3. 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

积分的应用

微元法

几何与物理学中，许多整体性质的量（记为$Q$）可以用积分来确定，如：曲线的弧长、非均匀物体的质心等。这里要求$Q$具有区域可加性

\[Q([a,c])+Q([c,b])=Q([a,b]), a\leq c\leq b \]

自然，取$c=a$，则有$Q(a)=0$。

用微元法来计算$Q$。

  1. 在区间$[a,b]$上任取一个长度为$dx$的小区间$[x,x+dx]$，救出局部量$\Delta Q=Q(x+dx)-Q(x)$的一个近似值$f(x)dx$，满足
    \[\Delta Q=f(x)dx+o(dx) \]
  2. 将所得的微元在$[a,b]$上积分，由Newton-Leibniz公式，
    \[\int_a^bf(x)dx=\int_a^bQ'(x)dx=Q(x)|_a^b=Q(b)-Q(a)=Q \]

微元法的关键在于确定$\Delta Q$的线性主要部分，即微元。问题的难点存在量$Q$是待求的，部分量$Q(x)$是未知的。

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}

曲线$AB$的长度定义为：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。

若$\displaystyle\max_{1\leq i\leq n}M_{i-1}M_i\to 0$时，折线的长度的极限存在，则这个极限就是曲线的长度。

弧长微元

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

\[\begin{aligned} |MM'|=&\sqrt{(dx)^2+(f(x+dx)-f(x))^2} \\ =&\sqrt{(dx)^2+(f'(x)dx+o(dx))^2} \end{aligned} \]
% 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] {$x+dx$}; \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

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

参数方程

设曲线$\widehat{AB}$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

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

因此，弧长为

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

极坐标

设曲线$\widehat{AB}$的极坐标方程 $\begin{cases}x=f(\theta)\cos\theta \\y=f(\theta)\sin\theta\end{cases}$， 则有

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

得到弧长微元

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

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

例 1. $9y^2=x(x-3)^2, y\geq0$，在$x=0$到$x=3$之间的长度

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

ex5

平面图形的面积

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

lemniscate

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。由

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

或

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

又$r^2=x^2+y^2$，所以有

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

例 4. 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

例 5. 求$x^2+xy+y^2=1$围成的面积

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] {$x+dx$}; \draw[dashed] (3,0) -- (3,1.5); \draw[dashed] (2,0)--(2,1.5); \end{tikzpicture}

空间中，某个立体的体积如左。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$[x,x+dx]$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

\[V=\int_a^b S(x)dx \]
% 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 left] {$a$}; \draw (4.0,0) node[below right] {$b$}; \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[dashed] (ia) to[out=60, in=-60] (a); \draw (ia) to[out=120, in=-120] (a); \draw[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。

点$x$处的截面的面积为

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

因此，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$dx$的区间$[x,x+dx]$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$x+dx$的两圆柱的体积差

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

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 6. 一个锥体，底面积为$Q$，高为$h$。求体积

例 7. 椭圆$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$，绕$x$轴一周，得到椭球体积

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 left] {$a$}; \draw (4.0,0) node[below right] {$b$}; \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] (ia) to[out=60, in=-60] (a); \draw (ia) to[out=120, in=-120] (a); \draw[dashed] (a) -- (\x,0); \draw (\x,0) node[below] {$x$}; \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] (ia) to[out=60, in=-60] (a); \draw (ia) to[out=120, in=-120] (a); \draw[dashed] (a) -- (\x,0); \draw (\x,0) node[below] {$x+dx$}; \draw (a) node[above] {$M'$}; \end{tikzpicture}

在区间上取长度$dx$的小段$[x,x+dx]$，对应的弧$\widehat{MM'}$得到的旋转体的侧面积为

\[\pi(y+(y+dy))ds \]

$ds=\sqrt{1+y'^2}dx$为$\widehat{MM'}$的弧长。

略去$dx$的高阶无穷小量$\pi dy ds$，侧面积微元为 $dS=2\pi yds=2\pi f(x)\sqrt{1+(f'(x))^2}dx$ 因此，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 $\begin{cases}x=\phi(t) \\ y=\psi(t)\end{cases}$

则弧长元

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

侧面积元

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

因此，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 8. 球面的面积

例 9. $y=\sqrt{x-1}$，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

变力做功和引力

谢谢

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

目录

本节读完

例 10. 谢

10.