第4章总复习题

第 1 题

证明: \(\left|\int_L\mathbf{V}\cdot\text{d}{\mathbf{r}}\right| \le \int_{L}\|\mathbf{V}\|\text{d}{l}\).

证明

设 \(\mathbf{n}\) 为 \(\mathbf{V}\) 的单位法向量, \(\mathbf{t}\) 为 \(\mathbf{r}\) 的切向量. 由于 \(\mathbf{V}\cdot \text{d}{\mathbf{r}} = \|\mathbf{V}\|\text{d}{l}\cdot\cos\left<\mathbf{n}, \mathbf{t}\right> \le \|\mathbf{V}\|\text{d}{l}\), 故 \(\left|\int_L\mathbf{V}\cdot\text{d}{\mathbf{r}}\right| \le \int_{L}\|\mathbf{V}\|\text{d}{l}\).

第 2 题

设曲面 \(S:z = f(x, y), (x, y) \in D\), \(S^{+}\) 的法方向向上, \(F \in C(\mathbb{R}^3)\), 求证:

(1) \(\iint\limits_{S^{+}}F(x, y, z)\text{d}{y}\wedge\text{d}{z} = -\iint\limits_{D}F(x, y, f(x, y))\dfrac{\partial f}{\partial x}\text{d}{x}\text{d}{y}\);

(2) \(\iint\limits_{S^{+}}F(x, y, z)\text{d}{z}\wedge\text{d}{x} = -\iint\limits_{D}F(x, y, f(x, y))\dfrac{\partial f}{\partial y}\text{d}{x}\text{d}{y}\).

证明

(1) \(\mathbf{n} = (-f_x, -f_y, 1)\) 为 \(S^{+}\) 的法向量.

\[ \begin{aligned} \iint\limits_{S^{+}}F(x, y, z)\text{d}{y}\wedge\text{d}{z} & = \iint\limits_{D}(F, 0, 0)\cdot \dfrac{\mathbf{n}}{\left|\mathbf{n}\right|}\text{d}{S} \\ & = \iint\limits_{D}F\cdot\dfrac{-f_x}{\left|\mathbf{n}\right|}\cdot \left|\mathbf{n}\right|\text{d}{x}\text{d}{y} \\ & = -\iint\limits_{D}F(x, y, f(x, y))\dfrac{\partial f}{\partial x}\text{d}{x}\text{d}{y} \end{aligned} \]

(2) 由 (1) 的对称性知命题得证.

第 3 题

设闭曲线 \(L:x=x(t), y = y(t), t\in[\alpha, \beta]\), \(L^{+}\) 的方向为 \(t\) 增大的方向, 证明: 由 \(L\) 围成区域的面积可以表示成 \(S = \dfrac{1}{2}\int_{\alpha}^{\beta}\begin{vmatrix}x(t) & y(t) \\ x'(t) & y'(t)\end{vmatrix}\text{d}{t}\).

证明

由于

\[ \begin{aligned} S & = \oint_{L^{+}}x\text{d}{y} - y\text{d}{x} \\ & = \int_{\alpha}^{\beta}(x(t)y'(t) - y(t)x'(t))\text{d}{t} \\ & = \dfrac{1}{2}\int_{\alpha}^{\beta}\begin{vmatrix}x(t) & y(t) \\ x'(t) & y'(t)\end{vmatrix}\text{d}{t} \end{aligned} \]

故命题得证.

第 4 题

设 \(L\subset \mathbb{R}^2\) 为光滑闭曲线, 逆时针为正向, \(\mathbf{n}\) 为 \(L\) 的外法线单位向量, \(\mathbf{a}\) 为一固定向量, 求证: \(\oint_{L}\cos\left<\mathbf{n}, \mathbf{a}\right>\text{d}{l} = 0\).

证明

不妨设 \(\mathbf{a} = (a_x, a_y)\) 为单位向量. 设 \(L\) 围成的区域为 \(D\). \(\mathbf{n} = \left(-\dfrac{\text{d}{y}}{\left|\text{d}{l}\right|}, \dfrac{text{d}{x}}{\left|\text{d}{l}\right|}\right)\). 由 Green 定理

\[ \begin{aligned} \oint_{L}\cos\left<\mathbf{n}, \mathbf{a}\right>\text{d}{l} & = \oint_{L}a_y\text{d}{x} - a_x\text{d}{y} \\ & = \iint\limits_{D}0\text{d}{x}\text{d}{y} \\ & = 0 \end{aligned} \]

第 5 题

设 \(S\) 为闭曲面, \(\mathbf{a}\) 为常向量, \(\mathbf{n} = (\cos\alpha, \cos\beta, \cos\gamma)\) 为 \(S\) 的单位法向量, 证明: \(\oint_{S}\cos\left<\mathbf{n}, \mathbf{a}\right>\text{d}{S} = 0\). % 这里符号有问题, 应该是 \(\oiint\)

证明

不妨设 \(\mathbf{a} = (a_x, a_y)\) 为单位向量. 设 \(S\) 为成的区域为 \(\Omega\). 由 Gauss 定理

\[ \begin{aligned} \oint_{S}\cos\left<\mathbf{n}, \mathbf{a}\right>\text{d}{S} & = \oint_{S}(a_x\cos\alpha, a_y\cos\beta, a_z\cos\gamma)\cdot\text{d}{S} \\ & = \iiint\limits_{\Omega}0\text{d}{x}\text{d}{y}\text{d}{z} \\ & = 0 \end{aligned} \]

第 8 题

设 \(\Omega \subset \mathbb{R}^3\) 是一空间区域, \(\partial\Omega\) 为逐段光滑曲线, \(\mathbf{n}\) 为 \(\Omega\) 的单位外法向, \(u, v \in C^2(\Omega)\), 证明:

(1) \(\oint\limits_{\partial\Omega}\dfrac{\partial u}{\partial \mathbf{n}}\text{d}{S} = \iiint\limits_{\Omega}\Delta u\text{d}{x}\text{d}{y}\text{d}{z}\); % 这里符号有问题, 应该是 \(\oiint\)

(2) \(\oint\limits_{\partial\Omega} v\dfrac{\partial u}{\partial \mathbf{n}}\text{d}{S} = \iiint\limits_{\Omega}v\Delta u\text{d}{x}\text{d}{y}\text{d}{z} + \iiint\limits_{\Omega}\nabla u\cdot \nabla v\text{d}{x}\text{d}{y}\text{d}{z}\) % 这里符号有问题, 应该是 \(\oiint\)

(3) \(\oint\limits_{\partial\Omega}\begin{vmatrix}\dfrac{\partial u}{\partial \mathbf{n}} & \dfrac{\partial v}{\partial \mathbf{n}} \\ u & v\end{vmatrix}\text{d}{S} = \iiint\limits_{\Omega}\begin{vmatrix}\Delta u & \Delta v \\ u & v\end{vmatrix}\text{d}{x}\text{d}{y}\text{d}{z}\), % 这里符号有问题, 应该是 \(\oiint\)

其中 \(\Delta = \dfrac{\partial^2}{\partial x^2} + \dfrac{\partial^2}{\partial y^2} + \dfrac{\partial^2}{\partial z^2}, \nabla = \dfrac{\partial}{\partial x}\mathbf{i} + \dfrac{\partial}{\partial y}\mathbf{j} + \dfrac{\partial}{\partial z}\mathbf{k}\).

证明 % 这里符号有问题, 应该是 \(\oiint\)

(1) 设 \(\mathbf{n} = (\cos\alpha, \cos\beta, \cos\gamma)\). 由 Gauss 定理,

\[ \begin{aligned} \oint_{\partial D}\dfrac{\partial u}{\partial \mathbf{n}}\text{d}{S} & = \oint_{\partial D}(\dfrac{\partial u}{\partial x}\cos\alpha, \dfrac{\partial u}{\partial y}\cos\beta, \dfrac{\partial u}{\partial z}\cos\gamma) \cdot\text{d}{S} \\ & = \iiint\limits_{D}\left(\dfrac{\partial^2u}{\partial x^2} + \dfrac{\partial^2u}{\partial y^2} + \dfrac{\partial^2u}{\partial z^2}\right)\text{d}{x}\text{d}{y}\text{d}{z} \\ & = \iiint\limits_{D}\Delta u\text{d}{x}\text{d}{y}\text{d}{z} \end{aligned} \]

(2) 由 Gauss 定理,

\[ \begin{aligned} \oint_{\partial D}v\dfrac{\partial u}{\partial \mathbf{n}}\text{d}{S} & = \oint_{\partial D}(v\dfrac{\partial u}{\partial x}\cos\alpha, v\dfrac{\partial u}{\partial y}\cos\beta, v\dfrac{\partial u}{\partial z}\cos\gamma) \cdot\text{d}{S} \\ & = \iint\limits_{D}\left(\dfrac{\partial v}{\partial x}\dfrac{\partial u}{\partial x} + v\dfrac{\partial^2u}{\partial x^2} + \dfrac{\partial v}{\partial y}\dfrac{\partial u}{\partial y} + v\dfrac{\partial^2u}{\partial y^2} + \dfrac{\partial v}{\partial z}\dfrac{\partial u}{\partial z} + v\dfrac{\partial^2u}{\partial z^2}\right)\text{d}{x}\text{d}{y}\text{d}{z} \\ & = \iiint\limits_{\Omega}v\Delta u\text{d}{x}\text{d}{y}\text{d}{z} + \iiint\limits_{\Omega}\nabla u\cdot \nabla v\text{d}{x}\text{d}{y}\text{d}{z} \end{aligned} \]

(3) 由 (2) 知,

\[ \begin{aligned} \oint_{\partial D}\begin{vmatrix}\dfrac{\partial u}{\partial \mathbf{n}} & \dfrac{\partial v}{\partial \mathbf{n}} \\ u & v\end{vmatrix}\text{d}{l} & = \oint_{\partial D}v\dfrac{\partial u}{\partial \mathbf{n}}\text{d}{l} - \oint_{\partial D}u\dfrac{\partial v}{\partial \mathbf{n}}\text{d}{l} \\ & = \left(\iint\limits_{D}v\Delta u\text{d}{x}\text{d}{y}\text{d}{z} + \iint\limits_{D}\nabla u\cdot \nabla v\text{d}{x}\text{d}{y}\text{d}{z}\right) - \left(\iint\limits_{D}u\Delta v\text{d}{x}\text{d}{y}\text{d}{z} + \iint\limits_{D}\nabla v\cdot \nabla u\text{d}{x}\text{d}{y}\text{d}{z}\right) \\ & = \iint\limits_{D}v\Delta u\text{d}{x}\text{d}{y}\text{d}{z} - \iint\limits_{D}u\Delta v\text{d}{x}\text{d}{y}\text{d}{z} \\ & = \iint\limits_{D}\begin{vmatrix}\Delta u & \Delta v \\ u & v\end{vmatrix}\text{d}{x}\text{d}{y}\text{d}{z} \end{aligned} \]