高等微积分（2） 第9次作业

Chasse_neige

1

(1)

设 \(S\) 是椭球面 $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 $$ 请用二重积分表示 \(S\) 的面积（不必计算该积分）。

对椭球面做如下参数化 $$ \begin{aligned} x &= a \sin \theta \cos \phi \\ y &= b \sin \theta \sin \phi \\ z &= c \cos \theta \end{aligned} $$ 所以 $$ \begin{aligned} x_{\theta} &= a \cos \theta \cos \phi, y_{\theta} = b \cos \theta \sin \phi, z_{\theta} = - c \sin \theta \\ x_{\phi} &= - a \sin \theta \sin \phi, y_{\phi} = b \sin \theta \cos \phi, z_{\phi} = 0 \end{aligned} $$

\[ \begin{aligned} S &= \iint d S = \iint \begin{vmatrix} \begin{pmatrix} a \cos \theta \cos \phi, b \cos \theta \sin \phi, - c \sin \theta \end{pmatrix} \times \begin{pmatrix} - a \sin \theta \sin \phi, b \sin \theta \cos \phi, 0 \end{pmatrix} \end{vmatrix} d \theta d \phi \\\\ &= \iint \sqrt{(bc \sin^{2} \theta \cos \phi)^{2} + (ac \sin^{2} \theta \sin \phi)^{2} + (ab \sin \theta \cos \theta)^{2}} d \theta d \phi \\\\ &= \iint \sqrt{b^{2} c^{2} \sin^{2} \theta \cos^{2} \phi + a^{2} c^{2} \sin^{2} \theta \sin^{2} \phi + a^{2} b^{2} \cos^{2} \theta} \sin \theta d \theta d \phi \end{aligned} \]

(2)

计算球面 \(x^2 + y^2 + z^2 = R^2\) 的面积。

取 \(a = b = c = R\) $$ S = R^{2} \iint \sin \theta d \theta d \phi = 2 \pi R^{2} \int_{-1}^{1} dx = 4 \pi R^{2} $$

2

设 \(\Sigma\) 为质量均匀的上半球面
$$ \Sigma = {(x, y, z) | x^2 + y^2 + z^2 = 1, z \geq 0} $$

求出 \(\Sigma\) 的质心位置，即计算 $$ \left( \frac{\iint_{\Sigma} x \, dS}{\iint_{\Sigma} dS}, \frac{\iint_{\Sigma} y \, dS}{\iint_{\Sigma} dS}, \frac{\iint_{\Sigma} z \, dS}{\iint_{\Sigma} dS} \right) $$ 由对称性，显然有 $$ \frac{\iint_{\Sigma} x \, dS}{\iint_{\Sigma} dS} = \frac{\iint_{\Sigma} y \, dS}{\iint_{\Sigma} dS} = 0 $$

\[ \frac{\iint_{\Sigma} z \, dS}{\iint_{\Sigma} dS} = \frac{1}{2 \pi} \iint \cos \theta \sin \theta d \theta d \phi = \int_{0}^{\frac{\pi}{2}} \cos \theta \sin \theta d \theta = \frac{1}{2} \]

3

给定 \(a > b > 0\)，定义曲面 \(T\) 为
$$ T = {(x, y, z) \in \mathbb{R}^3 | (\sqrt{x^2 + y^2} - a)^2 + z^2 = b^2} $$ 求 \(T\) 的面积。

显然，\(T\) 是双层曲面

做如下参数化 $$ \begin{aligned} x &= (b \cos \theta + a) \sin \phi \\ y &= (b \cos \theta + a) \cos \phi \\ z &= b \sin \theta \end{aligned} $$ 其中，参数取值范围 \(\theta \in [0, \pi], \phi \in [0, 2 \pi)\)所以 $$ \begin{aligned} x_{\theta} &= - b \sin \theta \sin \phi, y_{\theta} = - b \sin \theta \cos \phi, z_{\theta} = b \cos \theta \\ x_{\phi} &= (b \cos \theta + a) \cos \phi, y_{\phi} = - (b \cos \theta + a) \sin \phi, z_{\phi} = 0 \end{aligned} $$ 面积 $$ \begin{aligned} S &= \iint |(-b \sin \theta \sin \phi, -b \sin \theta \cos \phi, b \cos \theta) \times ((b \cos \theta + a) \cos \phi, - (b \cos \theta + a) \sin \phi, 0)| d \theta d \phi \\ &= \iint \sqrt{(b^{2} \cos^{2} \theta + ab \cos \theta)^{2} \sin^{2} \phi + (b^{2} \cos^{2} \theta + ab \cos \theta)^{2} \cos^{2} \phi + (b^{2} \sin \theta \cos \theta + ab \sin \theta)^{2}} d \theta d \phi \\ &= \iint \sqrt{b^{4} \cos^{2} \theta + 2 ab^{3} \cos \theta + a^{2} b^{2}} d \theta d \phi \\ &= \iint b (b \cos \theta + a) d \theta d \phi = 2 \pi b \int_{0}^{\pi} (b \cos \theta + a) d \theta = 2 \pi^{2} ab \end{aligned} $$

4

令 \(S\) 为单位球面 $$ S = {(x, y, z) | x^2 + y^2 + z^2 = 1} $$

设 \(p : [0, 1] \to S\) 是 \(C^1\) 光滑映射
$$ p(t) = (x(t), y(t), z(t)), \quad \forall t \in [0, 1] $$

假设 \(p(0) = (0, 0, -1), \, p(1) = (0, 0, 1)\)，且对任何 \(0 < t < 1\) 有 \(-1 < z(t) < 1\)。

(1)

证明：对任何 \(t \in [0, 1]\)，有
$$ x(t)x'(t) + y(t)y'(t) + z(t)z'(t) = 0 $$ 证明 $$ x^{2} (t) + y^{2} (t) + z^{2} (t) = 1 $$

\[ \frac{d}{dt} (x^{2} (t) + y^{2} (t) + z^{2} (t)) = 0 \]

所以 $$ x(t)x'(t) + y(t)y'(t) + z(t)z'(t) = 0 $$

(2)

证明：对任何 \(0 < t < 1\)，有
$$ x'(t)^2 + y'(t)^2 \geq \frac{z(t)^2}{x(t)^2 + y(t)^2} z'(t)^2 $$ 证明

由 Cauchy 不等式 $$ (x^{2} + y^{2}) (x'^{2} + y'^{2}) \geq (x x' + y y')^{2} = (- z z')^{2} $$ 所以 $$ x'(t)^2 + y'(t)^2 \geq \frac{z(t)^2}{x(t)^2 + y(t)^2} z'(t)^2 $$

(3)

定义 \(p\) 的弧长为 $$ L = \int_0^1 \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2} \, dt $$

证明：\(L \geq \pi\)

证明 $$ \begin{aligned} L &= \int_0^1 \sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2} \, dt \geq \int_{0}^{1} \sqrt{\frac{z(t)^2}{x(t)^2 + y(t)^2} z'(t)^2 + z'^{2} (t)} \, dt \\ &= \int_{0}^{1} \frac{|z' (t)|}{\sqrt{x^{2} (t) + y^{2} (t)}} dt = \int_{0}^{1} \sqrt{\frac{z'^{2} (t)}{1 - z^{2} (t)}} dt \\ \geq \int_{0}^{1} \frac{z' (t)}{\sqrt{1 - z^{2} (t)}} dt &= \int_{-1}^{1} \frac{dz}{\sqrt{1 - z^{2}}} = \pi \end{aligned} $$

5

给定正数 \(c < \sqrt{2}\)。设曲面 \(S = \{(x, y, z) | x^2 + y^2 + z^2 = 1, x + y \geq c\}\)。

(1)

计算第一型曲面积分 \(\iint_S x \, dS\)，其中 \(dS\) 表示面积微元。

作换元 $$ \begin{aligned} &\begin{pmatrix} u \\ v \\ w \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 \\ - \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} \end{aligned} $$ 由于变换矩阵是正交阵，所以面积元不变 $$ x = \frac{\sqrt{2}}{2} (u - v) $$ 再换到球坐标下 $$ \begin{aligned} u &= \cos \theta \\ v &= \sin \theta \sin \phi \\ w &= \sin \theta \cos \phi \end{aligned} $$ 所以积分范围为 \(\theta \in [0, \arccos \frac{\sqrt{2}}{2} c], \phi \in [0, 2 \pi]\) $$ \begin{aligned} \iint_{S} x dS &= \iint_{S} \frac{\sqrt{2}}{2} (\cos \theta - \sin \theta \sin \phi) \sin \theta d \theta d \phi \\ &= \frac{\sqrt{2}}{2} \cdot 2 \pi \int_{0}^{\arccos \frac{\sqrt{2}}{2} c} \cos \theta \sin \theta d \theta \\ &= \sqrt{2} \pi \cdot \frac{1}{2} (1 - \frac{c^{2}}{2}) = \frac{\sqrt{2}}{2} \pi (1 - \frac{c^{2}}{2}) \end{aligned} $$

(2)

计算第一型曲线积分 \(\int_{\partial S} x \, dl\)，其中 \(\partial S\) 表示 \(S\) 的边界，\(dl\) 表示弧长微元

参数化边界，由于边界上 $u = \frac{\sqrt{2}}{2} c $ $$ l: (\frac{\sqrt{2}}{2} c, \sqrt{1 - \frac{c^{2}}{2}} \sin \phi, \sqrt{1 - \frac{c^{2}}{2}} \cos \phi) $$ 所以 $$ \int_{\partial S} x \, dl = \int_{0}^{2 \pi} \frac{\sqrt{2}}{2} (\frac{\sqrt{2}}{2} c - \sqrt{1 - \frac{c^{2}}{2}} \sin \phi) \sqrt{1 - \frac{c^{2}}{2}} d \phi = \pi c \sqrt{1 - \frac{c^{2}}{2}} $$

6

设 \(C = \{(x, y) | x^2 + y^2 = 1\}\) 是平面上的单位圆周，取逆时针定向（方向）。

(1)

计算第二型曲线积分 $$ B(u) = \oint_C \frac{-y \, dx + x \, dy}{(x^2 + y^2 + u^2)^{3/2}} $$ 参数化 $$ \begin{aligned} x &= \cos \theta \\ y &= \sin \theta \end{aligned} $$

\[ B(u) = \oint_C \frac{-y \, dx + x \, dy}{(x^2 + y^2 + u^2)^{3/2}} = \int_{0}^{2 \pi} \frac{ \sin^{2} \theta + \cos^{2} \theta}{(1 + u^{2})^{\frac{3}{2}}} d \theta = \frac{2 \pi}{(1 + u^{2})^{\frac{3}{2}}} \]

(2)

计算极限 $$ \lim_{A \to +\infty} \int_{-A}^A B(u) \, du $$

\[ \begin{aligned} \lim_{A \to +\infty} \int_{-A}^A B(u) \, du &= \lim_{A \to \infty} \int_{-A}^{A} \frac{2 \pi}{\cosh^{3} \theta} d (\sinh \theta) \\\\ &= \lim_{A \to \infty} \int_{- A}^{A} \frac{2 \pi}{\cosh^{2} \theta} d \theta = 4 \pi \end{aligned} \]

7

计算 $$ \oint_L xy \, dx + yz \, dy + zx \, dz $$

其中 \(L\) 为曲线
$$ \begin{aligned} \begin{cases} x^2 + y^2 + z^2 &= 1 \\ x + y + z &= 1 \end{cases} \end{aligned} $$ 积分方向从 \((1, 0, 0)\) 沿着劣弧指向 \((0, 1, 0)\)

同样先做正交变换 $$ \begin{aligned} &\begin{pmatrix} u \\ v \\ w \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{3}}{3} & \frac{\sqrt{3}}{3} & \frac{\sqrt{3}}{3} \\ - \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} & 0 \\ \frac{\sqrt{6}}{6} & \frac{\sqrt{6}}{6} & - \frac{\sqrt{6}}{3} \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} \end{aligned} $$

\[ \begin{aligned} &\begin{pmatrix} x \\\\ y \\\\ z \end{pmatrix} = \begin{pmatrix} \frac{\sqrt{3}}{3} & - \frac{\sqrt{2}}{2} & \frac{\sqrt{6}}{6} \\\\ \frac{\sqrt{3}}{3} & \frac{\sqrt{2}}{2} & \frac{\sqrt{6}}{6} \\\\ \frac{\sqrt{3}}{3} & 0 & - \frac{\sqrt{6}}{3} \end{pmatrix} \begin{pmatrix} u \\\\ v \\\\ w \end{pmatrix} \end{aligned} \]

所以利用 Stokes 公式 $$ \nabla \times (xy, yz, zx) = - (y, z, x) $$

\[ \begin{aligned} \oint_L xy \, dx + yz \, dy + zx \, dz &= \iint_{S} - (y, z, x) \cdot \frac{\sqrt{3}}{3} (1, 1, 1) dS \\\\ &= - \frac{\sqrt{3}}{3} \iint_{S} d S = - \frac{2 \sqrt{3} \pi}{9} \end{aligned} \]

8

设 \(a, b, c\) 为给定的正数，\(S\) 为椭球面 $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 $$

取指向外面的定向。对于正整数 \(n\)，计算第二型曲面积分 $$ I_n = \iint_S (x^n \, dy \, dz + y^n \, dz \, dx + z^n \, dx \, dy) $$ 对椭球面做如下参数化 $$ \begin{aligned} x &= a \sin \theta \cos \phi \\ y &= b \sin \theta \sin \phi \\ z &= c \cos \theta \end{aligned} $$ 所以利用 Gauss 公式 $$ \begin{aligned} I_n &= \iint_S (x^n \, dy \, dz + y^n \, dz \, dx + z^n \, dx \, dy) \\ &= \iiint_{V} n (x^{n - 1} + y^{n - 1} + z^{n - 1}) dx dy dz \\ &= n abc \iiint_{V} (a^{n - 1} r^{n - 1} \sin^{n - 1} \theta \cos^{n - 1} \phi + b^{n - 1} r^{n - 1} \sin^{n - 1} \theta \sin^{n - 1} \phi + c^{n - 1} r^{n - 1} \cos^{n - 1} \theta) r^{2} \sin \theta dr d \theta d \phi \\ &= \frac{2n \pi}{n + 2} abc \int_{0}^{\pi} \int_{0}^{2 \pi} (a^{n - 1} \sin^{n - 1} \theta \cos^{n - 1} \phi + b^{n - 1} \sin^{n - 1} \theta \sin^{n - 1} \phi + c^{n - 1} \cos^{n - 1} \theta) \sin \theta d \theta d \phi \\ &= \frac{2n \pi}{n + 2} abc ((a^{n - 1} + b^{n - 1}) 2 \frac{(n - 2)!!}{n!!} + c^{n - 1} \frac{2}{n}) \\ &= \frac{4 \pi}{n + 2} abc (a^{n - 1} + b^{n - 1} + c^{n + 1}) \end{aligned} $$