1. 平面图形的面积
(1) 若平面域 D D D y = f ( x ) , y = g ( x ) ( f ( x ) ⩾ g ( x ) ) , x = a , x = b ( a < b ) y = f(x), y = g(x)(f(x) \geqslant g(x)), x = a, x = b(a < b) y = f ( x ) , y = g ( x ) ( f ( x ) ⩾
Dors Dors 是花野猫开发为知识工作者打造的数字花园应用，包含的博客、个人记事本、及其他实用功能。
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S = ∫ a b [ f ( x ) − g ( x ) ] d x . S = \int_a^b [f(x) - g(x)] \mathrm{d}x. S = ∫ a b [ f ( x ) − g ( x )] d x . (2) 若平面域 D D D r = r ( θ ) , θ = α , θ = β ( α < β ) r = r(\theta), \theta = \alpha, \theta = \beta (\alpha < \beta) r = r ( θ ) , θ = α , θ = β ( α < β ) S =
2. 旋转体体积 若区域 D D D y = f ( x ) ( f ( x ) ⩾ 0 ) y = f(x)(f(x) \geqslant 0) y = f ( x ) ( f ( x ) ⩾ 0 ) x = a , x = b ( 0 ⩽ a < b ) x = a, x = b(0 \leqslant a < b) x = a , x =
(1) 区域 D D D x x x V x = π ∫ a b f 2 ( x ) d x . V_x = \pi \int_a^b f^2(x) \mathrm{d}x. V x = π ∫ a b
(2) 区域 D D D y y y V y = 2 π ∫ a b x f ( x ) d x . V_y = 2\pi \int_a^b xf(x) \mathrm{d}x. V y = 2 π ∫ a
3. 曲线弧长 (1) C : y = y ( x ) , a ⩽ x ⩽ b C: y = y(x), a \leqslant x \leqslant b C : y = y ( x ) , a ⩽ x ⩽ b s = ∫ a b 1 + y ′ 2 d x . s = \int_a^b \sqrt{1 + y'^2} \mathrm{d}x. s = ∫
(2) C : { x = x ( t ) , y = y ( t ) , α ⩽ t ⩽ β . C: \begin{cases} x = x(t), \\ y = y(t), \end{cases} \quad \alpha \leqslant t \leqslant \beta. C : { x = x ( t ) , y = y ( t ) , α ⩽
(3) C : r = r ( θ ) , α ⩽ θ ⩽ β . C: r = r(\theta), \alpha \leqslant \theta \leqslant \beta. C : r = r ( θ ) , α ⩽ θ ⩽ β . s = ∫ α β r 2 + r ′ 2 d θ . s = \int_\alpha^\beta \sqrt{r^2 + r'^2} \mathrm{d}\theta. s =
4. 旋转体侧面积 曲线 y = f ( x ) ( f ( x ) ⩾ 0 ) y = f(x)(f(x) \geqslant 0) y = f ( x ) ( f ( x ) ⩾ 0 ) x = a , x = b ( 0 ⩽ a < b ) x = a, x = b(0 \leqslant a < b) x = a , x = b ( 0 ⩽
定积分的几何应用相关例题 【例 1】 (2014, 数三) 设 D D D x y + 1 = 0 xy + 1 = 0 x y + 1 = 0 y + x = 0 y + x = 0 y + x = 0 y = 2 y = 2 y = 2
解 所求面积为
S = ∬ D 1 d σ = ∫ 1 2 d y ∫ − y − 1 y d x = ∫ 1 2 ( y − 1 y ) d y = ( 1 2 y 2 − ln y ) ∣ 1 2 = 2 − ln 2 − 1 2 = 3 2 − ln 2. \begin{aligned} S &= \iint_D 1 \mathrm{d}\sigma = \int_1^2 \mathrm{d}y \int_{-y}^{-\frac{1}{y}} \mathrm{d}x \\ &= \int_1^2 \left(y - \frac{1}{y}\right) \mathrm{d}y = \left(\frac{1}{2}y^2 - \ln y\right)\bigg|_1^2 \\ &= 2 - \ln 2 - \frac{1}{2} = \frac{3}{2} - \ln 2. \end{aligned} S
【例 2】 (2013, 数二) 设封闭曲线 L L L r = cos 3 θ ( − π 6 ⩽ θ ⩽ π 6 ) r = \cos 3\theta (-\frac{\pi}{6} \leqslant \theta \leqslant \frac{\pi}{6}) r = cos 3 θ ( − 6 π ⩽ θ ⩽
解 【方法 1】
S = 1 2 ∫ − π 6 π 6 cos 2 3 θ d θ = ∫ 0 π 6 cos 2 3 θ d θ = 1 2 ∫ 0 π 6 ( 1 + cos 6 θ ) d θ = 1 2 ( θ + 1 6 sin 6 θ ) ∣ 0 π 6 = π 12 . \begin{aligned} S &= \frac{1}{2} \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \cos^2 3\theta \mathrm{d}\theta = \int_0^{\frac{\pi}{6}} \cos^2 3\theta \mathrm{d}\theta \\ &= \frac{1}{2} \int_0^{\frac{\pi}{6}} (1 + \cos 6\theta) \mathrm{d}\theta \\ &= \frac{1}{2} \left(\theta + \frac{1}{6}\sin 6\theta\right)\bigg|_0^{\frac{\pi}{6}} = \frac{\pi}{12}. \end{aligned}
【方法 2】
S = 1 2 ∫ − π 6 π 6 cos 2 3 θ d θ = ∫ 0 π 6 cos 2 3 θ d θ = 3 θ = t 1 3 ∫ 0 π 2 cos 2 t d t S = \frac{1}{2} \int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \cos^2 3\theta \mathrm{d}\theta = \int_0^{\frac{\pi}{6}} \cos^2 3\theta \mathrm{d}\theta \xlongequal{3\theta = t} \frac{1}{3} \int_0^{\frac{\pi}{2}} \cos^2 t \mathrm{d}t S = 2 1 ∫
【例 3】 (2015, 数二、三) 设 A > 0 , D A > 0, D A > 0 , D y = A sin x ( 0 ⩽ x ⩽ π 2 ) y = A\sin x (0 \leqslant x \leqslant \frac{\pi}{2}) y = A sin x ( 0 ⩽ x ⩽ 2
解
V 1 = π ∫ 0 π 2 A 2 sin 2 x d x = π A 2 ⋅ 1 2 ⋅ π 2 = π 2 A 2 4 . V_1 = \pi \int_0^{\frac{\pi}{2}} A^2 \sin^2 x \mathrm{d}x = \pi A^2 \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi^2 A^2}{4}. V 1 = π ∫ 0
【例 4】 (2012, 数二) 过点 ( 0 , 1 ) (0, 1) ( 0 , 1 ) L : y = ln x L: y = \ln x L : y = ln x A A A L L L x x x
【方法 1】 设切点 A A A ( x 0 , y 0 ) (x_0, y_0) ( x 0 , y 0 ) y − y 0 = 1 x 0 ( x − x 0 ) , y - y_0 = \frac{1}{x_0}(x - x_0), y −
【方法 2】 设过点 ( 0 , 1 ) (0, 1) ( 0 , 1 ) y − 1 = k x y - 1 = kx y − 1 = k x y = ln x y = \ln x y = ln x ln x − 1 = k x , k = 1 x \ln x - 1 = kx, k = \frac{1}{x}
所以切线方程为 y = x e 2 + 1 y = \frac{x}{e^2} + 1 y = e 2 x + 1
所求面积为
S = ∫ 1 e 2 ln x d x − 1 2 ( e 2 − 1 ) ⋅ 2 = x ln x ∣ 1 e 2 − ∫ 1 e 2 d x − e 2 + 1 = 2 e 2 − e 2 + 1 − e 2 + 1 = 2. \begin{aligned} S &= \int_1^{e^2} \ln x \mathrm{d}x - \frac{1}{2}(e^2 - 1) \cdot 2 \\ &= x\ln x \bigg|_1^{e^2} - \int_1^{e^2} \mathrm{d}x - e^2 + 1 \\ &= 2e^2 - e^2 + 1 - e^2 + 1 \\ &= 2. \end{aligned} S
所求体积为
V = π ∫ 1 e 2 ln 2 x d x − π 3 ⋅ 4 ⋅ ( e 2 − 1 ) = π ( x ln 2 x − 2 x ln x + 2 x ) ∣ 1 e 2 − 4 π 3 ( e 2 − 1 ) = 2 π 3 ( e 2 − 1 ) . \begin{aligned} V &= \pi \int_1^{e^2} \ln^2 x \mathrm{d}x - \frac{\pi}{3} \cdot 4 \cdot (e^2 - 1) \\ &= \pi (x\ln^2 x - 2x\ln x + 2x) \bigg|_1^{e^2} - \frac{4\pi}{3}(e^2 - 1) \\ &= \frac{2\pi}{3}(e^2 - 1). \end{aligned} V
心得笔记 2012 年数二题的这道定积分应用的题，是一道曲线在某个切点，与曲线上另一个点连成的割线，与曲线围成面积的题。做这道题，不论我是以 x 为积分变量，还是以 y 为积分变量，结果都算错。原因是：计算过程中完全没意识到，有两条直线！我计算出切线之后，就一直使用切线方程作为下函数来做定积分，但围成面积的，总是割线！割线！割线！切线和曲线围不成面积！！！。另外，如果图中出现了规则图形，例如下半部分是三角形，直接计算即可，写方程去求积分多费了事还容易积错。做这道题又令我搞清楚了一个不定积分： lnx 的不定积分直接用分部积分就可以求出：结果是 xlnx-x+C ！我连这个常识都不知道！！！一道大题做好几遍都是错，加油！
圆盘法算旋转体体积，积分前面的Π忘记乘，再这样下去就死翘翘啦。
旋转体的体积：积分变量与旋转轴一致的情况。我以为可以用面积乘以2π 来计算，但实际上这样就忽略了帕普斯定理，就是实际上不是乘以2π，而要用面积乘以图形扫过的周长的平均值，也就是质心走过的圆周周长。
旋转轴与积分变量不一致的情况，如果换积分变量不好积分时，实际上可以用柱壳法，先用周长乘高，再乘dx
旋转椭球体的体积只能算一次，抽上半部分来算，不要重复计算。椭圆以x轴旋转一周和以y轴旋转一周得到的体积并不相等。（柿子🍅和橄榄球🏈）
#柱壳法 #定积分的应用 #几何 #lnx的不定积分 #大题 #应用题 #立体图形 #割线 #切线
找到一个收录考研真题的网站 荒原之梦 zhaokaifeng.com/8699/
tikz 源码 \begin{tikzpicture}[scale=0.8]
%坐标轴
\draw[->] (0,-1) -- (0,5) node[above] {$y$};
\draw[->] (-1,0) -- (10,0) node[right] {$x$};
\node at (-0.3,-0.3) {$O$};
%曲线 y=lnx
\draw[thick] plot[domain=0.3:9,smooth] (\x,{ln(\x)});
\node at (1.1,-1.1) {$L$};
%B点(1,0)
\fill (1,0) circle (2pt);
\node[above right] at (0.55,0.17) {$B$};
%A点：切点(e²,2)，切线y=x/e²+1过(0,1)
\pgfmathsetmacro{\xa}{e^2}
\pgfmathsetmacro{\ya}{ln(\xa)}
\fill (\xa,\ya) circle (2pt);
\node[above right] at (\xa,\ya) {$A$};
%切线：y = x/e² + 1，过(0,1)与A(e²,2)
\draw[thick] (-0.8,{-0.8/(e^2)+1}) -- (9,{9/(e^2)+1});
%直线AB，连接A(e²,2)与B(1,0)
\draw[thick] (1,0) -- (\xa,\ya);
%阴影填充：曲线y=lnx 和直线AB围成区域
\fill[pattern=north west lines]
plot[domain=1:\xa,smooth] (\x,{ln(\x)})
-- plot[domain=\xa:1] (\x,{(2/(e^2-1))*(\x-1)})
-- cycle;
\end{tikzpicture}
\documentclass[tikz,border=10pt]{standalone}
\usepackage{amsmath}
\usetikzlibrary{arrows.meta}
\begin{document}
\begin{tikzpicture}[>=Stealth, scale=1.3]
% 坐标轴
\draw[->, thick] (-0.5, 0) -- (6.5, 0) node[right] {$x$};
\draw[->, thick] (0, -0.5) -- (0, 4.5) node[above] {$y$};
\node[below left] at (0, 0) {$O$};
% 曲线函数 y = f(x)
\draw[domain=0.8:5.5, smooth, variable=\x, thick]
plot ({\x}, {0.12*(\x-1.5)*(\x-1.5) + 0.8})
node[right] {$y = f(x)$};
% 积分区间 [a, b]
\def\xa{1.2}
\def\xb{5.0}
\def\ya{0.12*(\xa-1.5)*(\xa-1.5) + 0.8}
\def\yb{0.12*(\xb-1.5)*(\xb-1.5) + 0.8}
% 区间端点虚线与标注
\draw[dashed] (\xa, 0) -- (\xa, \ya);
\draw[dashed] (\xb, 0) -- (\xb, \yb);
\node[below] at (\xa, 0) {$a$};
\node[below] at (\xb, 0) {$b$};
\filldraw[black] (\xa, \ya) circle (1.5pt);
\filldraw[black] (\xb, \yb) circle (1.5pt);
% 微元位置 (x0, y0) 到 (x1, y1)
\def\xO{2.6}
\def\xI{3.8}
\def\yO{0.12*(\xO-1.5)*(\xO-1.5) + 0.8}
1 2 ∫ α β r 2 ( θ ) d θ . S = \frac{1}{2} \int_\alpha^\beta r^2(\theta) \mathrm{d}\theta. S = 2 1 ∫ α β r 2 ( θ ) d θ .
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S = 2 π ∫ a b f ( x ) 1 + f ′ 2 ( x ) d x . S = 2\pi \int_a^b f(x) \sqrt{1 + f'^2(x)} \mathrm{d}x. S = 2 π ∫ a b f ( x ) 1 + f ′2 ( x ) d x .
= ∬ D 1 d σ = ∫ 1 2 d y ∫ − y − y 1 d x = ∫ 1 2 ( y − y 1 ) d y = ( 2 = 2 − ln 2 − 2 1 = 2 3 − ln 2.
6 π
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= 2 1 ∫ − 6 π 6 π cos 2 3 θ d θ = ∫ 0 6 π cos 2 3 θ d θ = 2 1 ∫ 0 6 π ( 1 + cos 6 θ = 2 1 ( θ + 6 1 sin 6 θ ) − 6 π 6 π
cos 2
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3 1 ∫ 0 2 π cos 2 t d t
= 1 3 ⋅ 1 2 ⋅ π 2 = π 12 . = \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{12}. = 3 1 ⋅ 2 1 ⋅ 2 π = 12 π . π
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y = 0 , x = π 2 y = 0, x = \frac{\pi}{2} y = 0 , x = 2 π 2
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V 2 = 2 π ∫ 0 π 2 x ⋅ A sin x d x = − 2 π A ∫ 0 π 2 x d cos x = − 2 π A ( x cos x ∣ 0 π 2 − ∫ 0 π 2 cos x d x ) = − 2 π A ( 0 − 1 ) = 2 π A . \begin{aligned} V_2 &= 2\pi \int_0^{\frac{\pi}{2}} x \cdot A\sin x \mathrm{d}x = -2\pi A \int_0^{\frac{\pi}{2}} x \mathrm{d}\cos x \\ &= -2\pi A \left(x\cos x \bigg|_0^{\frac{\pi}{2}} - \int_0^{\frac{\pi}{2}} \cos x \mathrm{d}x\right) \\ &= -2\pi A (0 - 1) = 2\pi A. \end{aligned} V 2 = 2 π ∫ 0 2 π x ⋅ A sin x d x = − 2 π A π 2 A 2 4 = 2 π A , \frac{\pi^2 A^2}{4} = 2\pi A, 4 π 2 A 2 = 2 π A , A = 8 π A = \frac{8}{\pi} A = π 8 B B B
y 0
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x 0 1 ( x −
x 0 ) ,
x 0 = e 2 , y 0 = 2 x_0 = e^2, y_0 = 2 x 0 = e 2 , y 0 = 2 ln
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k x , k =
x 1
x = e 2 , y = 2 , ( e 2 , 2 ) x = e^2, y = 2, (e^2, 2) x = e 2 , y = 2 , ( e 2 , 2 ) k = 1 e 2 k = \frac{1}{e^2} k = e 2 1
= ∫ 1 e 2 ln x d x − 2 1 ( e 2 − 1 ) ⋅ 2 = x ln x 1 e 2 − ∫ 1 e = 2 e 2 − e 2 + 1 − e 2 + 1 = 2.
= π ∫ 1 e 2 ln 2 x d x − 3 π ⋅ 4 ⋅ ( e 2 − 1 ) = π ( x ln 2 x − 2 x ln x + 2 x ) 1 e 2 = 3 2 π ( e 2 − 1 ) . \def\yI{0.12*(\xI-1.5)*(\xI-1.5) + 0.8}
\draw[dashed] (\xO, \yO) -- (\xI, \yO) node[midway, below] {$\text{d}x$};
\draw[dashed] (\xI, \yO) -- (\xI, \yI) node[midway, right] {$\text{d}y$};
\draw (\xI-0.2, \yO) -- (\xI-0.2, \yO+0.2) -- (\xI, \yO+0.2);
% 斜弦长/切线微元近似 ds（用加粗线型进行视觉区分）
\draw[ultra thick] (\xO, \yO) -- (\xI, \yI)
node[above left, at end] {$\text{d}s \approx \sqrt{(\text{d}x)^2 + (\text{d}y)^2}$};
\filldraw[black] (\xO, \yO) circle (1.5pt) node[above left] {$P(x, y)$};
\filldraw[black] (\xI, \yI) circle (1.5pt) node[right] {$Q(x+\text{d}x, y+\text{d}y)$};
\draw[dotted] (\xO, 0) -- (\xO, \yO);
\draw[dotted] (\xI, 0) -- (\xI, \yI);
\node[below, font=\small] at (\xO, 0) {$x$};
\node[below, font=\small] at (\xI, 0) {$x+\text{d}x$};
\node[draw=black, rounded corners, align=left, font=\small] at (3.2, 3.8) {
$\text{d}s = \sqrt{1 + [f'(x)]^2}\,\text{d}x$\\[1mm]
$\displaystyle s = \int_a^b \sqrt{1 + [f'(x)]^2}\,\text{d}x$
9.1 二重积分的概念和性质 更新于2026-10-16
1
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=
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.
∫ 0 2 π
x
d
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= − 2 π A ( x cos x 0 2 π − ∫ 0 2 π cos x d x ) = − 2 π A ( 0 − 1 ) = 2 π A .
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