8.3 习题课讲解

8.3.1 第一型曲线和曲面积分

解¹ 椭圆$\gamma$的方程可写成$3x^2+4y^2=12$，于是$$\begin{array}{}\text{(1)}& {\oint }_{\gamma }(2xy+3x^2+4y^2)\mathrm{d}l={\oint }_L(12+2xy)\mathrm{d}l=12L+{\oint }_{\gamma }2xy\mathrm{d}l\end{array}$$ 由对称性，${\oint }_{\gamma }2xy\mathrm{d}l=0$。故${\oint }_{\gamma }(2xy+3x^2+4y^2)\mathrm{d}l=12L$。 $◻$

解² 椭圆$\gamma$的参数方程为$$\begin{array}{}\text{(2)}& x=2\cos \theta ,y=\sqrt{3}\sin \theta ,\theta \in [0,2\pi ]\end{array}$$ 于是$$\begin{array}{}\text{(3)}& \mathrm{d}l=\sqrt{4{\sin }^2\theta +3{\cos }^2\theta }\mathrm{d}\theta ,L={\int }_0^{2\pi }\sqrt{4{\sin }^2\theta +3{\cos }^2\theta }\mathrm{d}\theta \end{array}$$ 所求第一型曲线积分为$$\begin{array}{}\text{(4)}& \begin{array}{rl}& {\oint }_{\gamma }(2xy+3x^2+4y^2)\mathrm{d}l\\ & ={\int }_0^{2\pi }[2\cos \theta \cdot \sqrt{3}\sin \theta +3(2\cos \theta {)}^2+4(\sqrt{3}\sin \theta {)}^2]\sqrt{4{\sin }^2\theta +3{\cos }^2\theta }\mathrm{d}\theta \\ & ={\int }_0^{2\pi }[2\sqrt{3}\cos \theta \sin \theta +12]\sqrt{4{\sin }^2\theta +3{\cos }^2\theta }\mathrm{d}\theta \\ & =12L+\frac{\sqrt{3}}{\sqrt{2}}{\int }_0^{2\pi }\sin 2\theta \sqrt{7-\cos 2\theta }\mathrm{d}\theta =12L\end{array}\end{array}$$ $◻$

解¹ 计算可得$$\begin{array}{}\text{(5)}& {\mathit{v}}_r=\left(\begin{array}{c}\cos \phi \\ \sin \phi \\ \phi \end{array}\right),{\mathit{v}}_{\phi }=\left(\begin{array}{c}-r\sin \phi \\ r\cos \phi \\ r\end{array}\right),\{\begin{array}{l}E={\mathit{v}}_r\cdot {\mathit{v}}_r=1+{\phi }^2\\ G={\mathit{v}}_{\phi }\cdot {\mathit{v}}_{\phi }=2r^2\\ F={\mathit{v}}_r\cdot {\mathit{v}}_{\phi }=r\phi \end{array}\end{array}$$ 因此$$\begin{array}{}\text{(6)}& \begin{array}{rl}|S|& ={\int }_S\mathrm{d}S={\int }_0^{2\pi }\mathrm{d}\phi {\int }_0^R\sqrt{EG-F^2}\mathrm{d}r={\int }_0^{2\pi }\sqrt{2+{\phi }^2}\mathrm{d}\phi {\int }_0^{R}r\mathrm{d}r\\ & =\frac{R^2}{2}[\sqrt{2+4{\pi }^2}+\ln (\sqrt{2}\pi +\sqrt{1+2{\pi }^2})]\end{array}\end{array}$$ $◻$

解² 利用$$\begin{array}{}\text{(7)}& \mathrm{d}S=\sqrt{\det \left[{\left(\frac{\partial (x,y,z)}{\partial (r,\phi )}\right)}^{\mathrm{T}}\left(\frac{\partial (x,y,z)}{\partial (r,\phi )}\right)\right]}\mathrm{d}r\mathrm{d}\phi \end{array}$$ $◻$

解¹(利用第一类曲线积分的几何意义) 把$S$看成定义在平面圆周$L:x^2+y^2=R^2$上的函数$z=R^2-x^2$的图像到曲线$L$之间的柱面（图8.3.1中深蓝色部分），计算可得$$\begin{array}{}\text{(8)}& |S|={\int }_L(R^2-x^2)\mathrm{d}l={\int }_0^{2\pi }(R^2-R^2{\cos }^{2}t)R\mathrm{d}t=\pi R^3\end{array}$$ $◻$

解²(第一类曲面积分) 曲面在柱坐标系下的参数方程为$$\begin{array}{}\text{(9)}& x=R\cos t,y=R\sin t,z=s{R}^2{\sin }^{2}t,0\le t\le 2\pi ,0\le s\le 1\end{array}$$ 计算可得$$\begin{array}{}\text{(10)}& {\mathit{v}}_t=\left(\begin{array}{c}-R\sin t\\ R\cos t\\ s{R}^2\sin 2t\end{array}\right),{\mathit{v}}_s=\left(\begin{array}{c}0\\ 0\\ R^2{\sin }^{2}t\end{array}\right),\{\begin{array}{l}E=R^2+s^2{R}^4{\sin }^{2}2t\\ G=R^4{\sin }^{4}t\\ F=s{R}^4\sin 2t{\sin }^{2}t\end{array}\end{array}$$ 因此$$\begin{array}{}\text{(11)}& \mathrm{d}S=\sqrt{EG-F^2}\mathrm{d}t\mathrm{d}s=R^3{\sin }^{2}t\mathrm{d}t\end{array}$$ 于是$$\begin{array}{}\text{(12)}& |S|={\int }^{2\pi }{R}^3{\sin }^{2}t\mathrm{d}t=\pi R^3\end{array}$$ $◻$

解³(第一类曲面积分) 由对称性，只求曲面在$y>0$的部分，此时曲面可以看成函数$y=z(x,y)=\sqrt{R^2-x^2}$的图像，定义域为$\{(x,z)\mid 0\le z\le R^2-x^2\}$。计算可得$$\begin{array}{}\text{(13)}& \begin{array}{rl}|S|& ={\int }_S\mathrm{d}S=2{\int }_{0\le z\le R^2-x^2}\sqrt{1+\parallel \mathrm{\nabla }y{\parallel }^2}\mathrm{d}x\mathrm{d}z=2{\int }_{-R}^R\mathrm{d}x{\int }_0^{R^2-x^2}\sqrt{1+{\left(\frac{x}{\sqrt{R^2-x^2}}\right)}^2}\mathrm{d}z\\ & =4R{\int }_0^R\sqrt{R^2-x^2}\mathrm{d}x=4R^3{\int }_0^{\frac{\pi }{2}}{\sin }^{2}t\mathrm{d}t=\pi R^3\end{array}\end{array}$$ $◻$

8.3.2 第二型曲线积分

解 原积分可化为$$\begin{array}{}\text{(21)}& I:={\oint }_C\frac{ax\mathrm{d}y-by\mathrm{d}x}{|x|+|y|}=\frac{1}{2}{\oint }_C(ax\mathrm{d}y-by\mathrm{d}x)\end{array}$$ 分四段进行积分可得$$\begin{array}{}\text{(22)}& \begin{array}{rl}I& =\frac{1}{2}{\int }_2^0[ax\mathrm{d}(2-x)-b(2-x)\mathrm{d}x]+\frac{1}{2}{\int }_0^{-2}[ax\mathrm{d}(2+x)-b(2+x)\mathrm{d}x]\\ & +\frac{1}{2}{\int }_{-2}^0[ax\mathrm{d}(-2-x)-b(-2-x)\mathrm{d}x]+\frac{1}{2}{\int }_0^2[ax\mathrm{d}(-2+x)-b(-2+x)\mathrm{d}x]\\ & =(a+b)+(a+b)+(a+b)+(a+b)=4(a+b)\end{array}\end{array}$$ $◻$

解 利用Stokes公式并不是一个好的做法，因为3段积分曲线分别在3个坐标平面上，直接计算并不麻烦。利用$x,y,z$的轮换对称性可得$$\begin{array}{}\text{(24)}& {\int }_{{\gamma }_x}(z^2\mathrm{d}y-y^2\mathrm{d}z)={\int }_{{\gamma }_y}(x^2\mathrm{d}z-z^2\mathrm{d}x)={\int }_{{\gamma }_z}(y^2\mathrm{d}x-x^2\mathrm{d}y)\end{array}$$ 其中${\gamma }_x,{\gamma }_y,{\gamma }_z$分别为$\gamma$在坐标平面$x=0,y=0,z=0$的部分。对于${\gamma }_z:\theta \mapsto (\cos \theta ,\sin \theta ),0\le \theta \le \frac{\pi }{2}$，计算可得$$\begin{array}{}\text{(25)}& \frac{I}{3}={\int }_0^{\pi /2}({\sin }^2\theta \mathrm{d}\cos \theta -{\cos }^2\theta \mathrm{d}\sin \theta )=-\frac{4}{3}⟹I=-4\end{array}$$ $◻$

解 边界曲线${\gamma }_1:x=y,1\le y\le 2$：$$\begin{array}{}\text{(27)}& {\int }_{{\gamma }_1}\frac{f(xy)}{y}\mathrm{d}y={\int }_1^2\frac{f\left(y^2\right)}{y}\mathrm{d}y\end{array}$$ 边界曲线${\gamma }_2:x=\frac{4}{y},2\le y\le 4$：$$\begin{array}{}\text{(28)}& {\int }_{{\gamma }_2}\frac{f(xy)}{y}\mathrm{d}y={\int }_2^4\frac{f(4)}{y}\mathrm{d}y=f(4)\ln \frac{4}{2}=f(1)\ln 2\end{array}$$ 边界曲线${\gamma }_3:x=\frac{y}{4},y:4\to 2$：$$\begin{array}{}\text{(29)}& {\int }_{{\gamma }_3}\frac{f(xy)}{y}\mathrm{d}y={\int }_4^2\frac{f\left(\frac{y^2}{4}\right)}{y}\mathrm{d}y=-{\int }_2^4\frac{f\left(\frac{y^2}{4}\right)}{y}\mathrm{d}y\stackrel{u=y/2}{=}-{\int }_1^2\frac{f\left(u^2\right)}{u}\mathrm{d}u\end{array}$$ 边界曲线${\gamma }_4:x=\frac{1}{y},y:2\to 1$：$$\begin{array}{}\text{(30)}& {\int }_{{\gamma }_4}\frac{f(xy)}{y}\mathrm{d}y={\int }_2^1\frac{f(1)}{y}\mathrm{d}y=-f(1)\ln 2\end{array}$$ 所以$$\begin{array}{}\text{(31)}& {\oint }_{\partial D}\frac{f(xy)}{y}\mathrm{d}y=0\end{array}$$ 这个证明只需要$f$连续。 $◻$

解 (1) 在第二型曲线曲面积分中，换元$(x,y)\mapsto (y,x)$的行列式为$-1$，这导致曲线、曲面定向发生改变，所以$$\begin{array}{}\text{(32)}& \begin{array}{rl}{\oint }_{\partial D}[xf(y)\mathrm{d}y-\frac{y}{f(x)}\mathrm{d}x]& ={\oint }_{-\partial D}[yf(x)\mathrm{d}x-\frac{x}{f(y)}\mathrm{d}y]=-{\oint }_{\partial D}[yf(x)\mathrm{d}x-\frac{x}{f(y)}\mathrm{d}y]\\ & ={\oint }_{\partial D}[-yf(x)\mathrm{d}x+\frac{x}{f(y)}\mathrm{d}y]\end{array}\end{array}$$

(2) 由(1)的证明知$$\begin{array}{}\text{(33)}& \begin{array}{rl}{\oint }_{\partial D}[xf(y)\mathrm{d}y-\frac{y}{f(x)}\mathrm{d}x]& =\frac{1}{2}{\int }_D[f(x)+f(y)+\frac{1}{f(x)}+\frac{1}{f(y)}]\mathrm{d}x\mathrm{d}y\\ & \ge \frac{1}{2}{\int }_{D}4\sqrt[4]{f(x)f(y)\frac{1}{f(x)}\frac{1}{f(y)}}\mathrm{d}x\mathrm{d}y=2\pi \end{array}\end{array}$$ $◻$