定积分的应用

单变量函数的积分学

张瑞
中国科学技术大学数学科学学院

定积分的应用

实际问题的解决，往往需要通过数学建模过程，建立它的数学模型，然后再来求解。

微元分析法是用定积分建立数学模型的重要方法。它需要表达的量$Q$满足两个特点：

  1. $Q$依赖于某个区间$[a,b]$上的变量$x$
  2. $Q$在区间$[a,b]$上具有可加性。即若将区间$[a,b]$分成若干个小区间后， 所求的量$Q$是对应的各个小区间上的局部量$\Delta Q$之和。

对于一个具体问题

  1. 首先选定积分变量，并弄清楚它的变化范围(积分区间)
  2. 取小区间$[x,x+\Delta x]$上的局部量$\Delta Q$，并表达成
    \[\begin{aligned} \Delta Q=&f(x)\Delta x+o(\Delta x) \\ %dQ=&f(x)dx \end{aligned} \]
    进而可以得到未知量$Q$的微分$dQ=f(x)dx$
  3. 最后得到$\displaystyle Q=\int_a^bdQ=\int_a^bf(x)dx$

问题的关键是寻找到正确的微元表达式$dQ=f(x)dx$， 也就是验证$f(x)\Delta x$与$\Delta Q$之间是否相关一个高阶无穷小量$o(\Delta x)$。 因此，这种分析方法叫作微元分析法

平面图形的面积

直角坐标系下

$y=f(x)$在$[a,b]$上连续，则 曲线$y=f(x)$，$x$轴，及直线$x=a$和$x=b$所围成的曲边梯形的面积为

$ f(x)\geq 0$时

\begin{tikzpicture}[x=2cm, y=2cm, global scale=1] % 上面，用 x=2cm, y=2cm 来设置x,y方向的单位长度，缺省是1cm %\fill[fill=yellow!80!black] (0.8,0)--(0.8,0.7225)--(0.9,0.7225)--(0.9,0); \fill [blue!50!white, domain=-1:1, variable=\x] (-1, 0) -- plot ({\x}, {0.5*\x*\x+0.5}) -- (1, 0) -- cycle; \draw[domain=-1:1, color=blue, thick] plot (\x,{0.5*(\x)^2+0.5}); % node[right] {$x^2$}; %\draw[thick] (1,0) node[below] {$1$}--(1,1); \draw[->] (-1.1,0) -- (1.2,0) node[right] {$x$}; \draw[->] (0,-0.1) -- (0,1.1) node[right] {$y$}; \end{tikzpicture}
\[A=\int_a^b f(x)dx \]

$f(x)\leq 0$时，

\begin{tikzpicture}[x=2cm, y=2cm, global scale=1] % 上面，用 x=2cm, y=2cm 来设置x,y方向的单位长度，缺省是1cm %\fill[fill=yellow!80!black] (0.8,0)--(0.8,0.7225)--(0.9,0.7225)--(0.9,0); \fill [blue!50!white, domain=-1:1, variable=\x] (-1, 0) -- plot ({\x}, {0.5*\x*\x-1}) -- (1, 0) -- cycle; \draw[domain=-1:1, color=blue, thick] plot (\x,{0.5*(\x)^2-1}); % node[right] {$x^2$}; %\draw[thick] (1,0) node[below] {$1$}--(1,1); \draw[->] (-1.1,0) -- (1.2,0) node[right] {$x$}; \draw[->] (0,-1.1) -- (0,0.1) node[right] {$y$}; \end{tikzpicture}
\[A=-\int_a^b f(x) dx= \int_a^b |f(x)| dx \]

若不管$f(x)$的符号，则有面积

\[A=\int_a^b |f(x)|dx \]
\begin{tikzpicture}[x=2cm, y=2cm, global scale=1] % 上面，用 x=2cm, y=2cm 来设置x,y方向的单位长度，缺省是1cm %\fill[fill=yellow!80!black] (0.8,0)--(0.8,0.7225)--(0.9,0.7225)--(0.9,0); \fill [blue!50!white, domain=-1:1, variable=\x] (-1, 0) -- plot ({\x}, {\x*\x-0.5}) -- (1, 0) -- cycle; \draw[domain=-1:1, color=blue, thick] plot (\x,{(\x)^2-0.5}); % node[right] {$x^2$}; \fill[domain=-0.707:0.707, color=red!80!white, thick] plot (\x,{-(\x)^2+0.5}); % node[right] {$x^2$}; %\draw[thick] (1,0) node[below] {$1$}--(1,1); \draw[->] (-1.1,0) -- (1.2,0) node[right] {$x$}; \draw[->] (0,-0.6) -- (0,0.6) node[right] {$y$}; \end{tikzpicture}

若图形由两个连续函数$y=f_1(x)$, $y=f_2(x)$及直线$x=a$, $x=b$围成，则它的面积为

\[A=\int_a^b |f_1(x)-f_2(x)|dx \]

例 1. 求函数$f(x)=\frac1x$, $f(x)=x$及$x=2$所围成的面积

例 2. [例4.6.1] 求由曲线$y=x^2$, $y=\frac{x^2}4$, $y=1$所围成的平面图形的面积

极坐标系下的图形面积

% 极坐标下的面积 \begin{tikzpicture}[scale=1.0,samples=200, >=latex] \draw[thick,->] (-0.5,0) -- (3.5,0); \draw (0,0) node[below left] {O};% 原点 \coordinate (A) at (0.5,1.5); \coordinate (B) at (3,1.0); \draw[name path=fx, thick] (A) .. controls (1.5,0.5) and (2.4,1.8) .. (B); \draw[thin] (0,0)--(A); \draw[thin] (0,0)--(B); \path[name path=theta] (0,0)--(30:4); \path[name path=theta-d] (0,0)--(40:4); \path[name intersections={of=fx and theta}] (intersection-1) coordinate (a); \path[name intersections={of=fx and theta-d}] (intersection-1) coordinate (ia); \filldraw[fill=yellow!80!black, dashed, thin,fill opacity=0.5] (0,0)--(a) [rotate=30 ]arc [start angle=0, delta angle=30, radius=1] -- (0,0); \path (1,0) ++(15:.75) node {$\alpha$}; \draw[thin] (2,0) arc [start angle=0, delta angle=38, radius=1]; \end{tikzpicture}

曲线的极坐标方程为

\[r=f(\theta), \theta\in[\alpha,\beta] \]

任取长度为$d\theta$的区间$[\theta,\theta+d\theta]$

在小区间上用圆弧代替曲线弧，得到面积微元

\[dS=\frac{1}2r^2d\theta=\frac{1}2f^2(\theta)d\theta \]

这样，可以得到面积为

\[S=\int_{\alpha}^{\beta}\frac{1}2f^2(\theta)d\theta \]

例 3. (例4.6.2) 求 双纽线 $r^2=a^2\cos(2\theta) (a>0)$ 围成的面积

\begin{tikzpicture}[x=2cm, y=2cm, global scale=1, declare function={% rhoone(\a,\t)=sqrt(\a^2*cos(2*\t));}, ] % 上面，用 x=2cm, y=2cm 来设置x,y方向的单位长度，缺省是1cm %\fill[fill=yellow!80!black] (0.8,0)--(0.8,0.7225)--(0.9,0.7225)--(0.9,0); \draw[domain=-45:45, color=blue, thick, smooth]%, samples=101] plot (\x:{rhoone(1.0,\x)}); \draw[domain=135:225, color=blue, thick, smooth]%, samples=101] plot (\x:{rhoone(1.0,\x)}); %\draw[thick] (1,0) node[below] {$1$}--(1,1); \draw[->] (-1.1,0) -- (1.2,0) node[right] {$x$}; \draw[->] (0,-0.5) -- (0,0.5) node[right] {$y$}; \end{tikzpicture}

解. 由对称性，

\[\begin{aligned} A =&4\int_0^{\frac{\pi}4}\dfrac12r^2d\theta =2\int_0^{\frac{\pi}4}a^2\cos(2\theta)d\theta \\ =&a^2\sin(2\theta)\Big|_0^{\frac{\pi}4} =a^2 \end{aligned} \]

1.

\[\begin{aligned} A =&4\int_0^{\frac{\pi}4}\dfrac12r^2d\theta =2\int_0^{\frac{\pi}4}a^2\cos(2\theta)d\theta \\ =&a^2\sin(2\theta)\Big|_0^{\frac{\pi}4} =a^2 \end{aligned} \]

参数方程

设有参数方程， $\begin{cases}x=\phi(t) \\y=\psi(t)\end{cases} , t\in[t_0,T]$。 从极坐标入手，计算它的面积。

先算$d\theta$。由

\[\theta=\begin{cases} \arctan(\dfrac{y}{x}), & x>0, y>0 \\ \dfrac{\pi}2-\arctan(\dfrac{x}{y}), & x<0, y>0 \\ \pi+\arctan(\dfrac{y}{x}), & x<0,y<0 \\ \dfrac{3\pi}2-\arctan(\dfrac{x}{y}), & x>0, y<0 \\ \end{cases} \]
\[\dfrac{d\theta}{dt}=\dfrac{1}{1+(\frac{y}{x})^2}\dfrac{d}{dt}(\dfrac{y}{x}) =\dfrac{x^2}{x^2+y^2}\dfrac{y'_tx-yx'_t}{x^2} \]

或

\[\dfrac{d\theta}{dt}=-\dfrac{1}{1+(\frac{x}{y})^2}\dfrac{d}{dt}(\dfrac{x}{y}) =-\dfrac{y^2}{x^2+y^2}\dfrac{yx'_t-y'_tx}{y^2} \]

即有

\[d\theta = \frac{y'_t x-yx'_t}{x^2+y^2}dt \]

又$r^2=x^2+y^2$，所以有极坐标下的面积微元

\[\begin{aligned} dA=& \frac{1}2r^2d\theta=\frac{1}2(y'_tx-x'_ty)dt \\ %A=&\displaystyle\frac{1}2\int_{t_0}^T(\psi'(t)\phi(t)-\phi'(t)\psi(t))dt \end{aligned} \]

参数方程 $\begin{cases}x=\phi(t) \\y=\psi(t)\end{cases} , t\in[t_0,T]$ 的面积公式为

\[A=\displaystyle\frac{1}2\int_{t_0}^T(\psi'(t)\phi(t)-\phi'(t)\psi(t))dt \]

例 4. (例4.3.3) 求椭圆 $\begin{cases}x=a\cos t \\y=b\sin t\end{cases}, t\in[0,2\pi]$ 围成的面积

2.

\[\begin{aligned} A=&\dfrac12\int_0^{2\pi}(b\cos t\cdot a\cos t-(-a\sin t)b\sin t)dt \\ =&\dfrac12\int_0^{2\pi}ab(\cos^2t+\sin^2t)dt \\ =&\dfrac122\pi ab=\pi ab \end{aligned} \]

隐函数的面积

例 5. 求$x^2+xy+y^2=1$围成的面积

解. 分段给出表达

\[\begin{aligned} y_{1,2}(x) =&\dfrac{-x\pm\sqrt{x^2-4(x^2-1)}}2 \\ =&-\dfrac{x}2\pm\sqrt{1-\dfrac34x^2}, x\in[-\dfrac2{\sqrt 3},\dfrac2{\sqrt 3}] \end{aligned} \]

解. 或者，用极坐标 $\begin{cases}x=r\cos\theta, \\y=r\sin\theta,\end{cases}$

3.（ 隐式方程，分段给出表达）

\[\begin{aligned} y_{1,2}(x) =&\dfrac{-x\pm\sqrt{x^2-4(x^2-1)}}2 \\ =&-\dfrac{x}2\pm\sqrt{1-\dfrac34x^2}, x\in[-\dfrac2{\sqrt 3},\dfrac2{\sqrt 3}] \end{aligned} \]

则有

\[\begin{aligned} S=&\int_{-\frac2{\sqrt 3}}^{\frac2{\sqrt 3}}(y_1(x)-y_2(x))dx \\ =&\int_{-\frac2{\sqrt 3}}^{\frac2{\sqrt 3}}2\sqrt{1-\dfrac34x^2}dx \\ \end{aligned} \]
\[\begin{aligned} S=&\int_{-\frac2{\sqrt 3}}^{\frac2{\sqrt 3}}2\sqrt{1-\dfrac34x^2}dx \\ (x=\dfrac2{\sqrt 3}\sin t)=&\dfrac4{\sqrt 3}\int_{-\frac{\pi}2}^{\frac{\pi}2}\cos^2tdt \\ =&\dfrac{2\pi}{\sqrt 3} \end{aligned} \]

或者，用极坐标 $\begin{cases}x=r\cos\theta, \\y=r\sin\theta,\end{cases}$ 则有

\[\begin{aligned} &r^2\cos^2\theta+r^2\cos\theta\sin\theta+r^2\sin^2\theta=1 \\ &r^2=\dfrac1{1+\sin\theta\cos\theta} \end{aligned} \]

得到

\[\begin{aligned} S =&\dfrac12\int_0^{2\pi}r^2d\theta =\dfrac12\int_0^{2\pi}\dfrac1{1+\sin\theta\cos\theta}d\theta \\ =&\dfrac12\int_0^{2\pi}\dfrac{1}{2+\sin(2\theta)}d(2\theta) =\dfrac{2\pi}{\sqrt 3} \end{aligned} \]

平面曲线的弧长

\usetikzlibrary{arrows} \usetikzlibrary{intersections, calc} % 曲线的长度 \begin{tikzpicture}[scale=0.8,samples=200, >=latex,thick] \draw[thick,->] (-0.5,0) -- (5,0) node[right] {$x$};% x軸 \draw[thick,->] (0,-0.1) -- (0,3) node[below right] {$y$};% y軸 \draw (0,0) node[below left] {O};% 原点 \coordinate (A) at (0.5,1.0); \coordinate (B) at (4,1.5); \draw[name path=fx, thick] (A) .. controls (2,0) and (3,2.0) .. (B); \draw (A) node[below] {$A$}; \draw (B) node[above right] {$B$}; { \def\x{1.0}; \def\pointname{P1}; \path[name path=main] (\x,0) -- (\x,3.0); \path[name intersections={of=fx and main}] (intersection-1) coordinate (\pointname); \draw (\pointname) node[below] {$M_1$}; } { \def\x{2.0}; \def\pointname{P2}; \path[name path=main] (\x,0) -- (\x,3.0); \path[name intersections={of=fx and main}] (intersection-1) coordinate (\pointname); } { \def\x{2.5}; \def\pointname{P3}; \path[name path=main] (\x,0) -- (\x,3.0); \path[name intersections={of=fx and main}] (intersection-1) coordinate (\pointname); \draw (\pointname) node[above left] {$M_i$}; } { \def\x{3.5}; \def\pointname{P4}; \path[name path=main] (\x,0) -- (\x,3.0); \path[name intersections={of=fx and main}] (intersection-1) coordinate (\pointname); %\draw[dotted] (\x,0)-- (\pointname); \draw (\pointname) node[above] {$M_{i+1}$}; } \draw[dashed, red] (A) \foreach \i in {1,2,3,4} {--(P\i)} -- (B); \end{tikzpicture}

在曲线$\overset{\huge\frown}{AB}$上任取分点

\[A=M_0,M_1,M_2,\cdots,M_n=B \]

连接这些分点得到曲线$AB$的一条内折线。 记$|M_{i-1}M_i|$为弦的长度。 记$\lambda=\displaystyle\max_{1\leq i\leq n}|M_{i-1}M_i|$

定义 1.
若$\lambda\to 0$时，折线的长度的极限存在，则这个极限就是曲线的长度， 或称为曲线$\overset{\huge\frown}{AB}$的弧长。此时，称曲线为可求长曲线。

弧长微元

在曲线上取两点$M$和$M'$，其横坐标分别为$x$与$x+\Delta x$， 则两点的距离为

\[\begin{aligned} |MM'|=&\sqrt{(\Delta x)^2+(f(x+\Delta x)-f(x))^2} \\ =&\sqrt{(\Delta x)^2+(f'(x)\Delta x+o(\Delta x))^2} \\ =&\sqrt{(\Delta x)^2+(f'(x)\Delta x)^2+o((\Delta x)^2)} \\ \end{aligned} \]
% 曲线长度的微元 \begin{tikzpicture}[scale=0.8,samples=200, >=latex,thick] \draw[thick,->] (-0.5,0) -- (5,0) coordinate (x axis) node[right] {$x$};% x軸 \draw[thick,->] (0,-0.1) -- (0,3) node[below right] {$y$};% y軸 \draw (0,0) node[below left] {O};% 原点 \coordinate (A) at (0.5,0.5); \coordinate (B) at (4,2.5); \draw[name path=fx, thick] (A) .. controls (2,2.0) .. (B); \def\xx{1.5}; \def\xdx{2.60}; \path[name path=dx] (\xdx,0) -- (\xdx,3); \path[name path=main] (\xx,0) -- (\xx,3); \path[name path=sub] ($(\xx,0)+(1pt,0)$) -- ($(\xx,3)+(1pt,0)$); \def\tangentlength{3cm}; \path[name intersections={of=fx and main}]; \coordinate (a) at (intersection-1); \path[name intersections={of=fx and sub}]; \coordinate (b) at (intersection-1); %\coordinate (c) at ($(a)!\tangentlength/2!(b)$); \path[draw, name path=tangent, color=blue] (a) -- ($(a)!\tangentlength/2!(b)$); \path[name intersections={of=tangent and dx}] (intersection-1) coordinate (c); \path[name intersections={of=fx and dx}] (intersection-1) coordinate (d); %\draw[dashed, blue] (tangent); \draw (a) node[above left] {$M$}; \draw (d) node[below right] {$M'$}; \draw[dashed] (a) -- (\xx,0); \draw[dotted, red] (a) -- (d); \draw (\xx,0) node[below] {$x$}; %\draw[red,very thick] (c) -- +(0,-3); \draw[dashed] (\xdx,0) -- (c); \draw (\xdx,0) node[below] {$x+\Delta x$}; \draw[dashed] (a) -| (c); \end{tikzpicture}

可以得到弧长的微元为

\[ds=\sqrt{1+(f'(x))^2}dx \]

或

\[ds=\sqrt{(dx)^2+(dy)^2} \]

当$f'(x)$连续时（此时，称$f(x)$连续可微），可得弧长为

\[s=\int_a^b\sqrt{1+(f'(x))^2}dx \]

例 6. $9y^2=x(x-3)^2, y\geq0$，在$x=0$到$x=3$之间的长度

解. 弧长为

\[s=\int_0^3 \sqrt{1+(y'(x))^2}dx \]

4.

\[18yy'=(x-3)^2+x2(x-3)=3(x-3)(x-1) \]

则有

\[y'=\dfrac1{6y}(x-3)(x-1) \]
\[\begin{aligned} \sqrt{1+(y')^2} =&\sqrt{1+\left(\dfrac{(x-3)(x-1)}{6y}\right)^2} \\ =&\dfrac{1}{6y}(x-3)(x+1) =\dfrac{(x-3)(x+1)}{2\sqrt{x}(x-3)} \end{aligned} \]
\[s=\int_0^3\dfrac{x+1}{\sqrt x}dx=\int_0^3(x+1)d(\sqrt x)=2\sqrt 3 \]

参数方程

设曲线$\overset{\huge\frown}{AB}$的参数方程为

\[\begin{cases} x=\phi(t) \\ y=\psi(t) \end{cases}, t\in[\alpha,\beta] \]

则弧长微元

\[\begin{aligned} ds=&\sqrt{(\phi'(t)dt)^2+(\psi'(t)dt)^2} \\ =&\sqrt{(\phi'(t))^2+(\psi'(t))^2}dt \end{aligned} \]

弧长为 $s=\displaystyle\int_{\alpha}^{\beta}\sqrt{(\phi'(t))^2+(\psi'(t))^2}dt$

极坐标

设曲线$\overset{\huge\frown}{AB}$的极坐标方程$r=f(\theta)$， 则有

\[\begin{cases} x=f(\theta)\cos\theta \\ y=f(\theta)\sin\theta \end{cases} \Rightarrow \begin{cases} x'(\theta)=f'(\theta)\cos\theta-f(\theta)\sin(\theta) \\ y'(\theta)=f'(\theta)\sin\theta+f(\theta)\cos(\theta) \end{cases} \]

得到弧长微元

\[\begin{aligned} ds=&\sqrt{(x'(\theta))^2+(y'(\theta))^2}d\theta \\ =&\sqrt{(f'(\theta))^2+(f(\theta))^2}d\theta \end{aligned} \]

弧长为 $S=\displaystyle\int_{\alpha}^{\beta}ds=\displaystyle\int_{\alpha}^{\beta}\sqrt{(f'(\theta))^2+(f(\theta))^2}d\theta$

例 7. 求弧长 $\begin{cases}x=a\cos^3t \\y=b\sin^3t\end{cases}, t\in[0,2\pi]$ (星形线)

\begin{tikzpicture}[x=2cm, y=2cm, global scale=0.7, declare function={% rhoone(\a,\t)=sqrt(\a^2*cos(2*\t));}, ] % 上面，用 x=2cm, y=2cm 来设置x,y方向的单位长度，缺省是1cm %\fill[fill=yellow!80!black] (0.8,0)--(0.8,0.7225)--(0.9,0.7225)--(0.9,0); \draw[domain=0:360, color=blue, thick, smooth]%, samples=101] plot ({2*cos(\x)^3},{sin(\x)^3}); %\draw[domain=135:225, color=blue, thick, smooth]%, samples=101] % plot (\x:{rhoone(1.0,\x)}); %\draw[thick] (1,0) node[below] {$1$}--(1,1); \draw[->] (-2.1,0) -- (2.1,0) node[right] {$x$}; \draw[->] (0,-1.1) -- (0,1.1) node[right] {$y$}; \end{tikzpicture}

5.

\[ds=\sqrt{(x'(t))^2+(y'(t))^2}dt \]
\[\begin{aligned} s=&4\int_0^{\frac{\pi}2}\sqrt{(3a\cos^2t\sin t)^2+(3b\sin^2t\cos t)^2}dt \\ =&4\int_0^{\frac{\pi}2}3\sin t\cos t\sqrt{(a\cos t)^2+(b\sin t)^2}dt \\ =&4\int_0^{\frac{\pi}2}\dfrac32\sqrt{(a\cos t)^2+(b\sin t)^2}d(\sin^2t) \\ =&6\int_0^{1}\sqrt{a^2+(b^2-a^2)t}dt =\dfrac{4(b^2+ab+a^2)}{b+a} \end{aligned} \]

旋转体的体积

% 曲面的体积 \usetikzlibrary{intersections} \begin{tikzpicture}[scale=0.8,samples=200, >=latex,thick] \draw[thick,->] (-0.5,0) -- (5,0) node[right] {$x$};% x軸 %\draw[thick,->] (0,-1) -- (0,2) node[below right] {$y$};% y軸 \draw (0,0) node[below left] {O};% 原点 \tikzset{ volume/.pic={ \coordinate (A) at (0.5,1.0); \coordinate (B) at (4,1.1); \coordinate (Ai) at (0.5,-1.0); \coordinate (Bi) at (4,-1.1); \draw[name path=fx, thick] (A) .. controls (2,0) and (3,1.5) .. (B); \draw[name path=ifx, thick] (Ai) .. controls (2,0) and (3,-1.5) .. (Bi); \draw[dashed] (A)--(Ai); \draw[dashed] (B)--(Bi); % \draw[dashed] (Ai) to[out=60, in=-60] (A); \draw (Ai) to[out=120, in=-120] (A); \draw (Bi) to[out=60, in=-60] (B); \draw (Bi) to[out=120, in=-120] (B); \foreach \x in {2.0, 3.0} { \path[name path=main] (\x,-3.0) -- (\x,3.0); \path[name intersections={of=fx and main}] (intersection-1) coordinate (a); \path[name intersections={of=ifx and main}] (intersection-1) coordinate (ia); \draw[dashed] (ia) to[out=60, in=-60] (a); \draw (ia) to[out=120, in=-120] (a); } } } \draw (0,1.5) pic[scale=0.8] {volume}; \draw (0.5,0) node[below] {$a$}; \draw (4,0) node[below] {$b$}; \draw[dashed] (4,0)--(4,1.5); \draw[dashed] (0.5,0)--(0.5,1.5); \draw (2,0) node[below] {$x$}; \draw (3,0) node[below] {$x+\Delta x$}; \draw[dashed] (3,0) -- (3,1.5); \draw[dashed] (2,0)--(2,1.5); \end{tikzpicture}

若过点$x$且垂直与$x$轴的平面截得的截面面积为$S(x)$。 则任意长度为$dx$的小区间$[x,x+\Delta x]$上的立体可以近似看作一个小圆柱， 因此体积微元为

\[dV=S(x)dx \]

则体积为

\[V=\int_a^b S(x)dx \]

例 8. (例4.6.7) 一个锥体，底面积为$Q$，高为$h$。求该锥体的体积。

\begin{tikzpicture}[x=2cm, y=2cm, global scale=0.5, >=latex,thick] \draw[thick,->] (-0.5,0) -- (5,0) node[right] {$x$};% x軸 \draw[thick,->] (0,-1) -- (0,2) node[above] {$y$};% y軸 \draw (0,0) node[below left] {O};% 原点 \tikzset{ volume/.pic={ \coordinate (A) at (0,0.0); \coordinate (B) at (4,1.0); \coordinate (Ai) at (0,0.0); \coordinate (Bi) at (4,-1.0); \draw[name path=fx, thick] (A) -- (B); \draw[name path=ifx, thick] (Ai) -- (Bi); \draw[dashed] (A)--(Ai); \draw[dashed] (B)--(Bi); % \draw[dashed] (Ai) to[out=60, in=-60] (A); \draw (Ai) to[out=120, in=-120] (A); \draw (Bi) to[out=60, in=-60] (B); \draw (Bi) to[out=120, in=-120] (B); \foreach \x in {2.0} { \path[name path=main] (\x,-3.0) -- (\x,3.0); \path[name intersections={of=fx and main}] (intersection-1) coordinate (a); \path[name intersections={of=ifx and main}] (intersection-1) coordinate (ia); \draw[dashed] (ia) to[out=60, in=-60] (a); \draw (ia) to[out=120, in=-120] (a); \fill[gray] (ia) to[out=60, in=-60] (a) to [out=-120, in=120] (ia); %\draw (ia) to[out=120, in=-120] (a); } } } \draw (0,1.0) pic {volume}; %\draw (0.5,0) node[below] {$a$}; \draw (4,0) node[below] {$h$}; \draw (4,2) node[above] {$Q$}; \draw[dashed] (4,0)--(4,1.5); %\draw[dashed] (0.5,0)--(0.5,1.5); \draw (2,0) node[below] {$x$}; \draw (2,1.5) node[above] {$S(x)$}; %\draw (3,0) node[below] {$x+\Delta x$}; %\draw[dashed] (3,0) -- (3,1.5); \draw[dashed] (2,0)--(2,1.5); \end{tikzpicture}

解. 注意到面积关系

\[\dfrac{S(x)}Q=\left(\dfrac{x}{h}\right)^2 \]

则

\[V=\int_0^h\dfrac{x^2}{h^2}Qdx =\dfrac{Q}{h^2}\dfrac{x^3}3\bigg|_0^h =\dfrac{Qh}3 \]
% 旋转体的体积 \begin{tikzpicture}[scale=0.7,samples=200, >=latex,thick] \draw[thick,->] (-0.5,0) -- (5,0) node[right] {$x$};% x軸 \draw[thick,->] (0,-1) -- (0,2) node[below right] {$y$};% y軸 \draw (0,0) node[below left] {O};% 原点 \coordinate (A) at (0.5,1.0); \coordinate (iA) at (0.5,-1.0); \coordinate (B) at (4,1.5); \coordinate (iB) at (4,-1.5); \draw[name path=fx, thick] (A) .. controls (2,0) and (3,2.0) .. (B); \draw[name path=ifx, thick] (0.5,-1.0) .. controls (2,0) and (3,-2.0) .. (4,-1.5); \draw (0.5,0) node[below] {$a$}; \draw (4.0,0) node[below] {$b$}; \draw[dashed] (iA) to[out=60, in=-60] (A); \draw (iA) to[out=120, in=-120] (A); \draw[dashed] (A) -- (0.5,0); \draw (iB) to[out=60, in=-60] (B); \draw (iB) to[out=120, in=-120] (B); \draw[dashed] (B) -- (4,0); \def\x{2.0}; \path[name path=main] (\x,-3.0) -- (\x,3.0); \path[name intersections={of=fx and main}] (intersection-1) coordinate (a); \path[name intersections={of=ifx and main}] (intersection-1) coordinate (ia); \draw[red,dashed] (ia) to[out=60, in=-60] (a); \draw[red] (ia) to[out=120, in=-120] (a); \draw[red,dashed] (a) -- (\x,0); \draw (\x,0) node[below] {$x$}; \draw (a) node[above] {$f(x)$}; \end{tikzpicture}

将函数$y=f(x)$， $x=a$, $x=b$以及$x$轴围成的曲边梯形，绕$x$轴旋转一周， 得到了一个旋转体。

点$x$处的截面的面积为

\[S(x)=\pi y^2=\pi f^2(x) \]

因此，旋转体的体积为

\[V=\pi\int_a^b f^2(x)dx \]
  • 对$a>0$，将函数$y=f(x)$， $x=a$, $x=b$以及$x$轴围成的曲边梯形，绕$y$轴旋转一周。

    对于长度为$\Delta x$的区间$[x,x+\Delta x]$对应的小曲边梯形，绕$y$轴旋转一周， 得到的旋转体的体积可以近似地看作高为$f(x)$， 底面半径为$x$和$x+\Delta x$的两圆柱的体积差

    \[f(x)\pi((x+\Delta x)^2-x^2)=2\pi x f(x)\Delta x+\pi f(x)(\Delta x)^2 \]

    因此，体积元为$dV=2\pi xf(x)dx$。得到体积为

    \[V=2\pi\int_a^b xf(x)dx \]

例 9. (例4.6.8) 椭圆$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$，绕$x$轴一周，得到椭球体积

7.

\[\begin{aligned} V=&\pi\int_{-a}^ay^2dx=\pi\int_{-a}^ab^2(1-\dfrac{x^2}{a^2})dx \\ =&\pi 2ab^2-\pi b^2\dfrac{x^3}{3a^2}\bigg|_{-a}^a =\dfrac43\pi ab^3 \end{aligned} \]

旋转体的侧面积

% 旋转体的表面积 \begin{tikzpicture}[scale=0.8,samples=200, >=latex,thick] \draw[thick,->] (-0.5,0) -- (5,0) node[right] {$x$};% x軸 \draw[thick,->] (0,-1) -- (0,2) node[below right] {$y$};% y軸 \draw (0,0) node[below left] {O};% 原点 \coordinate (A) at (0.5,1.0); \coordinate (iA) at (0.5,-1.0); \coordinate (B) at (4,1.5); \coordinate (iB) at (4,-1.5); \draw[name path=fx, thick] (A) .. controls (2,0) and (3,2.0) .. (B); \draw[name path=ifx, thick] (0.5,-1.0) .. controls (2,0) and (3,-2.0) .. (4,-1.5); \draw (0.5,0) node[below] {$a$}; \draw (4.0,0) node[below] {$b$}; \draw[dashed] (iA) to[out=60, in=-60] (A); \draw (iA) to[out=120, in=-120] (A); \draw[dashed] (A) -- (0.5,0); \draw (iB) to[out=60, in=-60] (B); \draw (iB) to[out=120, in=-120] (B); \draw[dashed] (B) -- (4,0); \def\x{1.5}; \path[name path=main] (\x,-3.0) -- (\x,3.0); \path[name intersections={of=fx and main}] (intersection-1) coordinate (a); \path[name intersections={of=ifx and main}] (intersection-1) coordinate (ia); \draw[dashed, blue] (ia) to[out=60, in=-60] (a); \draw[blue] (ia) to[out=120, in=-120] (a); \draw[dashed,blue] (a) -- (\x,0); \draw[blue] (\x,0) node[below, blue] {$x$}; \draw (a) node[above] {$M$}; \def\x{2.9}; \path[name path=main] (\x,-3.0) -- (\x,3.0); \path[name intersections={of=fx and main}] (intersection-1) coordinate (a); \path[name intersections={of=ifx and main}] (intersection-1) coordinate (ia); \draw[dashed, blue] (ia) to[out=60, in=-60] (a); \draw[blue] (ia) to[out=120, in=-120] (a); \draw[dashed,blue] (a) -- (\x,0); \draw[blue] (\x,0) node[below, blue] {$x+\Delta x$}; \draw (a) node[above] {$M'$}; \end{tikzpicture}

小段区间$[x,x+\Delta x]$对应的弧$\widehat{MM'}$ 所得到的旋转体的侧面积为

\[\Delta S=\pi(f(x)+f(x+\Delta x))\Delta s \]

$\Delta s$为$\widehat{MM'}$的弧长。

\[\begin{aligned} &\Delta S=\pi(f(x)+f(x+\Delta x))\Delta s \\ =&\pi(f(x)+f(x)+f'(x)\Delta x+o(\Delta x))(\sqrt{1+(f'(x))^2}\Delta x+o(\Delta x)) \end{aligned} \]

略去$\Delta x$的高阶无穷小量，

旋转体侧面积微元为

\[dS=2\pi f(x)\sqrt{1+(f'(x))^2}dx \]

因此，旋转体的侧面积为

\[S=2\pi\int_a^b f(x)\sqrt{1+(f'(x))^2}dx \]

参数曲线

若曲线的参数方程为 $\begin{cases}x=\phi(t) \\ y=\psi(t)\end{cases}, t\in[\alpha,\beta]$

则弧长元

\[ds=\sqrt{(\phi'(t))^2+(\psi'(t))^2}dt \]

侧面积元

\[dS=2\pi \psi(t) \sqrt{(\phi'(t))^2+(\psi'(t))^2}dt \]

侧面积

\[S=2\pi\int_{\alpha}^{\beta} \psi(t) \sqrt{(\phi'(t))^2+(\psi'(t))^2}dt \]

极坐标方程

若曲线的极坐标方程为

\[r=r(\theta), \theta\in[\alpha,\beta] \]

以$\theta$为参数，则弧长微元为

\[ds=\sqrt{r^2(\theta)+r'^2(\theta)}d\theta \]

可得侧面积为

\[S=2\pi\int_{\alpha}^{\beta} r(\theta)\sin(\theta)\sqrt{r^2(\theta)+r'^2(\theta)}d\theta \]

例 10. (例4.6.9) 半径为$R$的球面的面积

解. 将球面看成是曲线的旋转面，

\[y=\sqrt{R^2-x^2}, x\in[-R, R] \]

解. 将球面写成极坐标曲线的旋转面

\[\rho = R, \theta\in[0,\pi] \]

解. 将球面写成参数曲线的旋转面

\[\begin{cases} x=R \cos\theta \\ y=R \sin\theta \end{cases}, \theta\in[0,\pi] \]

8.

\[y=\sqrt{R^2-x^2}, x\in[-R,R] \]

则

\[y'=\dfrac{-x}{\sqrt{R^2-x^2}}=\dfrac{-x}y , |x|\leq R \]
\[y\sqrt{1+(y')^2}=y\sqrt{1+\dfrac{x^2}{y^2}}=\sqrt{x^2+y^2}=R \]

因此，侧面积

\[P=\int_{-R}^R 2\pi y\sqrt{1+(y')^2}dx=4\pi R^2 \]

则

\[S=2\pi\int_0^{\pi} R\sin\theta \sqrt{R^2} d\theta =2\pi R^2\int_0^{\pi}\sin\theta d\theta \]

则

\[\begin{aligned} S=&2\pi\int_0^{\pi} R\sin\theta \sqrt{R^2\sin^2\theta+R^2\cos^2\theta} d\theta \\ =&2\pi R^2\int_0^{\pi}\sin\theta d\theta =4\pi R^2 \end{aligned} \]

例 11. $y=\sqrt{x-1}$，过原点作其切线，求此曲线、切线，与$x$轴围成的面积，及绕$x$轴一周的侧面积

\begin{tikzpicture}[x=2cm, y=2cm, global scale=0.8, >=latex,thick] \draw[thick,->] (-0.2,0) -- (2.7,0) node[right] {$x$};% x軸 \draw[thick,->] (0,-0.1) -- (0,1.5) node[above] {$y$};% y軸 \draw[domain=1:2.5, color=blue, thick, smooth]%, samples=101] plot (\x,{sqrt(\x-1)}); \draw[domain=0:2.5, color=red, thick, smooth]%, samples=101] plot (\x,{0.5*\x}); \draw[dashed] (2,1)--(2,0) node[below] {$x_0$}; \draw (1,0) node[below] {$1$}; \draw (0,0) node[below left] {$O$}; \end{tikzpicture}

11. 先求切点。曲线上点$(x_0,y_0)$的切线方程为

\[\dfrac{y-y_0}{x-x_0}=(\sqrt{x-1})'|_{x=x_0}=\frac1{2\sqrt{x_0-1}} \]

当切线过原点，则

\[\dfrac{0-y_0}{0-x_0}=\dfrac1{2\sqrt{x_0-1}} \]

这样有，

\[\dfrac{\sqrt{x_0-1}}{x_0}=\dfrac1{2\sqrt{x_0-1}} \]

可得$x_0=2$

侧面积，由切线的侧面积和曲线的侧面积组成。

物体间的引力

考虑一个半径为$r$的球壳对离球心$R$的$B$点处的质点的万有引力。

\begin{tikzpicture}[x=2cm, y=2cm, global scale=0.8, >=latex,thick] %\draw[thick,->] (-0.5,0) -- (5,0) node[right] {$x$};% x軸 %\draw[thick,->] (0,-1) -- (0,2) node[above] {$y$};% y軸 \draw (0,0) node[below] {$O$};% 原点 \draw (0,0) circle(1); \draw (0,0) --node[below] {$R$} (2.5,0) node[below] {$B$}; \coordinate (A) at (35:1); \coordinate (Ai) at (-35:1); \draw[blue] (Ai) to[out=100, in=-100] (A); \draw[dashed,blue] (Ai) to[out=80, in=-80] (A); \coordinate (B) at (45:1.0); \coordinate (Bi) at (-45:1.0); \draw[dashed, blue] (Bi) to[out=80, in=-80] (B); \draw[blue] (Bi) to[out=100, in=-100] (B); \draw[blue] (A) arc(35:45:1) node[right] {$rd\theta$}; \draw (0,0) -- node[above] {$r$} (A); \draw (A) -- node[above] {$x$} (2.5, 0); \draw (0:0.2) to[out=90, in=-55] (35:0.2); \draw[blue] (0:0.2) node[above right] {$\theta$}; \draw (0:2.28) to[out=90, in=-95] (2:2.3); \draw[blue] (1.5:2.28) node[left] {$\alpha$}; \end{tikzpicture}

微元： 考虑球壳上$\theta$角度处取出的一个条带状圆环对$B$点的引力

\begin{tikzpicture}[x=2cm, y=2cm, global scale=0.6, >=latex,thick] %\draw[thick,->] (-0.5,0) -- (5,0) node[right] {$x$};% x軸 %\draw[thick,->] (0,-1) -- (0,2) node[above] {$y$};% y軸 \draw (0,0) node[below] {$O$};% 原点 \draw (0,0) circle(1); \draw (0,0) --node[below] {$R$} (2.5,0) node[below] {$B$}; \coordinate (A) at (35:1); \coordinate (Ai) at (-35:1); \draw[blue] (Ai) to[out=100, in=-100] (A); \draw[dashed,blue] (Ai) to[out=80, in=-80] (A); \coordinate (B) at (45:1.0); \coordinate (Bi) at (-45:1.0); \draw[dashed, blue] (Bi) to[out=80, in=-80] (B); \draw[blue] (Bi) to[out=100, in=-100] (B); \draw[blue] (A) arc(35:45:1) node[right] {$rd\theta$}; \draw (0,0) -- node[above] {$r$} (A); \draw (A) -- node[above] {$x$} (2.5, 0); \draw (0:0.2) to[out=90, in=-55] (35:0.2); \draw[blue] (0:0.2) node[above right] {$\theta$}; \draw (0:2.28) to[out=90, in=-95] (2:2.3); \draw[blue] (1.5:2.28) node[left] {$\alpha$}; \end{tikzpicture}

设$B$处质点质量为$m$，球壳的密度为$\rho$。则球壳上$\theta$角度处对$B$点的引力为

\[dF=\frac{Gm\cdot\rho\cdot 2\pi r\sin\theta\cdot r d\theta}{x^2}\cos\alpha \]

考虑到$\alpha$, $x$与$\theta$都有关系，需要进一步的运算。

\[x^2=R^2+r^2-2Rr\cos\theta \]
\[r\cos\theta+x\cos\alpha=R \]

由$x\cos\alpha = R-r\cos\theta$，及$x=\sqrt{R^2+r^2-2Rr\cos\theta}$，得到

\[\begin{aligned} dF=&\frac{Gm\cdot\rho\cdot 2\pi r\sin\theta\cdot r d\theta}{x^2}\cos\alpha\\ =&2\pi Gm\rho r^2\frac{x\cos\alpha}{x^3}\sin\theta d\theta \\ =&-2\pi Gm\rho r\frac{ R-r\cos\theta}{(\sqrt{R^2+r^2-2Rr\cos\theta})^3} d(r\cos\theta) \\ \end{aligned} \]

令$x=r\cos\theta$，则有

\[F=\int_0^{\pi} dF =2\pi Gm\rho r \int_{-r}^r \frac{ R- x}{(\sqrt{R^2+r^2-2R \cdot x})^3} d x \]

或者，由$x^2=R^2+r^2-2Rr\cos\theta$，得到

\[2xdx = 2R r\sin\theta d\theta \]

另外，有余弦公式$r^2=R^2+x^2-2Rx\cos\alpha$，得到

\[x\cos\alpha = \frac{R^2-r^2+x^2}{2R} \]

代入微元表达式，得到

\[\begin{aligned} dF=&\frac{Gm\cdot\rho\cdot 2\pi r\sin\theta\cdot r d\theta}{x^2}\cos\alpha\\ =&\frac{\pi Gm\rho r}{R^2}\frac{R^2-r^2+x^2}{x^2}dx \end{aligned} \]

积分后，有

\[\begin{aligned} F=&\int_{R-r}^{R+r} \frac{\pi Gm\rho r}{R^2}\frac{R^2-r^2+x^2}{x^2}dx \\ =&\frac{\pi Gm\rho r}{R^2} \left(\frac{R^2-r^2}x\bigg|_{R+r}^{R-r}+2r\right) \\ =&\frac{Gm \rho\cdot 4\pi r^2}{R^2} \end{aligned} \]

其中$4\pi r^2\rho$正好是球壳的质量。这和球壳的全部质量集中在球心时产生的引力是一样的。

目录

本节读完

例 12. 谢谢

12.