习题4.6

第 1 题

利用 Green 公式计算下列曲线积分.

(1) \(\oint_{L^{+}}(x+y)^2\text{d}{x} - (x^2 + y^2)\text{d}{y}\), 其中 \(L^{+}\) 是以 \((0, 0), (1, 0), (0, 1)\) 为顶点的三角形的边界, 逆时针为正;

(3) \(\oint_{L^{+}}(x^2+y)\text{d}{x} - (x-y^2)\text{d}{y}\), 其中 \(L\) 是椭圆 \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\) 的正向边界;

解

(1) \(X = (x + y)^2, Y = -(x^2 + y^2)\), 因此由 Green 定理

\[ \begin{aligned} \oint_{L^{+}}(x+y)^2\text{d}{x} - (x^2 + y^2)\text{d}{y} & = \oint_{L^{+}}X\text{d}{x}+Y\text{d}{y} \\ & = \iint\limits_{D}\left(\dfrac{\partial Y}{\partial x} - \dfrac{\partial X}{\partial y}\right)\text{d}{x}\text{d}{y} \\ & = \iint\limits_{D}(-2x - 2x - 2y)\text{d}{x}\text{d}{y} \\ & = -\int_{0}^{1}\text{d}{x}\int_{0}^{1 - x}(4x + 2y)\text{d}{y} \\ & = \int_{0}^{1}(3x^2 - 2x - 1)\text{d}{x} \\ & = -1 \end{aligned} \]

(3) \(X = x^2 + y, Y = y^2 - x\). 由 Green 定理知

\[ \oint_{L^{+}}(x^2+y)\text{d}{x} - (x-y^2)\text{d}{y} = \oint_{L^{+}}X\text{d}{x} + Y\text{d}{y} = \iint\limits_{D}\left(\dfrac{\partial Y}{\partial x} - \dfrac{\partial X}{\partial y}\right)\text{d}{x}\text{d}{y} = \iint\limits_{D}(-2)\text{d}{x}\text{d}{y} = -2\pi ab \]

第 2 题

计算 \(\int_{L^{+}}\dfrac{(x+y)\text{d}{x} + (y-x)\text{d}{y}}{x^2 + y^2}\), 其中 \(L^{+}\) 为

(1) 区域 \(D = \{(x, y) | a^2 \le x^2 + y^2 \le b^2\}\) 的正向边界 \((b > a > 0)\);

(3) 正方形 \(D = \{(x, y) | \left|x\right| + \left|y\right| \le 1\}\) 的逆时针方向;

解

(1) \(X = \dfrac{x + y}{x^2 + y^2}, Y = \dfrac{y - x}{x^2 + y^2}\), 因此 \(\dfrac{\partial Y}{\partial x} = \dfrac{x^2 - 2xy - y^2}{(x^2 + y^2)^2} = \dfrac{\partial X}{\partial y}\). 由 Green 定理知 \(\int_{L^{+}}\dfrac{(x+y)\text{d}{x} + (y-x)\text{d}{y}}{x^2 + y^2} = 0\).

(3) 考虑 \(D^* = \{(x, y) | x = \epsilon\cos\theta, y = \epsilon\sin\theta, 0 \le \theta \le 2\pi\}\), 正方向为 \(\theta\) 从 \(2\pi\) 到 \(0\). 因此 \(\int_{\partial D\cup D^{*}}\dfrac{(x+y)\text{d}{x} + (y-x)\text{d}{y}}{x^2 + y^2} = 0\). 而

\[ \begin{aligned} \int_{\partial D^{*}}\dfrac{(x+y)\text{d}{x} + (y-x)\text{d}{y}}{x^2 + y^2} & = \int_{2\pi}^{0}\dfrac{\epsilon^2(\sin\theta + \cos\theta)\cdot(-\sin\theta) + \epsilon^2(\sin\theta - \cos\theta)\cdot\cos\theta}{\epsilon^2}\text{d}{\theta} \\ & = \int_{2\pi}^{0}\text{d}{\theta} \\ & = 2\pi \end{aligned} \]

因此

\[ \int_{L^{+}}\dfrac{(x+y)\text{d}{x} + (y-x)\text{d}{y}}{x^2 + y^2} = \int_{D\cup D^*}\dfrac{(x+y)\text{d}{x} + (y-x)\text{d}{y}}{x^2 + y^2} - \int_{D^*}\dfrac{(x+y)\text{d}{x} + (y-x)\text{d}{y}}{x^2 + y^2} = -2\pi \]

第 3 题

计算下列曲线积分.

(1) \(\int_{L^{+}}(1+xe^{2y})\text{d}{x} + (x^2e^{2y}-y^2)\text{d}{y}\), 其中 \(L\) 是从点 \((0, 0)\) 经上半圆周 \((x - 2)^2 + y^2 = 4\) 到点 \((4, 0)\) 的弧段;

(3) \(\int_{L^{+}}\left(\ln\dfrac{y}{x} - 1\right)\text{d}{x} + \dfrac{x}{y}\text{d}{y}\), 其中 \(L\) 是从点 \((1, 1)\) 出发到点 \((3, 3e)\) 的任何一条不与坐标轴相交的简单曲线.

解

(1) \(X = 1 + xe^{2y}, Y = x^2e^{2y} - y^2\). 设 \(A(0, 0), B(4, 0)\), 正方向由 \(B\) 到 \(A\). 则 \(AB\) 和 \(L\) 围成的图形 \(D\) 满足: \(\iint\limits_{D}\left(\dfrac{\partial Y}{\partial x} - \dfrac{\partial X}{\partial y}\right)\text{d}{x}\text{d}{y} = \iint\limits_{D}(2xe^{2y} - 2xe^{2y})\text{d}{x}\text{d}{y} = 0\). 由于 \(\int_{BA}X\text{d}{x} = \int_{4}^{0}(1 + x)\text{d}{x} = -12\), 故由 Green 定理

\[ \int_{L^{+}}(1+xe^{2y})\text{d}{x} + (x^2e^{2y}-y^2)\text{d}{y} = \iint\limits_{D}\left(\dfrac{\partial Y}{\partial x} - \dfrac{\partial X}{\partial y}\right)\text{d}{x}\text{d}{y} - \int_{BA}X\text{d}{x} = 12 \]

(3) \(X = \ln\dfrac{y}{x} - 1, Y = \dfrac{x}{y}\), 因此 \(\dfrac{\partial Y}{\partial x} = \dfrac{1}{y} = \dfrac{\partial X}{\partial y}\). 所以这个曲线积分与路径无关, 此时可以先沿着 \(x\) 轴正方向, 再沿着 \(y\) 轴正方向的路径来计算这个积分.

\[ \begin{aligned} \int_{L^{+}}\left(\ln\dfrac{y}{x} - 1\right)\text{d}{x} + \dfrac{x}{y}\text{d}{y} & = \int_{1}^{3}\left(\ln\dfrac{1}{x} - 1\right)\text{d}{x} + \int_{1}^{3e}\dfrac{3}{y}\text{d}{y} \\ & = -x\ln x\bigg\vert_{1}^{3} + 3\ln y\bigg\vert_{1}^{3e} \\ & = 3 \end{aligned} \]

第 4 题

计算下列区域的面积.

(1) 星形线 \(\begin{cases}x = a\cos^3t, \\ y = a\sin ^3t,\end{cases} 0 \le t \le 2\pi\) 所围区域 \((a > 0)\);

解

\[ \begin{aligned} S & = 4\int_{L}y\text{d}{x} \\ & = 4\int_{0}^{\frac{\pi}{2}}(a\sin^3t)\cdot(-3a^2\cos^2t\sin t)\text{d}{t} \\ & = 12a^2\left(\int_{0}^{\frac{\pi}{2}}\sin^4t\text{d}{t} - \int_{0}^{\frac{\pi}{2}}\sin^6t\text{d}{t}\right) \\ & = 12a^2\cdot\left(\dfrac{3\cdot 1}{4\cdot 2} - \dfrac{5\cdot 3\cdot 1}{6\cdot 4\cdot 2}\right)\cdot\dfrac{\pi}{2} \\ & = \dfrac{3}{8}\pi a^2 \end{aligned} \]

第 5 题

已知 \(f(u)\) 连续可微, \(L\) 为任意一条分段光滑曲线, 证明:

(1) \(\oint_{L^{+}}f(xy)(y\text{d}{x} + x\text{d}{y}) = 0\);

(2) \(\oint_{L^{+}}f(x^2 + y^2)(x\text{d}{x} + y\text{d}{y}) = 0\).

证明

(1) \(X = yf(xy), Y = xf(xy)\), 因此 \(\dfrac{\partial Y}{\partial x} = f(xy) + xyf'(xy) = \dfrac{\partial X}{\partial y}\). 由定理 4.6.3 和定理 4.6.4 得 \(\oint_{L^{+}}f(xy)(y\text{d}{x} + x\text{d}{y}) = 0\).

(2) \(X = xf(x^2 + y^2), Y = yf(x^2 + y^2)\), 因此 \(\dfrac{\partial Y}{\partial x} = 2xyf'(x^2 + y^2) = \dfrac{\partial X}{\partial y}\). 由定理 4.6.3 和定理 4.6.4 得 \(\oint_{L^{+}}f(x^2 + y^2)(x\text{d}{x} + y\text{d}{y}) = 0\).

第 6 题

设 \(D\) 是平面区域, \(\partial D\) 为逐段光滑曲线, \((\bar{x}, \bar{y})\) 为 \(D\) 的形心, \(\sigma(D)\) 是 \(D\) 的面积, 证明:

(1) \(\oint_{\partial D}x^2\text{d}{y} = 2\sigma(D)\bar{x}\);

(2) \(\oint_{\partial D}xy\text{d}{y} = \sigma(D)\bar{y}\).

证明

(1) 由 Green 定理 \(\oint_{\partial D}x^2\text{d}{y} = \iint\limits_{D}2x\text{d}{x}\text{d}{y}\). 由定义 \(\bar{x} = \dfrac{\iint\limits_{D}x\text{d}{x}\text{d}{y}}{\iint\limits_{D}\text{d}{x}\text{d}{y}} = \dfrac{\iint\limits_{D}x\text{d}{x}\text{d}{y}}{\sigma(S)}\). 因此 \(\oint_{\partial D}x^2\text{d}{y} = 2\sigma(D)\bar{x}\).

(2) 由 Green 定理 \(\oint_{\partial D}xy\text{d}{y} = \iint\limits_{D}y\text{d}{x}\text{d}{y}\). 由定义 \(\bar{y} = \dfrac{\iint\limits_{D}y\text{d}{x}\text{d}{y}}{\iint\limits_{D}\text{d}{x}\text{d}{y}} = \dfrac{\iint\limits_{D}y\text{d}{x}\text{d}{y}}{\sigma(S)}\). 因此 \(\oint_{\partial D}xy\text{d}{y} = \sigma(D)\bar{y}\).

第 8 题

设 \(D\) 是平面区域, \(\partial D\) 为逐段光滑曲线, \(\mathbf{n}\) 为 \(D\) 的单位外法向, \(u, v \in C^2(D)\), 证明:

(1) \(\oint_{\partial D}\dfrac{\partial u}{\partial \mathbf{n}}\text{d}{l} = \iint\limits_{D}\Delta u\text{d}{x}\text{d}{y}\);

(2) \(\oint_{\partial D}v\dfrac{\partial u}{\partial \mathbf{n}}\text{d}{l} = \iint\limits_{D}v\Delta u\text{d}{x}\text{d}{y} + \iint\limits_{D}\nabla u\cdot \nabla v\text{d}{x}\text{d}{y}\);

(3) \(\oint_{\partial D}\begin{vmatrix}\dfrac{\partial u}{\partial \mathbf{n}} & \dfrac{\partial v}{\partial \mathbf{n}} \\ u & v\end{vmatrix}\text{d}{l} = \iint\limits_{D}\begin{vmatrix}\Delta u & \Delta v \\ u & v\end{vmatrix}\text{d}{x}\text{d}{y}\).

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

证明

\(\text{d}{\mathbf{l}} = (\text{d}{x}, \text{d}{y}), \mathbf{n} = \left(\dfrac{\text{d}{y}}{\left|\text{d}{\mathbf{l}}\right|}, -\dfrac{\text{d}{x}}{\left|\text{d}{\mathbf{l}}\right|}\right)\).

(1) 由 Green 定理,

\[ \begin{aligned} \oint_{\partial D}\dfrac{\partial u}{\partial \mathbf{n}}\text{d}{l} & = \oint_{\partial D}\dfrac{\partial u}{\partial x}\text{d}{y} - \dfrac{\partial u}{\partial y}\text{d}{x} \\ & = \iint\limits_{D}\left(\dfrac{\partial^2u}{\partial x^2} + \dfrac{\partial^2u}{\partial y^2}\right)\text{d}{x}\text{d}{y} \\ & = \iint\limits_{D}\Delta u\text{d}{x}\text{d}{y} \end{aligned} \]

(2) 由 Green 定理,

\[ \begin{aligned} \oint_{\partial D}v\dfrac{\partial u}{\partial \mathbf{n}}\text{d}{l} & = \oint_{\partial D}v\dfrac{\partial u}{\partial x}\text{d}{y} - v\dfrac{\partial u}{\partial y}\text{d}{x} \\ & = \iint\limits_{D}\left(\dfrac{\partial v}{\partial x}\dfrac{\partial u}{\partial x} + v\dfrac{\partial^2u}{\partial x^2} + v\dfrac{\partial^2u}{\partial y^2} + \dfrac{\partial v}{\partial y}\dfrac{\partial u}{\partial y}\right)\text{d}{x}\text{d}{y} \\ & = \iint\limits_{D}v\Delta u\text{d}{x}\text{d}{y} + \iint\limits_{D}\nabla u\cdot \nabla v\text{d}{x}\text{d}{y} \end{aligned} \]

(3) 由 (2) 知,

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

第 9 题

设 \(L\) 为逐段光滑曲线, \(\mathbf{n}\) 为 \(L\) 所围区域的单位外法向, \(\left<\mathbf{n}, \mathbf{i}\right>, \left<\mathbf{n}, \mathbf{j}\right>\) 分别表示 \(\mathbf{n}\) 与 \(x\) 轴、\(y\) 轴正向的夹角, 计算: \(\oint_{L}(x\cos\left<\mathbf{n}, \mathbf{i}\right> + y\cos\left<\mathbf{n}, \mathbf{j}\right>)\text{d}{l}\).

解

设 \(L\) 所围区域为 \(D\), 面积为 \(S\). \(\text{d}{\mathbf{l}} = (\text{d}{x}, \text{d}{y})\), 则 \(\mathbf{n} = \left(\dfrac{\text{d}{y}}{\text{d}{l}}, -\dfrac{\text{d}{x}}{\text{d}{l}}\right)\). 由 Green 定理,

\[ \begin{aligned} \oint_{L}(x\cos\left<\mathbf{n}, \mathbf{i}\right> + y\cos\left<\mathbf{n}, \mathbf{j}\right>)\text{d}{l} & = \oint_{L}x\text{d}{y} - y\text{d}{x} \\ & = \iint\limits_{D}2\text{d}{x}\text{d}{y} \\ & = 2S \end{aligned} \]

第 10 题

求解下列常微分方程.

(1) \((x^2 - y)\text{d}{x} - (x + \sin^2y)\text{d}{y} = 0\);

(2) \(e^{y}\text{d}{x} + (xe^{y}-2y)\text{d}{y} = 0\);

(3) \(\dfrac{x\text{d}{x} + y\text{d}{y}}{\sqrt{x^2 + y^2}} = \dfrac{y\text{d}{x} - x\text{d}{y}}{x^2}\);

(4) \(\left(\cos x + \dfrac{1}{y}\right)\text{d}{x} + \left(\dfrac{1}{y} - \dfrac{x}{y^2}\right)\text{d}{y} = 0\).

解

(1) 原方程可化为 \(\text{d}{\left(\dfrac{1}{3}x^3 - xy - \dfrac{1}{2}y + \dfrac{1}{4}\sin2y\right)} = 0\), 因此解为 \(\dfrac{1}{3}x^3 - xy - \dfrac{1}{2}y + \dfrac{1}{4}\sin2y = C\).

(2) 原方程可化为 \(\text{d}{(xe^y - y^2)} = 0\), 因此解为 \(xe^y - y^2 = C\).

(3) 原方程可化为 \(\text{d}{\sqrt{x^2 + y^2}} = -\text{d}{\left(\dfrac{y}{x}\right)}\), 因此解为 \(\sqrt{x^2 + y^2} + \dfrac{y}{x} = C\).

(4) 原方程可化为 \(\text{d}{\left(\sin x + \dfrac{x}{y} + \ln y\right)} = 0\), 因此解为 \(\sin x + \dfrac{x}{y} + \ln y = C\).

第 11 题

解下列方程.

(1) \((y\cos x - x\sin x)\text{d}{x} + (y\sin x + x\cos x)\text{d}{y} = 0\);

(3) \((3x^2 + y)\text{d}{x} + (2x^2y - x)\text{d}{y} = 0\);

(5) \((x^2 - \sin^2 y)\text{d}{x} + x\sin 2y\text{d}{y} = 0\).

解

(1) 有积分因子 \(e^{y}\). 原方程可化为 \(\text{d}{(e^{y}(y\sin x + x\cos x - \sin x))} = 0\). 因此解为 \(e^{y}(y\sin x + x\cos x - \sin x) = C\).

(3) 有积分因子 \(\dfrac{1}{x^2}\). 原方程可化为 \(\text{d}{\left(3x - \dfrac{y}{x} + y^2\right)} = 0\). 同时, \(x = 0\) 也是原方程的解. 因此解为 \(3x^3 - xy + x^2y^2 = Cx^2\).

(5) 有积分因子 \(\dfrac{1}{x^2}\). 原方程可化为 \(\text{d}{\left(x + \dfrac{\sin^2y}{x}\right)} = 0\). 同时, \(x = 0\) 也是原方程的解. 因此解为 \(x^3 + x^2\sin^2y = Cx^2\).