10.2 雨课堂作业

解 旋轮线的弧长微元为$$\begin{array}{}\text{(17)}& \mathrm{d}l=\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}\mathrm{d}t=\sqrt{(1-\cos t{)}^2+{\sin }^{2}t}\mathrm{d}t=2\sqrt{\frac{1-\cos t}{2}}\mathrm{d}t=2\sin \frac{t}{2}\mathrm{d}t,t\in [0,2\pi ]\end{array}$$ 与$x$轴所围成的平面区域的面积微元为$$\begin{array}{}\text{(18)}& \mathrm{d}S=y(t)x^{\prime }(t)\mathrm{d}t=(1-\cos t)(1-\cos t)\mathrm{d}t=(1-\cos t{)}^2\mathrm{d}t,t\in [0,2\pi ]\end{array}$$ 曲线的弧长为$$\begin{array}{}\text{(19)}& L={\int }_0^{2\pi }\mathrm{d}l={\int }_0^{2\pi }2\sin \frac{t}{2}\mathrm{d}t=8\end{array}$$ 曲线与$x$轴所围成的平面区域的面积为$$\begin{array}{}\text{(20)}& S={\int }_0^{2\pi }\mathrm{d}S={\int }_0^{2\pi }(1-\cos t{)}^2\mathrm{d}t=3\pi \end{array}$$

(1) 由对称性可得$\bar{x}=\pi$，计算可得$$\begin{array}{}\text{(21)}& {\bar{y}}_L=\frac{{\int }_0^{2\pi }y(t)\mathrm{d}l}{L}=\frac{{\int }_0^{2\pi }(1-\cos t)\cdot 2\sin \frac{t}{2}\mathrm{d}t}{8}=\frac{4}{3}\end{array}$$

(2) 由对称性可得$\bar{x}=\pi$，计算可得$$\begin{array}{}\text{(22)}& {\bar{y}}_S=\frac{{\int }_0^{2\pi }\frac{y(t)}{2}\mathrm{d}S}{S}=\frac{{\int }_0^{2\pi }\frac{1-\cos t}{2}(1-\cos t{)}^2\mathrm{d}t}{3\pi }=\frac{5}{6}\end{array}$$

(3) 由Pappus-Guldinus定理可得体积为$$\begin{array}{}\text{(23)}& V=2\pi {\bar{y}}_S\cdot S=2\pi \cdot \frac{5}{6}\cdot 3\pi =5{\pi }^2\end{array}$$

(4) 由Pappus-Guldinus定理可得面积为$$\begin{array}{}\text{(24)}& A=2\pi {\bar{y}}_L\cdot L=2\pi \cdot \frac{4}{3}\cdot 8=\frac{64\pi }{3}\end{array}$$ $◻$

解 设球壳的面密度为$\sigma$，质点的质量为$m$。不妨将质点放置于$(r,0)$，由对称性可得引力的方向沿$x$轴。对于球壳上一点$(x,y,z)$附近的面积微元$\mathrm{d}S$，其贡献的沿$x$轴方向的引力可表示为$$\begin{array}{}\text{(25)}& \mathrm{d}{F}_x=\frac{Gm\cdot \sigma \mathrm{d}S}{[(x-r{)}^2+y^2+z^2]}\cdot \frac{x-r}{\sqrt{(x-r{)}^2+y^2+z^2}}=\frac{Gm\sigma (x-r)\mathrm{d}S}{[(x-r{)}^2+y^2+z^2{]}^{3/2}}\end{array}$$ 对于$[x,x+\mathrm{d}x]$范围内的球壳${\mathrm{\Sigma }}_x(\mathrm{d}x)$，其上所有点与质点的距离均可视作相同，故可应用上式计算。${\mathrm{\Sigma }}_x(\mathrm{d}x)$与质点的距离可表示为$$\begin{array}{}\text{(26)}& r=\sqrt{(x-r{)}^2+(R^2-x^2)}=\sqrt{R^2-2rx+r^2}\end{array}$$ ${\mathrm{\Sigma }}_x(\mathrm{d}x)$的面积微元可视为由$y=\sqrt{R^2-x^2}$在$[x,x+\mathrm{d}x]$的部分绕$x$轴旋转的部分，其可表示为$$\begin{array}{}\text{(27)}& \mathrm{d}S=2\pi y\mathrm{d}l=2\pi y\sqrt{1+{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)}^2}\mathrm{d}x=2\pi R\mathrm{d}x\end{array}$$ 故${\mathrm{\Sigma }}_x(\mathrm{d}x)$贡献的沿$x$轴方向的引力为$$\begin{array}{}\text{(28)}& \mathrm{d}{F}_x=\frac{Gm\sigma (x-r)\cdot 2\pi R\mathrm{d}x}{(R^2-2rx+r^2{)}^{3/2}}\end{array}$$ 积分可得$$\begin{array}{}\text{(29)}& \begin{array}{rl}{F}_x& ={\int }_{-R}^R\mathrm{d}{F}_x=2\pi RGm\sigma {\int }_{-R}^R\frac{(x-r)\mathrm{d}x}{(R^2-2rx+r^2{)}^{3/2}}\\ & \stackrel{t=\sqrt{R^2-2rx+r^2}}{=}2\pi RGm\sigma {\int }_{R+r}^{R-r}\frac{(\frac{R^2+r^2-t^2}{2r}-r)\frac{-t}{r}\mathrm{d}t}{t^3}\\ & =\frac{\pi RGm\sigma }{r^2}{\int }_{R-r}^{R+r}(\frac{R^2-r^2}{t^2}-1)\mathrm{d}t=\frac{\pi RGm\sigma }{r^2}[-{\frac{R^2-r^2}{t}|}_{R-r}^{R+r}-2r]=0\end{array}\end{array}$$ 即球壳对球内质点不产生引力。 $◻$

解 (1) 设区域$\mathrm{\Omega }$的面积为$S$，则柱体的体积为$Sh$。记$$\begin{array}{}\text{(31)}& S(t):=\{(tx,ty)\mid (x,y)\in \mathrm{\Omega }\}⟹|S(t)|={\oint }_{\partial \mathrm{\Omega }}(tx\mathrm{d}(ty)-ty\mathrm{d}(tx))=t^2{\oint }_{\partial \mathrm{\Omega }}(x\mathrm{d}y-y\mathrm{d}x)=t^{2}S\end{array}$$ 则$z\in h[t,t+\mathrm{d}t]$范围内的锥的体积微元可表示为$\mathrm{d}V(t)=|S(t)|\cdot h\mathrm{d}t$，故锥$V$的体积可表示为$$\begin{array}{}\text{(32)}& |V|={\int }_0^t|S(t)|\cdot h\mathrm{d}t={\int }_0^1{t}^{2}S\cdot h\mathrm{d}t=\frac{1}{3}Sh\end{array}$$

(2) 设$\mathrm{\Omega }$的质心位置为$(\bar{x}(1),\bar{y}(1))$，计算可得$S(t)$的质心位置为$$\begin{array}{}\text{(33)}& \bar{x}(t)=\frac{{\int }_{\mathrm{\Omega }}tx\cdot ty\mathrm{d}(tx)}{{\int }_{\mathrm{\Omega }}ty\mathrm{d}(tx)}=t\frac{{\int }_{\mathrm{\Omega }}x\cdot y\mathrm{d}x}{{\int }_{\mathrm{\Omega }}y\mathrm{d}x}=t{\bar{x}}_0,\bar{y}(t)=t{\bar{y}}_0\end{array}$$ 故锥的质心位置为$$\begin{array}{}\text{(34)}& \begin{array}{rl}\bar{x}& =\frac{{\int }_0^1\bar{x}(t)|S(t)|\cdot h\mathrm{d}t}{|V|}=\frac{{\int }_0^{1}t{\bar{x}}_0\cdot t^{2}S\cdot h\mathrm{d}t}{\frac{1}{3}Sh}=\frac{3}{4}{\bar{x}}_0\\ \bar{y}& =\frac{{\int }_0^1\bar{y}(t)|S(t)|\cdot h\mathrm{d}t}{|V|}=\frac{{\int }_0^{1}t{\bar{y}}_0\cdot t^{2}S\cdot h\mathrm{d}t}{\frac{1}{3}Sh}=\frac{3}{4}{\bar{y}}_0\\ \bar{z}& =\frac{{\int }_0^{1}th\cdot |S(t)|\cdot h\mathrm{d}t}{|V|}=\frac{{\int }_0^{1}th\cdot t^{2}S\cdot h\mathrm{d}t}{\frac{1}{3}Sh}=\frac{3}{4}h\end{array}\end{array}$$ 故锥的质心位置为$(\frac{3}{4}{\bar{x}}_0,\frac{3}{4}{\bar{y}}_0,\frac{3}{4}h)$。 $◻$

解 曲线的参数方程可表为$$\begin{array}{}\text{(35)}& x=\frac{3t}{1+t^3},y=\frac{3t^2}{1+t^3},t\in \mathbb{R}\setminus \{-1\}\end{array}$$

(1) $x,y\ge 0$当且仅当$t\ge 0$，因此$$\begin{array}{}\text{(36)}& A_1=-{\int }_{\gamma }y\mathrm{d}x=-{\int }_0^{+\mathrm{\infty }}\frac{9t^2(1-2t^3)}{(1+t^3{)}^3}\mathrm{d}t=\frac{3}{2}\end{array}$$

(2) 曲线的渐近线为$y=-x-1$，因此$$\begin{array}{}\text{(37)}& \begin{array}{rl}{A}_2& ={\int }_{-\mathrm{\infty }}^{-1}[y(t)+x(t)+1]x^{\prime }(t)\mathrm{d}t+{\int }_{-1}^0[y(t)+x(t)+1]x^{\prime }(t)\mathrm{d}t\\ & ={\int }_{-\mathrm{\infty }}^0\frac{3(1-2t^3)}{(1-t+t^2{)}^3}\mathrm{d}t={\frac{6t^2+3}{2(t^2-t+1)}|}_{-\mathrm{\infty }}^0=\frac{3}{2}\end{array}\end{array}$$ $◻$

解 (1) $$\begin{array}{}\text{(38)}& \begin{array}{rl}\int \frac{x\ln x}{(1+x^2{)}^3}\mathrm{d}x& =-\frac{\ln x}{4(1+x^2{)}^2}+\int \frac{\mathrm{d}x}{4x(1+x^2{)}^2}\\ & =-\frac{\ln x}{4(1+x^2{)}^2}+\frac{1}{8(1+x^2)}+\frac{1}{4}\ln x-\frac{1}{8}\ln (1+x^2)\end{array}\end{array}$$ 从而$$\begin{array}{}\text{(39)}& \begin{array}{rl}\underset{x\to +\mathrm{\infty }}{lim}[-\frac{\ln x}{4(1+x^2{)}^2}+\frac{1}{8(1+x^2)}+\frac{1}{8}\ln \frac{x^2}{1+x^2}]& =0\\ \underset{x\to 0}{lim}[\frac{x^2(2+x^2)\ln x}{4(1+x^2{)}^2}+\frac{1}{8(1+x^2)}-\frac{1}{8}\ln (1+x^2)]& =\frac{1}{8}\end{array}\end{array}$$ 故有$$\begin{array}{}\text{(40)}& {\int }_0^{+\mathrm{\infty }}\frac{x\ln x}{(1+x^2{)}^{3/2}}\mathrm{d}x=0-\frac{1}{8}=-\frac{1}{8}\end{array}$$

(2) $$\begin{array}{}\text{(41)}& \int \frac{\arctan x}{(1+x^2{)}^{3/2}}\mathrm{d}x\stackrel{x=\tan t}{=}\int t\cos t\mathrm{d}t=t\sin t+\cos t=\frac{x\arctan x+1}{\sqrt{1+x^2}}\end{array}$$ 从而$$\begin{array}{}\text{(42)}& {\int }_0^{+\mathrm{\infty }}\frac{\arctan x}{(1+x^2{)}^{3/2}}\mathrm{d}x=\underset{x\to +\mathrm{\infty }}{lim}\frac{x\arctan x+1}{\sqrt{1+x^2}}-1=\frac{\pi }{2}-1\end{array}$$

(3) $$\begin{array}{}\text{(43)}& {\int }_0^1\ln x\mathrm{d}x={[x\ln x-x]}_0^1=-1\end{array}$$ $◻$

解 (1) 记$I_1={\int }_0^1\frac{\arctan x}{x^{\alpha }}\mathrm{d}x$、$I_2={\int }_1^{+\mathrm{\infty }}\frac{\arctan x}{x^{\alpha }}\mathrm{d}x$，则$$\begin{array}{}\text{(44)}& I_1\sim {\int }_0^1\frac{\mathrm{d}x}{x^{\alpha -1}}(x\to 0^{+}),I_2\sim {\int }_1^{+\mathrm{\infty }}\frac{\mathrm{d}x}{x^{\alpha }}(x\to +\mathrm{\infty })\end{array}$$ 原积分收敛当且仅当$I_1,I_2$均收敛，即$\alpha -1<1$且$\alpha >1$，即$\alpha \in (1,2)$。

(2) 记$I_1={\int }_0^1\frac{\ln (1+x)}{x^{\alpha }}\mathrm{d}x$、$I_2={\int }_1^{+\mathrm{\infty }}\frac{\ln (1+x)}{x^{\alpha }}\mathrm{d}x$，则$$\begin{array}{}\text{(45)}& I_1\sim {\int }_0^1\frac{\mathrm{d}x}{x^{\alpha -1}}(x\to 0^{+}),I_2\sim {\int }_1^{+\mathrm{\infty }}\frac{\mathrm{d}x}{x^{\alpha -\epsilon }}(x\to +\mathrm{\infty })\end{array}$$ 原积分收敛当且仅当$I_1,I_2$均收敛，即$\alpha -1<1$且$\alpha -\epsilon >1$，即$\alpha \in (1,2)$。

(3) 记$I_1={\int }_2^3\frac{\mathrm{d}x}{\sqrt{x(x-1)(x-2)}}$、$I_2={\int }_3^{+\mathrm{\infty }}\frac{\mathrm{d}x}{\sqrt{x(x-1)(x-2)}}$，则$$\begin{array}{}\text{(46)}& I_1\sim {\int }_2^3\frac{\sqrt{2}\mathrm{d}x}{\sqrt{x-2}}(x\to 2^{+}),I_2\sim {\int }_3^{+\mathrm{\infty }}\frac{\mathrm{d}x}{x^{3/2}}(x\to +\mathrm{\infty })\end{array}$$ 由于$\frac{1}{2}<1$且$\frac{3}{2}>1$，故原积分收敛。

(4) 记$I_1={\int }_0^1\frac{x\mathrm{d}x}{\sqrt{{\mathrm{e}}^x-1}}$、$I_2={\int }_1^{+\mathrm{\infty }}\frac{x\mathrm{d}x}{\sqrt{{\mathrm{e}}^x-1}}$，则$$\begin{array}{}\text{(47)}& I_1\sim {\int }_0^1\sqrt{x}\mathrm{d}x(x\to 0^{+}),I_2\sim {\int }_1^{+\mathrm{\infty }}x{\mathrm{e}}^{-x/2}\mathrm{d}x(x\to +\mathrm{\infty })\end{array}$$ 由于$-\frac{1}{2}<1$，故原积分收敛。

(5) 记$x={\mathrm{e}}^{-t}$，则$$\begin{array}{}\text{(48)}& {\int }_0^1\frac{\mathrm{d}x}{\ln x}={\int }_{+\mathrm{\infty }}^0\frac{-{\mathrm{e}}^{-t}\mathrm{d}t}{-t}=-{\int }_0^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{-t}}{t}\mathrm{d}t=-{\int }_0^1\frac{{\mathrm{e}}^{-t}}{t}\mathrm{d}t-{\int }_1^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{-t}}{t}\mathrm{d}t=:-I_1-I_2\end{array}$$ 其中$$\begin{array}{}\text{(49)}& I_1\sim {\int }_0^1\frac{\mathrm{d}t}{t}(t\to 0^{+})\end{array}$$ 故原积分发散。

(6) 记$x={\mathrm{e}}^{-t}$，则$$\begin{array}{}\text{(50)}& {\int }_0^1\frac{x^{\beta -1}-x^{\alpha -1}}{\ln x}\mathrm{d}x={\int }_{+\mathrm{\infty }}^0\frac{{\mathrm{e}}^{-(\beta -1)t}-{\mathrm{e}}^{-(\alpha -1)t}}{-t}(-{\mathrm{e}}^{-t}\mathrm{d}t)={\int }_0^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{-\alpha t}-{\mathrm{e}}^{-\beta t}}{t}\mathrm{d}t\end{array}$$ 记$I_1={\int }_0^1\frac{{\mathrm{e}}^{-\alpha t}-{\mathrm{e}}^{-\beta t}}{t}\mathrm{d}t$、$I_2={\int }_1^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{-\alpha t}-{\mathrm{e}}^{-\beta t}}{t}\mathrm{d}t$，则$$\begin{array}{}\text{(51)}& I_1\sim {\int }_0^1(\beta -\alpha )\mathrm{d}t(t\to 0^{+}),I_2\sim {\int }_1^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{-\alpha t}-{\mathrm{e}}^{-\beta t}}{t}\mathrm{d}t(t\to +\mathrm{\infty })\end{array}$$ 故原积分收敛当且仅当$I_2$收敛，即$\alpha >0$且$\beta >0$。

(7) 原积分可表示为$$\begin{array}{}\text{(52)}& I\sim {\int }_0^{\pi /2}\ln x\mathrm{d}x(x\to 0^{+})\end{array}$$ 故原积分收敛。为了计算这个广义积分，利用$$\begin{array}{}\text{(53)}& \begin{array}{rl}I& ={\int }_0^{\pi /2}\ln \cos x\mathrm{d}x=\frac{1}{2}{\int }_0^{\pi /2}\ln (\sin x\cos x)\mathrm{d}x=\frac{1}{2}{\int }_0^{\pi /2}\ln \frac{\sin 2x}{2}\mathrm{d}x\\ & =\frac{1}{2}{\int }_0^{\pi /2}\ln \sin 2x\mathrm{d}x-\frac{\pi }{4}\ln 2\stackrel{t=2x}{=}\frac{1}{4}{\int }_0^{\pi }\ln \sin t\mathrm{d}t-\frac{\pi }{4}\ln 2=\frac{1}{2}I-\frac{\pi }{4}\ln 2\end{array}\end{array}$$ 故$I=-\frac{\pi }{2}\ln 2$。 $◻$