高等数学（第七版）同济大学习题10-4 （前7题）个人解答

高等数学（第七版）同济大学习题10-4（前7题）

函数作图软件：Mathematica

$\begin{aligned}&1. \ 求球面x^2+y^2+z^2=a^2含在圆柱面x^2+y^2=ax内部的那部分面积.&\end{aligned}$

解：

$\begin{aligned} &\ \ 上半球面的方程为z=\sqrt{a^2-x^2-y^2}，\frac{\partial z}{\partial x}=\frac{-x}{\sqrt{a^2-x^2-y^2}}，\frac{\partial z}{\partial y}=\frac{-y}{\sqrt{a^2-x^2-y^2}}，\\\\ &\ \ \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2}=\frac{a}{\sqrt{a^2-x^2-y^2}}，根据曲面对称性得所求面积为\\\\ &\ \ A=4\iint_{D}\sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2}dxdy=4\iint_{D}\frac{a}{\sqrt{a^2-x^2-y^2}}dxdy=4a\iint_{D}\frac{1}{\sqrt{a^2-\rho^2}}\rho d\rho d\theta=\\\\ &\ \ 4a\int_{0}^{\frac{\pi}{2}}d\theta \int_{0}^{acos\ \theta}\frac{\rho}{\sqrt{a^2-\rho^2}}d\rho=4a^2\int_{0}^{\frac{\pi}{2}}(1-sin\ \theta)d\theta=2a^2(\pi-2). & \end{aligned}$
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$\begin{aligned}&2. \ 求锥面z=\sqrt{x^2+y^2}被柱面z^2=2x所割下部分的曲面面积.&\end{aligned}$

解：

$\begin{aligned} &\ \ 由\begin{cases}z=\sqrt{x^2+y^2}\\\\z^2=2x\end{cases}，解得x^2+y^2=2x，因此曲面在xOy面上的投影区域D=\{(x, \ y)\ |\ x^2+y^2 \le 2x\}，\\\\ &\ \ 被割曲面的方程为z=\sqrt{x^2+y^2}，\sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2}=\sqrt{1+\frac{x^2+y^2}{x^2+y^2}}=\sqrt{2}，\\\\ &\ \ 所求曲面的面积为A=\iint_{D}\sqrt{2}dxdy=\sqrt{2}\pi. & \end{aligned}$
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$\begin{aligned}&3. \ 求底圆半径相等的两个直交圆柱面x^2+y^2=R^2及x^2+z^2=R^2所围立体的表面积.&\end{aligned}$

解：

$\begin{aligned} &\ \ 设第一卦限内的相交部分的表面积为A，根据对称性可知，全部表面积为16A，\\\\ &\ \ A=\iint_{D}\sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2}dxdy=\iint_{D}\sqrt{1+\frac{x^2}{R^2-x^2}+0}dxdy=\iint_{D}\frac{R}{\sqrt{R^2-x^2}}dxdy=\\\\ &\ \ R\int_{0}^{R}dx\int_{0}^{\sqrt{R^2-x^2}}\frac{1}{\sqrt{R^2-x^2}}dy=R\int_{0}^{R}dx=R^2，因此全部表面积为16R^2. & \end{aligned}$
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$\begin{aligned}&4. \ 设薄片所占的闭区域D如下，求均匀薄片的质心:&\end{aligned}$

$\begin{aligned} &\ \ (1)\ \ D由y=\sqrt{2px}，x=x_0，y=0所围成；\\\\ &\ \ (2)\ \ D是半椭圆形闭区域\left\{(x, \ y)\ |\ \frac{x^2}{a^2}+\frac{y^2}{b^2} \le 1，y \ge 0\right\}；\\\\ &\ \ (3)\ \ D是界于两个圆\rho=acos\ \theta，\rho=bcos\ \theta\ (0 \lt a \lt b)之间的闭区域. & \end{aligned}$

解：

$\begin{aligned} &\ \ (1)\ 设质心为(\overline{x}, \ \overline{y})，A=\iint_{D}dxdy=\int_{0}^{x_0}dx\int_{0}^{\sqrt{2px}}dy=\int_{0}^{x_0}\sqrt{2px}dx=\frac{2}{3}\sqrt{2px_0^3}，\\\\ &\ \ \ \ \ \ \ \ \iint_{D}xdxdy=\int_{0}^{x_0}xdx\int_{0}^{\sqrt{2px}}dy=\int_{0}^{x_0}\sqrt{2p}x^{\frac{3}{2}}dx=\frac{2}{5}\sqrt{2px_0^5}，\\\\ &\ \ \ \ \ \ \ \ \iint_{D}ydxdy=\int_{0}^{x_0}dx\int_{0}^{\sqrt{2px}}ydy=\int_{0}^{x_0}pxdx=\frac{1}{2}px_0^2，因此\\\\ &\ \ \ \ \ \ \ \ \overline{x}=\frac{1}{A}\iint_{D}xdxdy=\frac{3}{5}x_0，\overline{y}=\frac{1}{A}\iint_{D}ydxdy=\frac{3}{8}\sqrt{2px_0}=\frac{3}{8}y_0，所求质心为\left(\frac{3}{5}x_0, \ \frac{3}{8}y_0\right).\\\\ &\ \ (2)\ 因为D关于y轴对称，所以质心位于y轴上，因此\overline{x}=0，\overline{y}=\frac{1}{A}\iint_{D}ydxdy=\frac{1}{A}\int_{-a}^{a}dx\int_{0}^{\frac{b}{a}\sqrt{a^2-x^2}}ydy=\\\\ &\ \ \ \ \ \ \ \ \frac{1}{A}\int_{-a}^{a}\frac{b^2}{2a^2}(a^2-x^2)dx=\frac{1}{\frac{\pi ab}{2}}\cdot \frac{2}{3}ab^2=\frac{4b}{3\pi}，因此质心为\left(0, \ \frac{4b}{3\pi}\right).\\\\ &\ \ (3)\ 因为D关于x轴对称，所以质心位于x轴上，因此\overline{y}=0，A=\pi \left(\frac{b}{2}\right)^2-\pi \left(\frac{a}{2}\right)^2=\frac{\pi}{4}(b^2-a^2)，\\\\ &\ \ \ \ \ \ \ \ \iint_{D}xdxdy=\iint_{D}\rho cos\ \theta \cdot \rho d\rho d\theta=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}cos\ \theta d\theta \int_{acos\ \theta}^{bcos\ \theta}\rho^2d\rho=\frac{2}{3}(b^3-a^3)\int_{0}^{\frac{\pi}{2}}cos^4\ \theta d\theta=\\\\ &\ \ \ \ \ \ \ \ \frac{2}{3}(b^3-a^3)\cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}=\frac{\pi}{8}(b^3-a^3)，因此\overline{x}=\frac{1}{A}\iint_{D}xdxdy=\frac{a^2+ab+b^2}{2(a+b)}，所求质心为\left(\frac{a^2+ab+b^2}{2(a+b)}, \ 0\right). & \end{aligned}$

$\begin{aligned}&5. \ 设平面薄片所占的闭区域D由抛物线y=x^2及直线y=x所围成，它在点(x, \ y)处的\\\\&\ \ \ \ 面密度\mu(x, \ y)=x^2y，求该薄片的质心.&\end{aligned}$

解：

$\begin{aligned} &\ \ M=\iint_{D}x^2ydxdy=\int_{0}^{1}x^2dx\int_{x^2}^{x}ydy=\int_{0}^{1}\frac{1}{2}(x^4-x^6)dx=\frac{1}{35}，\\\\ &\ \ M_x=\iint_{D}y\mu(x, \ y)dxdy=\iint_{D}x^2y^2dxdy=\int_{0}^{1}x^2dx\int_{x^2}^{x}y^2dy=\int_{0}^{1}\frac{1}{3}(x^5-x^8)dx=\frac{1}{54}，\\\\ &\ \ M_y=\iint_{D}x\mu(x, \ y)dxdy=\iint_{D}x^3ydxdy=\int_{0}^{1}x^3dx\int_{x^2}^{x}ydy=\int_{0}^{1}\frac{1}{2}(x^5-x^7)dx=\frac{1}{48}，\\\\ &\ \ 因此\overline{x}=\frac{M_y}{M}=\frac{35}{48}，\overline{y}=\frac{M_x}{M}=\frac{35}{54}，所求质心为\left(\frac{35}{48}, \ \frac{35}{54}\right). & \end{aligned}$

$\begin{aligned}&6. \ 设有一等腰直角三角形薄片，腰长为a，各点处的面密度等于该点到直角顶点的距离的平方，\\\\&\ \ \ \ 求这薄片的质心.&\end{aligned}$

解：

$\begin{aligned} &\ \ 面密度\mu(x, \ y)=x^2+y^2，根据对称性可知\overline{x}=\overline{y}，M=\iint_{D}(x^2+y^2)dxdy=\int_{0}^{a}dx\int_{0}^{a-x}(x^2+y^2)dy=\\\\ &\ \ \int_{0}^{a}\left[x^2(a-x)+\frac{(a-x)^3}{3}\right]dx=\frac{1}{6}a^4，M_y=\iint_{D}x(x^2+y^2)dxdy=\int_{0}^{a}xdx\int_{0}^{a-x}(x^2+y^2)dy=\\\\ &\ \ \int_{0}^{a}\left[x^3(a-x)+\frac{x(a-x)^3}{3}\right]dx=\int_{0}^{a}\left(-\frac{4}{3}x^4+2ax^3-a^2x^2+\frac{a^3}{3}x\right)dx=\frac{1}{15}a^5，\\\\ &\ \ 因此\overline{x}=\frac{M_y}{M}=\frac{2}{5}a，\overline{y}=\overline{x}=\frac{2}{5}a，所求质心为\left(\frac{2}{5}a, \ \frac{2}{5}a\right). & \end{aligned}$

$\begin{aligned}&7. \ 利用三重积分计算下列由曲面所围立体的质心（设密度\rho=1）:&\end{aligned}$

$\begin{aligned} &\ \ (1)\ \ z^2=x^2+y^2，z=1；\\\\ &\ \ (2)\ \ z=\sqrt{A^2-x^2-y^2}，z=\sqrt{a^2-x^2-y^2}\ (A \gt a \gt 0)，z=0；\\\\ &\ \ (3)\ \ z=x^2+y^2，x+y=a，x=0，y=0，z=0. & \end{aligned}$

解：

$\begin{aligned} &\ \ (1)\ 曲面所围立体为圆锥体，顶点在原点，关于z轴对称，又因是匀质，因此质心位于z轴上，即\overline{x}=\overline{y}=0，\\\\ &\ \ \ \ \ \ \ \ 所围立体体积为V=\frac{1}{3}\pi，\overline{z}=\frac{1}{V}\iiint_{\Omega}zdv=\frac{1}{V}\iint_{x^2+y^2 \le 1}dxdy\int_{\sqrt{x^2+y^2}}^{1}zdz=\frac{1}{V}\iint_{x^2+y^2 \le 1}\frac{1}{2}(1-x^2-y^2)dxdy=\\\\ &\ \ \ \ \ \ \ \ \frac{1}{V}\int_{0}^{2\pi}d\theta \int_{0}^{1}\frac{1}{2}(1-\rho^2)\rho d\rho=\frac{3}{\pi}\cdot 2\pi \cdot \frac{1}{2}\left[\frac{\rho^2}{2}-\frac{\rho^4}{4}\right]_{0}^{1}=\frac{3}{4}，所求质心为\left(0, \ 0, \ \frac{3}{4}\right).\\\\ &\ \ (2)\ 立体由两个同心的上半球面和xOy面围成，关于z轴对称，因为是匀质，质心位于z轴上，即\overline{x}=\overline{y}=0，\\\\ &\ \ \ \ \ \ \ \ 体积为V=\frac{2}{3}\pi(A^3-a^3)，\overline{z}=\frac{1}{V}\iiint_{\Omega}zdv=\frac{1}{V}\iiint_{\Omega}rcos\ \varphi \cdot r^2sin\ \varphi dr d\varphi d\theta=\\\\ &\ \ \ \ \ \ \ \ \frac{1}{V}\int_{0}^{2\pi}d\theta \int_{0}^{\frac{\pi}{2}}sin\ \varphi cos\ \varphi d\varphi \int_{a}^{A}r^3dr=\frac{3}{2\pi(A^3-a^3)}\cdot 2\pi \cdot \frac{1}{2} \cdot \frac{A^4-a^4}{4}=\frac{3(A^4-a^4)}{8(A^3-a^3)}，\\\\ &\ \ \ \ \ \ \ \ 所求质心为\left(0, \ 0, \ \frac{3(A^4-a^4)}{8(A^3-a^3)}\right).\\\\ &\ \ (3)\ \Omega=\{(x, \ y, \ z)\ |\ 0 \le x \le a，0 \le y \le a-x，0 \le z \le x^2+y^2\}，V=\iiint_{\Omega}dv=\int_{0}^{a}dx\int_{0}^{a-x}dy\int_{0}^{x^2+y^2}dz=\\\\ &\ \ \ \ \ \ \ \ \int_{0}^{a}dx\int_{0}^{a-x}(x^2+y^2)dy=\int_{0}^{a}\left[x^2(a-x)+\frac{1}{3}(a-x)^3\right]dx=\int_{0}^{a}\left[ax^2-x^3+\frac{1}{3}(a-x)^3\right]dx=\frac{1}{6}a^4，\\\\ &\ \ \ \ \ \ \ \ \overline{z}=\frac{1}{V}\iiint_{\Omega}zdv=\frac{1}{V}\int_{0}^{a}dx\int_{0}^{a-x}dy\int_{0}^{x^2+y^2}zdz=\frac{1}{V}\int_{0}^{a}dx\int_{0}^{a-x}\frac{1}{2}(x^4+2x^2y^2+y^4)dy=\\\\ &\ \ \ \ \ \ \ \ \frac{1}{2V}\int_{0}^{a}\left[x^4(a-x)+\frac{2}{3}x^2(a-x)^3+\frac{1}{5}(a-x)^5\right]dx=\frac{3}{a^4}\cdot \frac{7a^6}{90}=\frac{7}{30}a^2，\\\\ &\ \ \ \ \ \ \ \ \overline{x}=\frac{1}{V}\iiint_{\Omega}xdv=\frac{1}{V}\int_{0}^{a}xdx\int_{0}^{a-x}dy\int_{0}^{x^2+y^2}dz=\frac{1}{V}\int_{0}^{a}x\left[x^2(a-x)+\frac{1}{3}(a-x)^3\right]dx=\frac{6}{a^4}\cdot \frac{a^5}{15}=\frac{2}{5}a，\\\\ &\ \ \ \ \ \ \ \ 因为立体匀质，并且关于平面y=x对称，所以\overline{y}=\overline{x}=\frac{2}{5}a，所求质心为\left(\frac{2}{5}a, \ \frac{2}{5}a，\ \frac{7}{30}a^2\right). & \end{aligned}$
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