9.1 第7次作业评讲

9.1.1 解答和证明部分

解 弧长微元为$$\begin{array}{}\text{(2)}& \mathrm{d}l=\sqrt{{\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)}^2+{\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)}^2+{\left(\frac{\mathrm{d}z}{\mathrm{d}t}\right)}^2}\mathrm{d}t=\sqrt{4+1+4}\mathrm{d}t=3\mathrm{d}t\end{array}$$ 代入积分中可得$$\begin{array}{}\text{(3)}& I={\int }_0^1[2(2t)+4t+(2-2t{)}^2-4]\cdot 3\mathrm{d}t=12{\int }_0^1{t}^2\mathrm{d}t=4\end{array}$$ $◻$

解 曲面$\mathrm{\Sigma }$的一个合适的参数化方案就是选用极坐标系，亦即$$\begin{array}{}\text{(4)}& \mathit{x}(r,\theta )=\left(\begin{array}{c}r\cos \theta \\ r\sin \theta \\ r^2\cos \theta \sin \theta \end{array}\right),(r,\theta )\in [0,1]\times (-\pi ,\pi ]\end{array}$$ 计算可得$$\begin{array}{}\text{(5)}& \frac{\partial \mathit{x}}{\partial r}=\left(\begin{array}{c}\cos \theta \\ \sin \theta \\ r\sin 2\theta \end{array}\right),\frac{\partial \mathit{x}}{\partial \theta }=\left(\begin{array}{c}-r\sin \theta \\ r\cos \theta \\ r^2\cos 2\theta \end{array}\right)\end{array}$$ 进一步计算可得$$\begin{array}{}\text{(6)}& E={\Vert \frac{\partial \mathit{x}}{\partial r}\Vert }^2=1+r^2{\sin }^{2}2\theta ,G={\Vert \frac{\partial \mathit{x}}{\partial \theta }\Vert }^2=r^2+r^4{\cos }^{2}2\theta ,F=⟨\frac{\partial \mathit{x}}{\partial r},\frac{\partial \mathit{x}}{\partial \theta }⟩=r^3\sin 2\theta \cos 2\theta \end{array}$$ 故面积微元为$$\begin{array}{}\text{(7)}& \begin{array}{rl}EG-F^2& =r^2[(1+r^2{\sin }^{2}2\theta )(1+r^2{\cos }^{2}2\theta )-r^4{\sin }^{2}2\theta {\cos }^{2}2\theta ]=r^2(1+r^2)\\ \mathrm{d}S& =\sqrt{EG-F^2}\mathrm{d}r\mathrm{d}\theta =r\sqrt{1+r^2}\mathrm{d}r\mathrm{d}\theta \end{array}\end{array}$$ 因此$$\begin{array}{}\text{(8)}& A={\int }_0^1\mathrm{d}r{\int }_{-\pi }^{\pi }\sqrt{EG-F^2}\mathrm{d}\theta =2\pi {\int }_0^{1}r\sqrt{1+r^2}\mathrm{d}r={\frac{2\pi }{3}{(1+r^2)}^{3/2}|}_0^1=\frac{2\pi }{3}(2\sqrt{2}-1)\end{array}$$ $◻$

解 由于对称性，我们只需要计算曲线在第一象限中的质量，随后乘以$4$即可。$L$在第一象限中可化为$L_1:x+y=\sqrt{2}$，则弧长微元为$$\begin{array}{}\text{(9)}& \mathrm{d}l=\sqrt{1+{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}^2}\mathrm{d}x=\sqrt{2}\mathrm{d}x\end{array}$$ 因此$$\begin{array}{}\text{(10)}& M=4{\int }_{L_1}\mu \mathrm{d}l=4{\int }_0^{\sqrt{2}}[3+x^2-(\sqrt{2}-x{)}^2]\sqrt{2}\mathrm{d}x=4\sqrt{2}{\int }_0^{\sqrt{2}}(2\sqrt{2}x+1)\mathrm{d}x=24\end{array}$$ $◻$

解 计算可得$$\begin{array}{}\text{(12)}& \frac{\partial \mathit{x}}{\partial r}=\left(\begin{array}{c}\cos \theta \\ \sin \theta \\ 0\end{array}\right),\frac{\partial \mathit{x}}{\partial \theta }=\left(\begin{array}{c}-r\sin \theta \\ r\cos \theta \\ 1\end{array}\right)\end{array}$$ 进一步计算可得$$\begin{array}{}\text{(13)}& E={\Vert \frac{\partial \mathit{x}}{\partial r}\Vert }^2=1,G={\Vert \frac{\partial \mathit{x}}{\partial \theta }\Vert }^2=r^2+1,F=⟨\frac{\partial \mathit{x}}{\partial r},\frac{\partial \mathit{x}}{\partial \theta }⟩=0\end{array}$$ 故面积微元为$$\begin{array}{}\text{(14)}& \mathrm{d}S=\sqrt{EG-F^2}\mathrm{d}r\mathrm{d}\theta =\sqrt{1+r^2}\mathrm{d}r\mathrm{d}\theta \end{array}$$ 因此$$\begin{array}{}\text{(15)}& I={\int }_0^1\mathrm{d}r{\int }_0^{2\pi }\frac{\sqrt{1+r^2}}{\pi \sqrt{r^2+1}}\mathrm{d}\theta =2\end{array}$$ $◻$

解 设$(L_1,L_2,L_3)=(AB,BC,CA)$，则$$\begin{array}{}\text{(16)}& \begin{array}{rl}{L}_1& :(x,y,z)=(a-t,t,0),0\le t\le a\\ L_2& :(x,y,z)=(0,a-t,t),0\le t\le a\\ L_3& :(x,y,z)=(t,0,a-t),0\le t\le a\end{array}\end{array}$$ 弧长微元为$$\begin{array}{}\text{(17)}& \mathrm{d}{l}_1=\mathrm{d}{l}_2=\mathrm{d}{l}_3=\sqrt{1+1}\mathrm{d}t=\sqrt{2}\mathrm{d}t\end{array}$$ 计算可得$$\begin{array}{}\text{(18)}& \begin{array}{rl}{I}_1& ={\int }_0^a[(a-t)+t]\sqrt{2}\mathrm{d}t=\sqrt{2}{\int }_0^{a}a\mathrm{d}t=\sqrt{2}{a}^2\\ I_2& ={\int }_0^a[0+(a-t)]\sqrt{2}\mathrm{d}t=\sqrt{2}\cdot \frac{1}{2}{a}^2=\frac{\sqrt{2}}{2}{a}^2\\ I_3& ={\int }_0^a[t+0]\sqrt{2}\mathrm{d}t=\sqrt{2}\cdot \frac{1}{2}{a}^2=\frac{\sqrt{2}}{2}{a}^2\end{array}\end{array}$$ 因此$$\begin{array}{}\text{(19)}& I=I_1+I_2+I_3=2\sqrt{2}{a}^2=4\end{array}$$ $◻$

解 注意到$$\begin{array}{}\text{(21)}& \omega =(1-x^2)\mathrm{d}x+2(1-y^2)\mathrm{d}y+3(1-z^2)\mathrm{d}z=\mathrm{d}(x+2y+3z-\frac{1}{3}{x}^3-\frac{2}{3}{y}^3-z^3)\end{array}$$ 故原积分可化为$$\begin{array}{}\text{(22)}& I={\int }_{L^{+}}\omega ={[x+2y+3z-\frac{1}{3}{x}^3-\frac{2}{3}{y}^3-z^3]}_{S=(0,0,-1)}^{N=(0,0,1)}=4\end{array}$$ $◻$

解 注意到$$\begin{array}{}\text{(24)}& \omega =(-x-2y)\mathrm{d}x+(2x+y)\mathrm{d}y+(y-x)\mathrm{d}(-x-y)=3(x\mathrm{d}y-y\mathrm{d}x)\end{array}$$ 设$S=2$为$|x|+|y|=1$围成的平面区域$D$的面积，利用Green公式可得$$\begin{array}{}\text{(25)}& I={\oint }_{L^{+}}\omega =6{\oint }_{L^{+}}\frac{1}{2}(x\mathrm{d}y-y\mathrm{d}x)=6{\int }_D\mathrm{d}x\mathrm{d}y=6S=12\end{array}$$ $◻$

解 本题等价于：对任意有向（简单）闭曲线$L^{+}$，都满足$$\begin{array}{}\text{(26)}& {\oint }_{L^{+}}[(2x^2+axy)\mathrm{d}x+(x^2+3y^2)\mathrm{d}y]=0\end{array}$$ 设$L^{+}$为$D$的边界，由Green公式可得$$\begin{array}{}\text{(27)}& {\int }_D[\frac{\partial }{\partial x}(x^2+3y^2)-\frac{\partial }{\partial y}(2x^2+axy)]\mathrm{d}x\mathrm{d}y={\int }_D(2-a)x\mathrm{d}x\mathrm{d}y=0\end{array}$$ 为了使上式对任意区域$D$成立，必须有$2-a=0⟹a=2$。 $◻$

解 直接代入计算可得$$\begin{array}{}\text{(29)}& I={\int }_0^1(9t^2\cdot \mathrm{d}t-3t\cdot 2t\mathrm{d}t+4t^4\cdot 4t^3\mathrm{d}t)=3-2+2=3\end{array}$$ $◻$

解 凑全微分可得$$\begin{array}{}\text{(31)}& (\sin x+y+{\mathrm{e}}^y)\mathrm{d}x+(3x+x{\mathrm{e}}^y)\mathrm{d}y=\mathrm{d}(-\cos x+xy+x{\mathrm{e}}^y)+2x\mathrm{d}y\end{array}$$ 故有$$\begin{array}{}\text{(32)}& I=\frac{1}{\pi {\lambda }^2}[{\oint }_{L_{\lambda }^{+}}\mathrm{d}(-\cos x+xy+x{\mathrm{e}}^y)+2{\oint }_{L_{\lambda }^{+}}x\mathrm{d}y]=\frac{0+2\pi {\lambda }^2}{\pi {\lambda }^2}=2\end{array}$$ $◻$