ds\text{为弧微分；}\\ & \text{特殊情况：若为直线型物体，此时}ds=dx\text{。}\\ & \\ & 2）\text{若为曲面型物体：}dm=\rho dS\text{，其中}\rho \text{为面密度，}dS\text{为面积元；}\\ & \text{特殊情况：若为平面型物体，此时}dS=dxdy\text{。}\\ & \\ & \text{若是求形心，令质心公式中的}\rho =1\text{即可。}\\ & \\ & 3）\text{若为几何体：}dm=\rho dV\text{，其中}\rho \text{为体密度，}dV\text{为体积微元。}\end{array}$$

$$\begin{array}{rl}\text{解:}\overline{x}& =\frac{{\int }_0^{1}xdm}{{\int }_0^{1}dm}=\frac{{\int }_0^{1}x(-x^2+2x+1)dx}{{\int }_0^1(-x^2+2x+1)dx}\\ & =\frac{{\int }_0^1(-x^3+2x^2+x)dx}{{\int }_0^1(-x^2+2x+1)dx}=\frac{-\frac{1}{4}+\frac{2}{3}+\frac{1}{2}}{-\frac{1}{3}+1+1}\\ & =\frac{\frac{11}{12}}{\frac{5}{3}}=\frac{11}{20}\end{array}$$
$$\begin{array}{rl}& \text{【小结】1、判断是直线型物体求质心，本质公式是}\bar{x}=\frac{\int xdm}{\int dm},\bar{y}=\frac{\int ydm}{\int dm}；\\ & \\ & 2、\text{计算质量元 }dm=\rho (x)ds=\rho (x)dx(\text{直线}ds=dx)；\\ & \\ & 3、\text{计算定积分，算出}\bar{x}=\frac{{\int }_0^{1}x\rho (x)dx}{{\int }_0^1\rho (x)dx}\text{即可。}\end{array}$$

$$\begin{array}{rl}\text{解: (1)}l& ={\int }_1^e\sqrt{1+(y^{\prime }{)}^2}dx={\int }_1^e\sqrt{1+{\left(\frac{x}{2}-\frac{1}{2x}\right)}^2}dx\\ & ={\int }_1^e\sqrt{1+\frac{x^2}{4}-\frac{1}{2}+\frac{1}{4x^2}}dx={\int }_1^e\sqrt{{\left(\frac{x}{2}\right)}^2+2\cdot \frac{x}{2}\cdot \frac{1}{2x}+{\left(\frac{1}{2x}\right)}^2}dx\\ & ={\int }_1^e\sqrt{{\left(\frac{x}{2}+\frac{1}{2x}\right)}^2}dx={\int }_1^e\left(\frac{x}{2}+\frac{1}{2x}\right)dx\\ & =\frac{e^2-1}{4}+\frac{1}{2}\cdot 1=\frac{e^2}{4}+\frac{1}{4}=\frac{e^2+1}{4}\end{array}$$
$$\begin{array}{rl}(2)\overline{x}& =\frac{\iint xdxdy}{\iint dxdy}=\frac{{\int }_1^{e}dx{\int }_0^{\frac{x^2}{4}-\frac{1}{2}\ln x}xdy}{{\int }_1^{e}dx{\int }_0^{\frac{x^2}{4}-\frac{1}{2}\ln x}dy}\\ & =\frac{{\int }_1^e\left(\frac{x^3}{4}-\frac{1}{2}x\ln x\right)dx}{{\int }_1^e\left(\frac{x^2}{4}-\frac{1}{2}\ln x\right)dx}=\frac{\frac{e^4-1}{16}-\frac{1}{2}\left(\frac{1}{2}{x}^2\ln x-\frac{x^2}{4}\right){|}_1^e}{\frac{e^3-1}{12}-\frac{1}{2}\left(x\ln x-x\right){|}_1^e}\\ & =\frac{\frac{1}{16}{e}^4-\frac{1}{16}-\frac{1}{2}\left(\frac{e^2}{4}+\frac{1}{4}\right)}{\frac{1}{12}{e}^3-\frac{1}{12}-\frac{1}{2}}=\frac{e^4-1-2e^2-2}{e^3-7}\frac{12}{16}=\frac{3}{4}\cdot \frac{e^4-2e^2-3}{e^3-7}\end{array}$$
$$\begin{array}{rl}& \text{【小结】1、判断是平面型求形心横坐标，本质公式是}\bar{x}=\frac{\int xdm}{\int dm}；\\ & \\ & 2、\text{计算质量元}dm=dxdy；\\ & \\ & 3、\text{计算二重积分算出}\bar{x}=\frac{{\iint }_{D}xdxdy}{{\iint }_{D}dxdy}\text{即可。}\end{array}$$

$$\begin{array}{rl}\text{解:}V& ={\int }_{-1}^{\frac{1}{2}}\pi x_1^2(y)dy+{\int }_{\frac{1}{2}}^2\pi x_2^2(y)dy\\ & ={\int }_{-1}^{\frac{1}{2}}\pi (1-y^2)dy+{\int }_{\frac{1}{2}}^2\pi (2y-y^2)dy\\ & =\pi \left(y-\frac{y^3}{3}\right){|}_{-1}^{\frac{1}{2}}+\pi \left(y^2-\frac{y^3}{3}\right){|}_{\frac{1}{2}}^2\\ & =\pi \left(\frac{11}{24}-\left(-\frac{2}{3}\right)\right)+\pi \left(\frac{4}{3}-\frac{5}{24}\right)\\ & =\pi \cdot \frac{27}{24}+\pi \cdot \frac{27}{24}=\frac{9}{4}\pi \end{array}$$
$$\begin{array}{rl}(\text{II})W& ={\int }_{-1}^2(2-y)g\rho \cdot \pi x^{2}dy\\ & ={\int }_{-1}^{\frac{1}{2}}(2-y)g\rho \pi (1-y^2)dy+{\int }_{\frac{1}{2}}^2(2-y)g\rho \pi (2y-y^2)dy\\ & =\rho g\pi \left({\int }_{-1}^{\frac{1}{2}}(y^3-2y^2-y+2)dy+{\int }_{\frac{1}{2}}^2(y^3-2y^2-2y^2+4y)dy\right)\\ & =\rho g\pi \left({\int }_{-1}^2(y^3-2y^2)dy+{\int }_{-1}^{\frac{1}{2}}(2-y)dy+{\int }_{\frac{1}{2}}^2(4y-2y^2)dy\right)\\ & =\rho g\pi \left({\left(\frac{y^4}{4}-\frac{2}{3}{y}^3\right)|}_{-1}^2+{\left(2y-\frac{y^2}{2}\right)|}_{-1}^{\frac{1}{2}}+{\left(2y^2-\frac{2}{3}{y}^3\right)|}_{\frac{1}{2}}^2\right)\\ & =\rho g\pi \left(\left(-\frac{4}{3}-\frac{11}{12}\right)+\left(\frac{7}{8}+\frac{5}{2}\right)+\left(\frac{8}{3}-\frac{5}{12}\right)\right)\\ & =\rho g\pi \left(-\frac{9}{4}+\frac{27}{8}+\frac{9}{4}\right)=\frac{27}{8}\rho g\pi =3375\pi g(J)\end{array}$$
$$\begin{array}{rl}(\text{II})& \text{因水密度恒定，质心即形心，}\\ & \text{故质心坐标为}\left(0,\frac{1}{2}\right)\\ & \\ W& =\frac{9}{4}\pi \rho \cdot g\cdot \left(2-\frac{1}{2}\right)=\frac{27}{8}\pi \rho g=3375\pi g\end{array}$$
$$\begin{array}{rl}& \text{【小结】1.抽水做功的求解思路有两个：}\\ & \\ & \text{一是利用微元法：抽水做功物理学公式为}W=FS，\text{其中}F=mg=\rho gV，S\text{为提升高}\\ & \text{度，利用微元法，在}x\text{轴上截取一个长度为}dx\text{的小区间，由于}dx\text{很小，所以该区间段}\\ & \text{面积函数}A(x)\text{近似不变，该区间上容器近似一个柱体，体积为}A(x)dx，\text{对应的质量}\\ & \rho A(x)dx，\text{将该部分水提升到容器瓶口高度应为}b-x，\text{所以需要克服重力做功}\\ & \rho g(b-x)\cdot A(x)dx，\text{对其积分即为容器的水全部抽出需要做的功。}\\ & \\ & \text{二是计算出液体的质心（就是形心）的位置，再用质心提升的高度乘以液体的总重力。}\\ & \\ & 2.\text{当物体的形状比较规则时，形心的位置就在它的几何中心。}\end{array}$$

$$\begin{array}{rl}D& ={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}d\theta {\int }_0^{2\cos \theta }\left(r\cos \theta +r\sin \theta +\sqrt{1-(r\cos \theta -1{)}^2-(r\sin \theta {)}^2}\right)rdr\\ & ={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left(\frac{8}{3}{\cos }^4\theta +\frac{8}{3}{\cos }^3\theta \sin \theta \right)d\theta +{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}d\theta {\int }_0^{2\cos \theta }\sqrt{2r\cos \theta -r^2}\cdot rdr\end{array}$$
$$\begin{array}{rl}& =\frac{16}{3}\cdot \frac{3\times 1}{4\times 2}\cdot \frac{\pi }{2}+0+{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{\pi }{2}{\cos }^3\theta d\theta \\ & =\pi +\pi \cdot {\int }_0^{\frac{\pi }{2}}{\cos }^3\theta d\theta =\pi +\frac{2}{3}\pi =\frac{5}{3}\pi \end{array}$$
$$\begin{array}{rl}{\int }_0^{2\cos \theta }\sqrt{2r\cos \theta -r^2}\cdot rdr& ={\int }_0^{2\cos \theta }\sqrt{{\cos }^2\theta -(r-\cos \theta {)}^2}\cdot rdr\\ & \stackrel{\cos \theta =a}{=}{\int }_0^{2a}\sqrt{a^2-(r-a{)}^2}\cdot rdr\\ & \stackrel{r-a=a\sin t}{=}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}a\cos t\cdot (a\sin t+a)d(a\sin t+a)\\ & ={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{a}^2{\cos }^{2}t\cdot a\sin tdt+{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{a}^3{\cos }^{2}tdt\\ & =2a^3{\int }_0^{\frac{\pi }{2}}{\cos }^{2}tdt=\frac{a^3\pi }{2}=\frac{\pi }{2}{\cos }^3\theta \end{array}$$
$$\begin{array}{rl}\text{解:}& \text{由}\overline{x}=\frac{\iint xdxdy}{\iint dxdy}\text{可知}\\ & \iint xdxdy=\overline{x}\cdot \iint dxdy=1\cdot \pi \cdot 1^2=\pi \\ & \text{同理}\iint ydxdy=\overline{y}\cdot \iint dxdy=0\cdot \pi \cdot 1^2=0\\ & \\ & \text{令}z=\sqrt{1-(x-1{)}^2-y^2},z^2+(x-1{)}^2+y^2=1\text{，球心为}(1,0,0)\text{、半径为1的上半球面，}\\ & \iint zdxdy\text{即半球体积，}\frac{2}{3}\pi \cdot 1^3=\frac{2}{3}\pi \\ & \\ & \text{故原式}=\pi +\frac{2}{3}\pi =\frac{5}{3}\pi \end{array}$$

液体的静压力

$$\begin{array}{rl}& \text{对于液体的静压力，计算公式为}F=PS，P=\rho gh（P\text{为压强，}S\text{为面积，}h\text{是水下}\\ & \text{深度）。比如一个薄板水平放在水中受到的压强就是一个恒定的值，算液体对板的压力就}\\ & \text{可以直接用}F=PS\text{计算即可。但我们要考虑的是薄板竖直插入水中时受到的压力，此}\\ & \text{时不同深度处压强是不同的，不能直接应用压力的计算公式，需要借助“微元法”。具体}\\ & \text{思路如下：}\\ & \\ & \text{建立坐标轴，以水平面为坐标原点，向下为正方向；}\\ & \text{微元法：}\\ & 1）\text{分割；}\\ & 2）\text{近似：分割的比较细时近似看作压强是恒定不变的，分割出的小块近似看作长方形；}\\ & \text{故}{P}_i=\rho g{x}_i,\Delta S=f(x_i)\Delta x_i，\text{故压力为}\Delta F(x_i)=P_i\cdot \Delta S_i=\rho g{x}_i\cdot f(x_i)\Delta x_i，\text{其中}f(x_i)\\ & \text{为深度}{x}_i\text{处物体的宽度；}\\ & 3）\text{求和：}\sum _{i=1}^n\rho g{x}_i\cdot f(x_i)\Delta x_i；\\ & 4）\text{取极限：}\underset{\lambda \to 0}{\lim }\sum _{i=1}^n\rho g{x}_i\cdot f(x_i)\Delta x_i={\int }_a^b\rho gx\cdot f(x)dx。\\ & \\ & \text{可以看出，关键是需要求出深度}x\text{处物体的宽度}f(x)。\end{array}$$

$$\begin{array}{rl}\text{解:}F& ={\int }_0^a\rho gy\cdot 2(a-y)dy\\ & =2\rho g\left(\frac{a{y}^2}{2}-\frac{y^3}{3}\right){|}_0^a\\ & =\frac{\rho g{a}^3}{3}\end{array}$$

$$\begin{array}{rl}\text{解:}& \text{设上下部分压力为}{F}_1,F_2\\ F_1& ={\int }_0^h\rho gy\cdot 2dy=\rho g{h}^2\\ & \\ & \text{如图建立坐标系，抛物线过}(0,0),(1,1),\\ & \text{对称轴是}y\text{轴，方程为}{x}^2=y\\ & \\ F_2& ={\int }_0^1\rho g(h+1-y)\cdot 2\sqrt{y}dy\\ & =2\rho g(h+1)\cdot \frac{2}{3}-2\rho g\cdot \frac{2}{5}\\ & =\frac{4}{3}\rho gh+\frac{8}{15}\rho g\\ & \\ & \text{由}{F}_1:F_2=5:4,\frac{h^2}{\frac{4}{3}h+\frac{8}{15}}=\frac{5}{4}⟹h=2\end{array}$$