习题3.5

第 1 题

求下列曲面的面积.

(1) 柱面 \(x^2 + z^2 = a^2\) 在柱面 \(x^2 + y^2 = a^2\) 内的部分;

(2) 锥面 \(z = \sqrt{x^2 + y^2}\) 在柱面 \(z^2 = 2x\) 内的部分;

(3) 由曲面 \(x^2 + y^2 = az\) 与 \(z = 2a - \sqrt{x^2 + y^2}\) 所包围的空间几何体的表面积.

解

(1) 只考虑 \(z \ge 0\) 部分的面积 \(S_1\). 由于 \(\dfrac{\partial z}{\partial x} = -\dfrac{x}{\sqrt{a^2-x^2}}, \dfrac{\partial z}{\partial y} = 0\), 因此 \(\sqrt{1 + \left(\dfrac{\partial z}{\partial x}\right)^2 + \left(\dfrac{\partial z}{\partial y}\right)^2} = \sqrt{1 + \dfrac{x^2}{a^2-x^2}} = \dfrac{a}{\sqrt{a^2-x^2}}\).

\[ \begin{aligned} S_1 & = \iint\limits_{x^2+y^2\le a^2}\dfrac{a}{\sqrt{a^2-x^2}}\text{d}{x}\text{d}{y} \\ & = \int_{-a}^{a}\text{d}{x}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\dfrac{a}{\sqrt{a^2-x^2}}\text{d}{y} \\ & = \int_{-a}^{a}2a\text{d}{x} \\ & = 4a^2 \end{aligned} \]

由对称性知 \(S = 2S_1 = 8a^2\).

(2) 在 \(xy\) 面上的投影为 \(D_{xy} = \{(x, y) | x^2 + y^2 \le 2x\}\), 是一个以 \((1, 0)\) 为圆心, \(1\) 为半径的圆, 面积为 \(\pi\). 由于 \(\dfrac{\partial z}{\partial x} = \dfrac{x}{\sqrt{x^2 + y^2}}, \dfrac{\partial z}{\partial y} = \dfrac{y}{\sqrt{x^2 + y^2}}\), 故 \(\sqrt{1 + \left(\dfrac{\partial z}{\partial x}\right)^2 + \left(\dfrac{\partial z}{\partial y}\right)^2} = \sqrt{2}\). 因此

\[ S = \iint\limits_{D_{xy}}\sqrt{2}\text{d}{x}\text{d}{y} = \sqrt{2}\pi \]

(3) 两者的交线为 \(x^2 + y^2 = a(2a-\sqrt{x^2+y^2})\). 积分区域为 \(\{(x, y) | x^2+y^2 \le a(2a - \sqrt{x^2+y^2})\}\), 化简后即为 \(\{(x, y) | x^2 + y^2 \le a^2\}\). 先考虑 \(z = 2a - \sqrt{x^2 + y^2}\) 带来的上表面面积 \(S_1\). 由于 \(\dfrac{\partial z}{\partial x} = -\dfrac{x}{\sqrt{x^2 + y^2}}, \dfrac{\partial z}{\partial y} = -\dfrac{y}{\sqrt{x^2 + y^2}}\), 故 \(\sqrt{1 + \left(\dfrac{\partial z}{\partial x}\right)^2 + \left(\dfrac{\partial z}{\partial y}\right)^2} = \sqrt{2}\).

\[ \begin{aligned} S_1 & = \iint\limits_{x^2+y^2\le a^2}\sqrt{1 + \left(\dfrac{\partial z}{\partial x}\right)^2 + \left(\dfrac{\partial z}{\partial y}\right)^2}\text{d}{x}\text{d}{y} \\ & = \sqrt{2}\iint\limits_{x^2+y^2\le a^2}\text{d}{x}\text{d}{y} \\ & = \sqrt{2}\pi a^2 \end{aligned} \]

再考虑 \(x^2 + y^2 = az\) 带来的下表面面积 \(S_2\). 由于 \(\dfrac{\partial z}{\partial x} = \dfrac{2x}{a}, \dfrac{\partial z}{\partial y} = \dfrac{2y}{a}\), 故 \(\sqrt{1 + \left(\dfrac{\partial z}{\partial x}\right)^2 + \left(\dfrac{\partial z}{\partial y}\right)^2} = \sqrt{\dfrac{a^2 + 4(x^2 + y^2)}{a^2}}\). 极坐标变换后, 有

\[ \begin{aligned} S_2 & = \iint\limits_{x^2 + y^2 \le a^2}\sqrt{1 + \left(\dfrac{\partial z}{\partial x}\right)^2 + \left(\dfrac{\partial z}{\partial y}\right)^2}\text{d}{x}\text{d}{y} \\ & = \int_{0}^{2\pi}\text{d}{\theta}\int_{0}^{a}r\cdot\sqrt{1+\dfrac{4r^2}{a^2}}\text{d}{r} \\ & = \dfrac{2\pi}{a}\int_{0}^{a}r\sqrt{a^2 + 4r^2}\text{d}{r} \\ & = \dfrac{\pi}{4a}\int_{a^2}^{5a^2}\sqrt{u}\text{d}{u} \\ & = \dfrac{\pi}{6}(5\sqrt{5} - 1)a^2 \end{aligned} \]

因此表面积为 \(S = S_1 + S_2 = \dfrac{\pi}{6}(6\sqrt{2} + 5\sqrt{5} - 1)a^2\).

第 2 题

求下列曲面所包围的均匀物体的质心.

(1) \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} = 1, x \ge 0, y \ge 0, z \ge 0\);

解

(1) 积分区域为 \(\Omega = \{(x, y, z) | \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} \le 1, x \ge 0, y \ge 0, z \ge 0\}\). 做坐标变换 \(\begin{cases}x = ar\sin\varphi\cos\theta, \\ x = br\sin\varphi\sin\theta, \\ z = cr\cos\varphi, \end{cases}\), 则 \(\left|\dfrac{D(x, y, z)}{D(r, \varphi, \theta)}\right| = abcr^2\sin\varphi\). 因此

\[ \begin{aligned} M & = \iiint\limits_{\Omega}\text{d}{x}\text{d}{y}\text{d}{z} \\ & = \int_{0}^{\frac{\pi}{2}}\text{d}{\theta}\int_{0}^{\frac{\pi}{2}}\text{d}{\varphi}\int_{0}^{1}\left|\dfrac{D(x, y, z)}{D(r, \varphi, \theta)}\right|\text{d}{r} \\ & = \int_{0}^{\frac{\pi}{2}}\text{d}{\theta}\int_{0}^{\frac{\pi}{2}}\text{d}{\varphi}\int_{0}^{1}abcr^2\sin\varphi\text{d}{r} \\ & = \int_{0}^{\frac{\pi}{2}}\text{d}{\theta}\int_{0}^{\frac{\pi}{2}}\sin\varphi\text{d}{\varphi}\int_{0}^{1}abcr^2\text{d}{r} \\ & = \dfrac{\pi}{2}\cdot 1\cdot \dfrac{1}{3}abc \\ & = \dfrac{\pi}{6}abc \end{aligned} \]

对 \(xy\) 平面的静力矩为

\[ \begin{aligned} M_{xy} & = \iiint\limits_{\Omega}z\text{d}{x}\text{d}{y}\text{d}{z} \\ & = \int_{0}^{\frac{\pi}{2}}\text{d}{\theta}\int_{0}^{\frac{\pi}{2}}\text{d}{\varphi}\int_{0}^{1}\left|\dfrac{D(x, y, z)}{D(r, \varphi, \theta)}\right|cr\cos\varphi\text{d}{r} \\ & = \int_{0}^{\frac{\pi}{2}}\text{d}{\theta}\int_{0}^{\frac{\pi}{2}}\text{d}{\varphi}\int_{0}^{1}abc^2r^3\sin\varphi\cos\varphi\text{d}{r} \\ & = \int_{0}^{\frac{\pi}{2}}\text{d}{\theta}\int_{0}^{\frac{\pi}{2}}\sin\varphi\cos\varphi\text{d}{\varphi}\int_{0}^{1}abc^2r^3\text{d}{r} \\ & = \dfrac{\pi}{2}\cdot \dfrac{1}{2}\cdot \dfrac{1}{4}abc^2 \\ & = \dfrac{\pi}{16}abc^2 \end{aligned} \]

由对称性知 \(M_{yz} = \dfrac{\pi}{16}a^2bc, M_{xz} = \dfrac{\pi}{16}ab^2c\). 因此质心坐标为 \(\left(\dfrac{3a}{8}, \dfrac{3b}{8}, \dfrac{3c}{8}\right)\).

第 3 题

求解下列问题.

(1) 物体在 \(P(x, y, z)\) 的点密度为 \(\rho = \dfrac{1}{\sqrt{x^2+y^2+z^2}}, \Omega = \{(x, y, z) | 0 \le x, y, z \le 1\}\), 求物体的质心;

解

(1) 总质量为

\[ \begin{aligned} M & = \iiint\limits_{\Omega}(x+y+z)\text{d}{x}\text{d}{y}\text{d}{z} \\ & = \int_{0}^{1}\text{d}{x}\int_{0}^{1}\text{d}{y}\int_{0}^{1}(x+y+z)\text{d}{z} \\ & = \int_{0}^{1}\text{d}{x}\int_{0}^{1}(x+y+\dfrac{1}{2})\text{d}{y} \\ & = \int_{0}^{1}(x+1)\text{d}{x} \\ & = \dfrac{3}{2} \end{aligned} \]

对 \(xy\) 平面的静力矩为

\[ \begin{aligned} M_{xy} & = \iiint\limits_{\Omega}z(x+y+z)\text{d}{x}\text{d}{y}\text{d}{z} \\ & = \int_{0}^{1}\text{d}{x}\int_{0}^{1}\text{d}{y}\int_{0}^{1}z(x+y+z)\text{d}{z} \\ & = \int_{0}^{1}\text{d}{x}\int_{0}^{1}\left(\dfrac{1}{2}(x+y)+\dfrac{1}{3}\right)\text{d}{y} \\ & = \int_{0}^{1}\left(\dfrac{1}{2}x + \dfrac{7}{12}\right)\text{d}{x} \\ & = \dfrac{5}{6} \end{aligned} \]

因此 \(\bar{z} = \dfrac{M_{xy}}{M} = \dfrac{5}{9}\). 由对称性知 \(\bar{x} = \bar{y} = \bar{z} = \dfrac{5}{9}\). 因此质心坐标为 \(\left(\dfrac{5}{9}, \dfrac{5}{9}, \dfrac{5}{9}\right)\).

第 4 题

求下列曲面所包围的均匀物体关于 \(z\) 轴的转动惯量.

(1) \(z = x^2+y^2, x+y = \pm 1, x - y = \pm 1, z = 0\);

解

(1) 设积分区域为 \(\Omega = \{(x, y, z) | z \le x^2 + y^2, x+y\le 1, x\ge 0, y \ge 0, z \ge 0\}\).

\[ \begin{aligned} J_z' & = \iiint\limits_{\Omega}(x^2+y^2)\text{d}{x}\text{d}{y}\text{d}{z} \\ & = \int_{0}^{1}\text{d}{x}\int_{0}^{1-x}\text{d}{y}\int_{0}^{x^2+y^2}(x^2+y^2)\text{d}{z} \\ & = \int_{0}^{1}\text{d}{x}\int_{0}^{1-x}(x^2+y^2)^2\text{d}{y} \\ & = \int_{0}^{1}\left(-\dfrac{28}{15}x^5+4x^4-4x^3+\dfrac{8}{3}x^2-x+\dfrac{1}{5}\right)\text{d}{x} \\ & = \dfrac{7}{90} \end{aligned} \]

由对称性知 \(J_z = 4J_z' = 4\cdot\dfrac{7}{90} = \dfrac{14}{45}\).