10.1 第七次作业参考答案

10.1.1 习题7.5

解 记$\gamma$的弧长参数为$l$，则$$\begin{array}{}\text{(3)}& \parallel {\mathit{x}}^{\prime }(l)\parallel =1,\parallel {\mathit{x}}^{″}(l)\parallel =\kappa (l),\mathrm{d}l=\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}\mathrm{d}t\end{array}$$ 不妨设${\mathit{x}}^{\prime }(l)=(\cos \theta (l),\sin \theta (l))$，则${\mathit{x}}^{″}(l)=(-\sin \theta (l),\cos \theta (l)){\theta }^{\prime }(l)$，故$$\begin{array}{}\text{(4)}& \kappa (l)=\frac{det\left(\begin{array}{cc}\cos \theta (l)& -\sin \theta (l){\theta }^{\prime }(l)\\ \sin \theta (l)& \cos \theta (l){\theta }^{\prime }(l)\end{array}\right)}{1}={\theta }^{\prime }(l)\end{array}$$ 于是$$\begin{array}{}\text{(5)}& {\int }_a^b\kappa (t)\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}\mathrm{d}t={\int }_0^{L_{\gamma }}{\theta }^{\prime }(l)\mathrm{d}l=\theta (L_{\gamma })-\theta (0)=2N\pi \end{array}$$ 其中$N$为曲线$\gamma$绕自身切线旋转的圈数¹（Turning Number）。 $◻$

解 设星形线的参数方程为$(x,y)=(a{\cos }^{3}t,a{\sin }^{3}t)$，$t\in [0,2\pi ]$。

(1) $$\begin{array}{}\text{(6)}& \begin{array}{rl}\mathrm{d}l& =\sqrt{\mathrm{d}{x}^2+\mathrm{d}{y}^2}=\sqrt{9a^2{\cos }^{4}t{\sin }^{2}t+9a^2{\sin }^{4}t{\cos }^{2}t}\mathrm{d}t\\ & =3a|\cos t\sin t|\mathrm{d}t=\frac{3a}{2}|\sin 2t|\mathrm{d}t\end{array}\end{array}$$ 从而$$\begin{array}{}\text{(7)}& L={\int }_{\gamma }\mathrm{d}l={\int }_0^{2\pi }\frac{3a}{2}|\sin 2t|\mathrm{d}t=6a\end{array}$$

(2) $$\begin{array}{}\text{(8)}& S=-{\int }_{\gamma }y\mathrm{d}x=-{\int }_0^{2\pi }a{\sin }^{3}t\cdot 3a{\cos }^{2}t(-\sin t)\mathrm{d}t=\frac{3}{8}\pi a^2\end{array}$$

(3) 记${\gamma }_{+}$表示$\gamma$在$x$轴上方的部分，方向与$\gamma$的自然正向相同，则$$\begin{array}{}\text{(9)}& V=-{\int }_{{\gamma }_{+}}\pi y^2\mathrm{d}x={\int }_0^{\pi }\pi a^2{\sin }^{6}t\cdot 3a{\cos }^{2}t\sin t\mathrm{d}t=\frac{32\pi }{105}{a}^3\end{array}$$

(4) $$\begin{array}{}\text{(10)}& A={\int }_{{\gamma }^{+}}2\pi y\mathrm{d}l={\int }_0^{\pi }2\pi a{\sin }^{3}t\cdot \frac{3a}{2}|\sin 2t|\mathrm{d}t=\frac{12\pi }{5}{a}^2\end{array}$$ $◻$

解 参考例10.3.1。 $◻$

解 (1) $$\begin{array}{}\text{(11)}& \mathrm{d}l=\sqrt{\mathrm{d}{\rho }^2+{\rho }^2\mathrm{d}{\theta }^2}=\sqrt{a^2{\sin }^2\theta \mathrm{d}{\theta }^2+a^2(1+\cos \theta {)}^2\mathrm{d}{\theta }^2}=2a|\cos \frac{\theta }{2}|\mathrm{d}\theta \end{array}$$ 为了保持对称性，我们选择积分区域为$[-\pi ,\pi ]$，从而$$\begin{array}{}\text{(12)}& L={\int }_{\gamma }\mathrm{d}l={\int }_{-\pi }^{\pi }2a\cos \frac{\theta }{2}\mathrm{d}\theta =8a\end{array}$$

(2) 同样为了保持对称性，我们选择弧长参数$l:[-\pi ,\pi ]\to [-4a,4a]$为$$\begin{array}{}\text{(13)}& l(\theta )={\int }_0^{\theta }2a\cos \frac{t}{2}\mathrm{d}t=4a\sin \frac{\theta }{2}⟹\theta (l)=2\arcsin \frac{l}{4a}\end{array}$$ 从而$$\begin{array}{}\text{(14)}& \begin{array}{rl}\cos \theta (l)& =1-2{\sin }^2\frac{\theta (l)}{2}=1-\frac{l^2}{8a^2}\\ \sin \theta (l)& =2\sin \frac{\theta (l)}{2}\cos \frac{\theta (l)}{2}=\frac{l}{2a}\sqrt{1-\frac{l^2}{16a^2}}\end{array}\end{array}$$ 故心脏线的参数方程为$$\begin{array}{}\text{(15)}& \begin{array}{rl}x(l)& =a(1+\cos \theta (l))\cos \theta (l)=a(2-\frac{l^2}{8a^2})(1-\frac{l^2}{8a^2})\\ y(l)& =a(1+\cos \theta (l))\sin \theta (l)=\frac{l}{2}(2-\frac{l^2}{8a^2})\sqrt{1-\frac{l^2}{16a^2}}\end{array}\end{array}$$ 求导可得$$\begin{array}{}\text{(16)}& \kappa (l)=\parallel {\mathit{x}}^{″}(l)\parallel =\frac{3}{\sqrt{16a^2-l^2}}\end{array}$$

(3) 由对称性知质心在$x$轴上，故$\overline{y}=0$，而$$\begin{array}{}\text{(17)}& \overline{x}=\frac{1}{L}{\int }_{-\pi }^{\pi }x(l)\mathrm{d}l=\frac{1}{8a}{\int }_{-\pi }^{\pi }a(1+\cos \theta )\cos \theta \cdot 2a\cos \frac{\theta }{2}\mathrm{d}\theta =\frac{4}{5}a\end{array}$$ 即质心为$(\frac{4}{5}a,0)$。

(4) $$\begin{array}{}\text{(18)}& S=\frac{1}{2}{\int }_{-\pi }^{\pi }\rho (\theta {)}^2\mathrm{d}\theta =\frac{1}{2}{\int }_{-\pi }^{\pi }{a}^2(1+\cos \theta {)}^2\mathrm{d}\theta =\frac{3}{2}\pi a^2\end{array}$$ 由对称性知质心在$x$轴上，故$\overline{y}=0$，而$$\begin{array}{}\text{(19)}& \overline{x}=-\frac{1}{S}{\int }_{\gamma }xy\mathrm{d}x=-\frac{2}{3\pi a^2}{\int }_{-\pi }^{\pi }{a}^2(1+\cos \theta {)}^2\cos \theta \sin \theta \cdot a(-\sin \theta )(1+2\cos \theta )\mathrm{d}\theta =\frac{5}{6}a\end{array}$$ 即质心为$(\frac{5}{6}a,0)$。

(5) 记${\gamma }_{+}$表示$\gamma$在$x$轴上方的部分，方向与$\gamma$的自然正向相同，则$$\begin{array}{}\text{(20)}& A={\int }_{{\gamma }^{+}}2\pi y\mathrm{d}l={\int }_0^{\pi }2\pi a(1+\cos \theta )\sin \theta \cdot 2a\cos \frac{\theta }{2}\mathrm{d}\theta =\frac{32}{5}\pi a^2\end{array}$$ 由对称性知质心在$x$轴上，故$\overline{y}=0$，而$$\begin{array}{}\text{(21)}& \overline{x}=\frac{1}{A}{\int }_{{\gamma }^{+}}2\pi xy\mathrm{d}l=\frac{5}{32\pi a^2}{\int }_0^{\pi }2\pi a^2(1+\cos \theta {)}^2\sin \theta \cos \theta \cdot 2a\cos \frac{\theta }{2}\mathrm{d}\theta =\frac{50}{63}a\end{array}$$ 即质心为$(\frac{50}{63}a,0)$。

(6) $$\begin{array}{}\text{(22)}& V=-{\int }_{{\gamma }^{+}}\pi y^2\mathrm{d}x={\int }_0^{\pi }\pi a^2(1+\cos \theta {)}^2{\sin }^2\theta \cdot a\sin \theta (1+2\cos \theta )\mathrm{d}\theta =\frac{3}{8}\pi a^3\end{array}$$ 由对称性知质心在$x$轴上，故$\overline{y}=0$，而$$\begin{array}{}\text{(23)}& \overline{x}=-\frac{1}{V}{\int }_{{\gamma }^{+}}\pi x{y}^2\mathrm{d}x=\frac{8}{3\pi a^3}{\int }_0^{\pi }\pi a^3(1+\cos \theta {)}^3{\sin }^2\theta \cos \theta \cdot a\sin \theta (1+2\cos \theta )\mathrm{d}\theta =\frac{4}{5}a\end{array}$$ 即质心为$(\frac{4}{5}a,0)$。 $◻$

10.1.2 习题7.6

解 参考例11.2.9。 $◻$

解 (1) $$\begin{array}{}\text{(24)}& \begin{array}{rl}\int \frac{x\ln x}{(1+x^2{)}^3}\mathrm{d}x& =-\frac{\ln x}{4(1+x^2{)}^2}+\int \frac{\mathrm{d}x}{4x(1+x^2{)}^2}\\ & =-\frac{\ln x}{4(1+x^2{)}^2}+\frac{1}{8(1+x^2)}+\frac{1}{4}\ln x-\frac{1}{8}\ln (1+x^2)\end{array}\end{array}$$ 从而$$\begin{array}{}\text{(25)}& \begin{array}{rl}\underset{x\to +\mathrm{\infty }}{lim}[-\frac{\ln x}{4(1+x^2{)}^2}+\frac{1}{8(1+x^2)}+\frac{1}{8}\ln \frac{x^2}{1+x^2}]& =0\\ \underset{x\to 0}{lim}[\frac{x^2(2+x^2)\ln x}{4(1+x^2{)}^2}+\frac{1}{8(1+x^2)}-\frac{1}{8}\ln (1+x^2)]& =\frac{1}{8}\end{array}\end{array}$$ 故有$$\begin{array}{}\text{(26)}& {\int }_0^{+\mathrm{\infty }}\frac{x\ln x}{(1+x^2{)}^{3/2}}\mathrm{d}x=0-\frac{1}{8}=-\frac{1}{8}\end{array}$$

(2) $$\begin{array}{}\text{(27)}& \int \frac{\arctan x}{(1+x^2{)}^{3/2}}\mathrm{d}x\stackrel{x=\tan t}{=}\int t\cos t\mathrm{d}t=t\sin t+\cos t=\frac{x\arctan x+1}{\sqrt{1+x^2}}\end{array}$$ 从而$$\begin{array}{}\text{(28)}& {\int }_0^{+\mathrm{\infty }}\frac{\arctan x}{(1+x^2{)}^{3/2}}\mathrm{d}x=\underset{x\to +\mathrm{\infty }}{lim}\frac{x\arctan x+1}{\sqrt{1+x^2}}-1=\frac{\pi }{2}-1\end{array}$$

(3) $$\begin{array}{}\text{(29)}& {\int }_0^1\ln x\mathrm{d}x={[x\ln x-x]}_0^1=-1\end{array}$$ $◻$