MIT_单变量微积分_28

古城微笑少年丶 2022-02-19 15:36 67阅读 0赞

### 1.弧长 ###

**引入：** 如图所示：图中有条线没画出来， S 0 − a S\_0 - a S0−a.

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 Δ S = S i − S i − 1 ( Δ S ) ≈ ( Δ x ) 2 + ( Δ y ) 2 ( D x ) 2 = ( d x ) 2 + ( d y ) 2 ( 极 限 情 况 ) d s = d x 2 + d y 2 d s = 1 + ( d y d x ) d x 弧 长 = S n − S 0 = ∫ a b 1 + ( d y d x ) 2 d x = ∫ d s ∫ a b 1 + ( f ′ ( x ) ) 2 d x ( y = f ( x ) ) \\Delta S = S\_i-S\_\{i-1\}\\\\ (\\Delta S) \\approx(\\Delta x)^2+(\\Delta y)^2\\\\ (Dx)^2=(dx)^2+(dy)^2(极限情况)\\\\ ds=\\sqrt\{dx^2+dy^2\}\\\\ ds=\\sqrt\{1+(\\frac\{dy\}\{dx\})\} dx\\\\ 弧长=S\_n-S\_0\\\\ =\\int\_a^b\\sqrt\{1+(\\frac\{dy\}\{dx\})^2\}dx=\\int ds\\\\ \\int\_a^b\\sqrt\{1+(f&\#x27;(x))^2\}dx(y=f(x)) ΔS=Si−Si−1(ΔS)≈(Δx)2\+(Δy)2(Dx)2=(dx)2\+(dy)2(极限情况)ds=dx2\+dy2ds=1\+(dxdy)dx弧长=Sn−S0=∫ab1\+(dxdy)2dx=∫ds∫ab1\+(f′(x))2dx(y=f(x))  
`Ex1:` y = m x y=mx y=mx.  
**解：**  
 y ′ = m , d s = 1 + ( y ′ ) 2 d x = 1 + m 2 d x L e n g t h ： 0 ≤ x ≤ 10 ∫ 0 10 1 + m 2 d x = 10 1 + m 2 y&\#x27;=m,ds=\\sqrt\{1+(y&\#x27;)^2\}dx=\\sqrt\{1+m^2\}dx\\\\ Length： 0 \\leq x \\leq 10\\\\ \\int\_0^\{10\}\\sqrt\{1+m^2\}dx=10\\sqrt\{1+m^2\} y′=m,ds=1\+(y′)2dx=1\+m2dxLength：0≤x≤10∫0101\+m2dx=101\+m2  
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如图所示，就可以得出此函数的弧长…  
`Ex2:` y = 1 − x 2 y=\\sqrt\{1-x^2\} y=1−x2  
 y ′ = − x 1 − x 2 d s = 1 + ( y ′ ) 2 = 1 1 − x 2 l = ∫ 0 a d x 1 − x 2 = s i n − 1 x ∣ 0 a l = s i n − 1 a , 同 理 得 ， s i n l = a y&\#x27;=\\frac\{-x\}\{\\sqrt\{1-x^2\}\}\\\\ ds=\\sqrt\{1+(y&\#x27;)^2\}=\\sqrt\{\\frac\{1\}\{1-x^2\}\}\\\\ l=\\int\_0^a\\frac\{dx\}\{\\sqrt\{1-x^2\}\}=sin^\{-1\}x|\_0^a\\\\ l=sin^\{-1\}a,同理得，sin l = a y′=1−x2−xds=1\+(y′)2=1−x21l=∫0a1−x2dx=sin−1x∣0al=sin−1a,同理得，sinl=a

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`Ex3:`抛物线的长度：图形如图所示：  
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 y = x 2 , y ′ = 2 x d s = 1 + ( 2 x ) 2 d x a r c ( 0 ≤ x ≤ a ) = ∫ 0 a 1 + 4 x 2 d x = 1 2 ∫ 0 a s e c 3 u d u = \[ 1 4 l n ( 2 x + 1 + 4 x 2 ) + 1 2 ∗ 1 + 4 x 2 \] ∣ 0 a y=x^2,y&\#x27;=2x\\\\ ds=\\sqrt\{1+(2x)^2\}dx\\\\ arc(0 \\leq x \\leq a)=\\int\_0^a\\sqrt\{1+4x^2\}dx\\\\ =\\frac\{1\}\{2\}\\int\_0^asec^3udu\\\\ =\[\\frac\{1\}\{4\}ln(2x+\\sqrt\{1+4x^2\})+\\frac\{1\}\{2\}\*\\sqrt\{1+4x^2\}\]|\_0^a y=x2,y′=2xds=1\+(2x)2dxarc(0≤x≤a)=∫0a1\+4x2dx=21∫0asec3udu=\[41ln(2x\+1\+4x2)\+21∗1\+4x2\]∣0a  
`Ex4`曲面面积： y = x 2 绕 x 轴 旋 转 。 y=x^2绕x轴旋转。 y=x2绕x轴旋转。  
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 d A = ( 2 π y ) d s 表 面 积 = ∫ 0 a 2 π x 2 1 + 4 x 2 d x dA=(2\\pi y)ds\\\\ 表面积=\\int\_0^a2 \\pi x^2\\sqrt\{1+4x^2\}dx dA=(2πy)ds表面积=∫0a2πx21\+4x2dx  
`Ex5:`球面面积:  
![在这里插入图片描述][20190416210026684.png_pic_center]  
 y = a 2 − x 2 , y ′ = − x a 2 − x 2 1 + ( − x a 2 − x 2 ) 2 = a 2 a 2 − x 2 a r e a = ∫ x 1 x 2 2 π y d s = ∫ x 1 x 2 2 π a 2 − x 2 a 2 a 2 − x 2 d x = ∫ x 1 x 2 2 π a d x = 2 π a d x = 2 π a ( x 2 − x 1 ) y=\\sqrt\{a^2-x^2\},y&\#x27;=\\frac\{-x\}\{\\sqrt\{a^2-x^2\}\}\\\\ 1+(\\frac\{-x\}\{\\sqrt\{a^2-x^2\}\})^2=\\frac\{a^2\}\{a^2-x^2\}\\\\ area=\\int\_\{x1\}^\{x2\}2 \\pi y ds\\\\ =\\int\_\{x\_1\}^\{x2\}2 \\pi \\sqrt\{a^2-x^2\}\\sqrt\{\\frac\{a^2\}\{a^2-x^2\}\}dx\\\\ =\\int\_\{x\_1\}^\{x2\}2\\pi adx\\\\ =2\\pi adx=2\\pi a(x\_2-x\_1) y=a2−x2,y′=a2−x2−x1\+(a2−x2−x)2=a2−x2a2area=∫x1x22πyds=∫x1x22πa2−x2a2−x2a2dx=∫x1x22πadx=2πadx=2πa(x2−x1)  
`Ex6:` \{ x = 2 s i n t y = c o s t \\left\\\{\\begin\{matrix\} x = 2 sint& \\\\ y = cost& \\end\{matrix\}\\right. \{ x=2sinty=cost,如图所示：  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM4Mzg2MzE2_size_16_color_FFFFFF_t_70 5]  
 1 4 x 2 + y 2 = s i n 2 t + c o s 2 t = 1 d s d t = ( 2 c o s t ) 2 + ( s i n t ) 2 弧 长 = ∫ 0 2 π 4 c o s 2 t + s i n 2 t d t 椭 球 的 表 面 积 ： 绕 y 轴 旋 转 d A = ∫ 0 π 2 π ( 2 s i n t ) 4 c o s 2 t + s i n 2 t d t \\frac\{1\}\{4\}x^2+y^2=sin^2t+cos^2t=1\\\\ \\frac\{ds\}\{dt\}=\\sqrt\{(2cost)^2+(sint)^2\}\\\\ 弧长=\\int\_0^\{2\\pi\}\\sqrt\{4cos^2t+sin^2t\}dt\\\\ 椭球的表面积：绕y轴旋转\\\\ dA=\\int\_0^\{\\pi\}\{2\\pi(2sint)\\sqrt\{4cos^2t+sin^2t\}\}dt 41x2\+y2=sin2t\+cos2t=1dtds=(2cost)2\+(sint)2弧长=∫02π4cos2t\+sin2tdt椭球的表面积：绕y轴旋转dA=∫0π2π(2sint)4cos2t\+sin2tdt

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[20190416210026684.png_pic_center]: /images/20220219/896800966a3046d8bd39036fe66bbef5.png
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