定积分的应用。
This commit is contained in:
unlockable
224
08定积分应用.tex
Normal file
224
08定积分应用.tex
Normal file
@@ -0,0 +1,224 @@
|
||||
\chapter{定积分的应用}
|
||||
\section{几何应用}
|
||||
\subsection{平面图形的面积}
|
||||
\subsubsection{直角坐标系下平面图形面积的计算}
|
||||
考虑由直线$x = a, x = b$及$x$轴和连续曲线$y = f(x)$所围成的曲边梯形的面积$A$。根据定积分的定义和几何意义可知，
|
||||
\[A = \int_a^b \abs{f(x)} \dif x\eqper\]
|
||||
|
||||
再考虑由曲线$y = f(x), y = g(x)$和直线$x = a, x = b$所围成的面积$A$。对于$g(x) \leq f(x), x \in [a,b]$，有
|
||||
\[\dif A = [f(x) - g(x)] \dif x\]
|
||||
因此
|
||||
\[A = \int_a^b [f(x) - g(x)] \dif x\]
|
||||
|
||||
进一步若$f(x), g(x)$大小关系不确定，有
|
||||
\[A = \int_a^b \abs{f(x) - g(x)} \dif x\eqper\]
|
||||
|
||||
最后考虑：设连续函数$\varphi(y), \psi(y)$满足
|
||||
\[0 \leq \psi(y) \leq \varphi(y), y \in [c,d]\]
|
||||
求由曲线$x = \varphi(y), x = \psi(y)$和直线$y = c, y = d$围成的面积$A$。
|
||||
仿照上面，有
|
||||
\[A = \int_c^d[\varphi(y) - \psi(y)] \dif y\eqper\]
|
||||
|
||||
\subsubsection{极坐标系下平面图形面积的计算}
|
||||
求曲线$\rho = \rho(\theta)$及射线$\theta = \alpha, \theta = \beta$所围成的面积$A$。
|
||||
|
||||
注意面积微元，即小扇形的面积为
|
||||
\[\dif A = \frac{1}{2}\rho^2(\theta \dif \theta)\]
|
||||
因此面积为
|
||||
\[A = \frac{1}{2} \int_\alpha^\beta \rho^2(\theta) \dif \theta\eqper\]
|
||||
|
||||
\subsubsection{利用参数方程求图形面积}
|
||||
\begin{example}
|
||||
求星形线：
|
||||
\(\begin{cases}
|
||||
x = a\cos^3 t\\
|
||||
y = a\sin^3 t
|
||||
\end{cases} t \in [0,2\pi]\)
|
||||
所围成的面积。
|
||||
\end{example}
|
||||
|
||||
\begin{proof}[解]
|
||||
利用对称性
|
||||
\begin{align*}
|
||||
A & = 4A_1 = 4\int_0^a y \dif x\\
|
||||
& = 4 \int_\frac{\pi}{2}^0 a \sin^3 t \cdot 3a\cos^2 t (-\sin t) \dif t\\
|
||||
& = 12\int_0^\frac{\pi}{2}a^2 \sin^4 t(1 - \sin^2 t) \dif t\\
|
||||
& = 12a^2 \left(\frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} - \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}\right)\\
|
||||
& = \frac{3}{8} \pi a^2 \eqper \qedhere
|
||||
\end{align*}
|
||||
\end{proof}
|
||||
|
||||
\subsection{空间体的体积}
|
||||
\subsubsection{已知平行截面面积的空间体的体积}
|
||||
若已知平行截面面积关于$x$的函数为$A(x)$，那么体积
|
||||
\[V = \int_a^b A(x) \dif x\eqper\]
|
||||
|
||||
\subsubsection{旋转体的体积}
|
||||
将$f(x), x = a, x = b$与$x$轴所围成的图形绕$x$轴旋转一周，求所得的空间体的体积。注意到每个截面的面积$A(x) = \pi y^2 = \pi f^2(x)$。因此
|
||||
\[V = \int_a^b \pi f^2(x) \dif x = \pi \int_a^b f^2(x) \dif x\eqper\]
|
||||
|
||||
\subsection{平面曲线的弧长}
|
||||
将曲线$AB$取细分的点$A = M_0, M_1, M_2, \dots, M_n = B$，曲线的长度定义为
|
||||
\[l = \tolim{\lambda}{0} \sum_{i=1}^n \abs{M_{i-1}M_i}\]
|
||||
即折线的长度的极限。
|
||||
|
||||
\begin{enumerate}
|
||||
\item 假设曲线方程由$y = f(x)$给出，且曲线是光滑的，即$\deriv{f}(x)$在$[a,b]$上连续。那么
|
||||
\[\abs{M_{i-1}M_i} = \sqrt{(\Delta x_i)^2 + (\Delta y_i)^2}(i = 1, 2, \dots, n)\]
|
||||
由Lagrange中值定理可以得到
|
||||
\[\Delta y_i = f(x_i) - f(x_{i-1}) = \deriv{f}(\xi_i) \cdot \Delta x_i (x_{i-1} < \xi_i < x_i)\]
|
||||
这导出
|
||||
\[\abs{M_{i-1}M_i} = \sqrt{1 + [\deriv{f}(\xi_i)]^2} \cdot \Delta x_i (i = 1, 2, \dots, n)\]
|
||||
因而
|
||||
\[\sum_{i=1}^n \abs{M_{i-1}M_i} = \sum_{i=1}^n \sqrt{1 + [\deriv{f}(\xi_i)]^2}\cdot \Delta x_i\]
|
||||
|
||||
那么令$\lambda = \max \limits_{1 \leq i \leq n} \abs{M_{i-1}M_i}$，$\mu = \max \limits_{1 \leq i \leq n} \{\Delta x_i\}$。$\Delta x_i \leq \abs{M_{i-1}M_i}$导出$\mu \leq \lambda$。因此$\lambda \to 0$时，有$\mu \to 0$。那么
|
||||
\begin{align*}
|
||||
l & = \tolim{\lambda}{0} \sum_{i=1}^n \abs{M_{i-1}M_i}\\
|
||||
& = \tolim{\mu}{0} \sum_{i=1}^n \sqrt{1 + [\deriv{f}(\xi_1)]^2} \cdot \Delta x_1\\
|
||||
& = \int_a^b \sqrt{1 + [\deriv{f}(x)]^2}\dif x
|
||||
\end{align*}
|
||||
|
||||
因此弧长为
|
||||
\[l = \int_a^b \sqrt{1 + [\deriv{f}(x)]^2} \dif x = \int_a^b \sqrt{1 + \left(\deriv{y}\right)^2} \dif x\]
|
||||
\item 假设曲线由参数方程
|
||||
\(\begin{cases}
|
||||
x = x(t)\\
|
||||
y = y(t)
|
||||
\end{cases} (\alpha \leq t \leq \beta)\)
|
||||
给出，$\deriv{x}(t), \deriv{y}(t) \in C[\alpha, \beta]$且不同时为零。$t = \alpha$对应起点$A$，$t = \beta$对应重点$B$，即当$\dif t > 0$时，由$\dif l > 0$。因而弧长公式为
|
||||
\[l = \int_\alpha^\beta \sqrt{\left(\deriv{x}\right)^2(t) + \left(\deriv{y}\right)^2(t)}\dif t\eqper\]
|
||||
\item 假设曲线段由极坐标方程$\rho = \rho(\theta)(\alpha \leq \theta \leq \beta)$给出，且$\deriv{\rho}(\theta)$在$[\alpha, \beta]$上连续。选择$\theta$作为参数，那么
|
||||
\[\begin{cases}
|
||||
x = \rho(\theta) \cos \theta\\
|
||||
y = \rho(\theta) \sin \theta
|
||||
\end{cases} (\alpha \leq \theta \leq \beta)\]
|
||||
因此弧长公式为
|
||||
\[l = \int_\alpha^\beta \sqrt{\rho^2(\theta) + \left(\deriv{\rho}\right)^2(\theta)}\dif \theta\eqper\]
|
||||
对应的弧微分公式为
|
||||
\[\dif l = \sqrt{\rho^2(\theta) + [\deriv{\rho}(\theta)]^2} \dif \theta\eqper\]
|
||||
\end{enumerate}
|
||||
|
||||
\subsection{旋转体的侧面积}
|
||||
将$y = f(x)$绕$x$轴旋转，求其侧面积。用每一点的切线旋转所得圆台的侧面积近似。侧面积为
|
||||
\[\text{圆台侧面积} = \pi [y + (y + \dif y)] \cdot \dif l = 2\pi y \dif l + \pi \dif y \cdot \dif l\]
|
||||
当$\dif x \to 0$时，$\dif y \cdot \dif l = o(\dif x)$，略去。
|
||||
|
||||
由此得到侧面积微元
|
||||
\[\dif S = 2\pi y \dif l = 2\pi y \sqrt{1 + (\deriv{y})^2} \dif x\]
|
||||
因此侧面积为
|
||||
\[S = 2\pi \int_a^b y \sqrt{1 + (\deriv{y})^2} \dif x\]
|
||||
|
||||
\subsection{曲率与曲率半径}
|
||||
曲率问题就是研究曲线的弯曲程度的问题。
|
||||
|
||||
\begin{definition}
|
||||
设$M_0, M$之间的弧长为$\Delta l$，$\dfrac{\Delta \alpha}{\Delta l}$为$M_0, M$之间的平均曲率。若
|
||||
\[\abs{\tolim{\Delta l}{0} \frac{\Delta \alpha}{\Delta l}}\]
|
||||
存在，则称$k = \abs{\tolim{\Delta l}{0} \dfrac{\Delta \alpha}{\Delta l}}$为曲线在$M_0$处的曲率。
|
||||
\end{definition}
|
||||
|
||||
设曲线$y = f(x)$二阶可导，那么
|
||||
\[k = \abs{\tolim{\Delta l}{0} \frac{\Delta \alpha}{\Delta l}} = \abs{\frac{\dif \alpha}{\dif l}}\]
|
||||
又因为
|
||||
\[\tan \alpha = \deriv{y}, \alpha = \arctan \deriv{y}\]
|
||||
可以得到
|
||||
\[\dif \alpha = \frac{1}{1 + (\deriv{y})^2} y^{\prime \prime} \dif x, \dif l = \sqrt{1 + (\deriv{y})^2}\dif x\]
|
||||
由此我们能得到
|
||||
|
||||
\begin{theorem}
|
||||
曲线的曲率公式为
|
||||
\[k = \frac{\abs{y^{\prime \prime}}}{(1+(\deriv{y})^2)^\frac{3}{2}}\eqper\]
|
||||
\end{theorem}
|
||||
|
||||
\begin{definition}
|
||||
$R = \dfrac{1}{k}$称为曲线$y = f(x)$在$M_0$处的曲率半径。
|
||||
\end{definition}
|
||||
|
||||
\section{物理应用}
|
||||
\subsection{引力问题}
|
||||
\begin{example}
|
||||
设有一均匀细杆长为$2l$，质量为$M$。另有一质量为$m$的质点，位于细杆所在的直线上，与杆的近端的距离为$a$。求细杆对质点的引力$F$。
|
||||
\end{example}
|
||||
|
||||
\begin{figure}[H]
|
||||
\centering
|
||||
\begin{tikzpicture}
|
||||
\draw[-{Stealth[width=5pt]}] (-3,0)--(3,0) node[below] {$x$};
|
||||
\node[below] at (-2.8,0) {$O$};
|
||||
\node[above] at (-2.8,0) {$m$};
|
||||
\node[draw=black, fill=black, circle, inner sep=0, minimum size=4pt] at (-2.8,0) {};
|
||||
\node[above] at (-1,0) {$a$};
|
||||
\node[above] at (0,0) {$x$};
|
||||
\node[above] at (1,0) {$x + \dif x$};
|
||||
\node[above] at (2.3,0) {$2l + a$};
|
||||
\draw[color=red, line width=2pt] (-1,0)--(2,0);
|
||||
\draw[color=black!50!white, line width=2pt] (0,0)--(0.6,0);
|
||||
\end{tikzpicture}
|
||||
\end{figure}
|
||||
|
||||
\begin{proof}[解]
|
||||
取一个小区间$[x, x + \dif x]$，视为质点质量$\dfrac{M}{2l}\dif x$。
|
||||
因此
|
||||
\[\dif F = k \frac{m\left(\frac{M}{2l}\dif x\right)}{x^2} = \frac{kmM}{2l}\cdot \frac{1}{x^2} \dif x\]
|
||||
因此合力应为
|
||||
\[F = \int_a^{a + 2l} \frac{kmM}{2l} \cdot \frac{1}{x^2} \dif x = \frac{kmM}{2l} \cdot \eval{\left(-\frac{1}{x}\right)}_a^{a+2l} = \frac{kmM}{a(a+2l)}\eqper \qedhere\]
|
||||
\end{proof}
|
||||
|
||||
\subsection{变力做功问题}
|
||||
问题：求物体从$x = a$移到$x = b$时变力$f(x)$所做的功。注意到功的微元$\dif W = f(x) \dif x$。因此
|
||||
\[W = \int_a^b f(x) \dif x \eqper\]
|
||||
|
||||
\subsection{静力矩和质心}
|
||||
\subsubsection{质点系的质心}
|
||||
设有若干个点$A_1, A_2, \dots, A_n$，它们的坐标分别为$(x_1, y_1), (x_2, y_2), \dots, (x_n, y_n)$，质量分别为$m_1, m_2, \dots, m_n$。
|
||||
|
||||
\begin{figure}[H]
|
||||
\centering
|
||||
\begin{tikzpicture}
|
||||
\draw[-{Stealth[width=5pt]}] (-0.5,0)--(4,0);
|
||||
\draw[-{Stealth[width=5pt]}] (0,-0.5)--(0,3);
|
||||
\node[draw=black, fill=black, circle, inner sep=0, minimum size=2pt] at (0.5,1) {};
|
||||
\node[draw=black, fill=black, circle, inner sep=0, minimum size=2pt] at (1.2,0.8) {};
|
||||
\node[draw=black, fill=black, circle, inner sep=0, minimum size=2pt] at (1.5,1.4) {};
|
||||
\node at (2.3,1.5) {$A_i(m_i)$};
|
||||
\node[draw=black, fill=black, circle, inner sep=0, minimum size=2pt] at (0.7,2) {};
|
||||
\node[draw=black, fill=black, circle, inner sep=0, minimum size=2pt] at (2.5,0.5) {};
|
||||
\node[draw=black, fill=black, circle, inner sep=0, minimum size=2pt] at (3.5,2.8) {};
|
||||
\node[draw=black, fill=black, circle, inner sep=0, minimum size=2pt] at (2,2.3) {};
|
||||
\draw[dashed] (0,1.4)--(1.5,1.4)--(1.5,0);
|
||||
\node[left] at (0,1.4) {$y_i$};
|
||||
\node[below] at (1.5,0) {$x_i$};
|
||||
\end{tikzpicture}
|
||||
\end{figure}
|
||||
|
||||
质点$A_i$对$x$轴的静力矩为$m_iy_i$，对$y$轴的静力矩为$m_ix_i$。因此质点系对$x$轴的静力矩
|
||||
\[M_x = \sum_{i=1}^n m_iy_i\]
|
||||
对$y$轴的静力矩
|
||||
\[M_y = \sum_{i=1}^n m_ix_i\]
|
||||
同时质点系的总质量
|
||||
\[M = \sum_{i=1}^n m_i\]
|
||||
我们可设质心的坐标为$(\overline{x}, \overline{y})$，那么
|
||||
\[M_x = M\overline{y}, M_y = M\overline{x}\]
|
||||
即
|
||||
\[\overline{x} = \frac{\displaystyle_{i=1}^n m_i x_i}{M}, \overline{y} = \frac{\displaystyle_{i=1}^n m_i y_i}{M}\eqper\]
|
||||
|
||||
\subsubsection{平面曲线的质心}
|
||||
设线密度$\rho$为常数。将弧长区间$[0,L]$分割。任取一个小区间$[l,l + \dif l]$。将其视为坐标为$(x,y)$的质点。因此$\dif M = \rho \dif l$。那么静力矩微元
|
||||
\[\dif M_x = y \rho \dif l, \dif M_y = x \rho \dif l\]
|
||||
于是有
|
||||
\[M_x = \int_0^L y \rho \dif l = \rho \int_0^L y \dif l, \dif M_y = \int_0^L x\rho \dif l = \rho \int_0^L x \dif l\]
|
||||
同时
|
||||
\[M = \int_0^L \rho \dif l = \rho \int_0^L \dif l = \rho L\]
|
||||
因此质心坐标为
|
||||
\[\overline{x} = \frac{\rho \int_0^L x \dif l}{\rho L} = \frac{\int_0^L x \dif l}{L}\]
|
||||
|
||||
\subsubsection{平面薄板的质心}
|
||||
设面密度$\mu$为常数。
|
||||
质量$M = \int_a^b \mu y \dif x$。静力矩
|
||||
\[M_x = \frac{1}{2} \mu \int_a^b y^2 \dif x, M_y = \mu \int_a^b xy \dif x\]
|
||||
因此质心坐标为
|
||||
\[\overline{x} = \frac{\int_a^b xy \dif x}{\int_a^b y \dif x}, \overline{y} = \frac{\frac{1}{2} \int_a^b y^2 \dif x}{\int_a^b y \dif x}\]
|
||||
|
||||
对极坐标方程$\rho = \rho(\theta)$，质量微元的质心坐标为
|
||||
\[\dif M_x = \frac{2}{3}\rho(\theta)\sin \theta \dif \theta, \dif M_y = \frac{2}{3} \rho(\theta) \cos \theta \dif \theta\eqper\]
|
||||
Reference in New Issue
Block a user