MIT_单变量微积分_28

1.弧长

引入： 如图所示：图中有条线没画出来，$S_0 - a$S0​−a.

$\Delta S = S_i-S_{i-1}\\ (\Delta S) \approx(\Delta x)^2+(\Delta y)^2\\ (Dx)^2=(dx)^2+(dy)^2(极限情况)\\ ds=\sqrt{dx^2+dy^2}\\ ds=\sqrt{1+(\frac{dy}{dx})} dx\\ 弧长=S_n-S_0\\ =\int_a^b\sqrt{1+(\frac{dy}{dx})^2}dx=\int ds\\ \int_a^b\sqrt{1+(f'(x))^2}dx(y=f(x))$ΔS=Si​−Si−1​(ΔS)≈(Δx)2+(Δy)2(Dx)2=(dx)2+(dy)2(极限情况)ds=dx2+dy2​ds=1+(dxdy​)​dx弧长=Sn​−S0​=∫ab​1+(dxdy​)2​dx=∫ds∫ab​1+(f′(x))2​dx(y=f(x))
Ex1:$y=mx$y=mx.
解：
$y'=m,ds=\sqrt{1+(y')^2}dx=\sqrt{1+m^2}dx\\ Length： 0 \leq x \leq 10\\ \int_0^{10}\sqrt{1+m^2}dx=10\sqrt{1+m^2}$y′=m,ds=1+(y′)2​dx=1+m2​dxLength：0≤x≤10∫010​1+m2​dx=101+m2​

如图所示，就可以得出此函数的弧长…
Ex2:$y=\sqrt{1-x^2}$y=1−x2​
$y'=\frac{-x}{\sqrt{1-x^2}}\\ ds=\sqrt{1+(y')^2}=\sqrt{\frac{1}{1-x^2}}\\ l=\int_0^a\frac{dx}{\sqrt{1-x^2}}=sin^{-1}x|_0^a\\ l=sin^{-1}a,同理得，sin l = a$y′=1−x2​−x​ds=1+(y′)2​=1−x21​​l=∫0a​1−x2​dx​=sin−1x∣0a​l=sin−1a,同理得，sinl=a

Ex3:抛物线的长度：图形如图所示：

$y=x^2,y'=2x\\ ds=\sqrt{1+(2x)^2}dx\\ arc(0 \leq x \leq a)=\int_0^a\sqrt{1+4x^2}dx\\ =\frac{1}{2}\int_0^asec^3udu\\ =[\frac{1}{4}ln(2x+\sqrt{1+4x^2})+\frac{1}{2}*\sqrt{1+4x^2}]|_0^a$y=x2,y′=2xds=1+(2x)2​dxarc(0≤x≤a)=∫0a​1+4x2​dx=21​∫0a​sec3udu=[41​ln(2x+1+4x2​)+21​∗1+4x2​]∣0a​
Ex4曲面面积：$y=x^2绕x轴旋转。$y=x2绕x轴旋转。

$dA=(2\pi y)ds\\ 表面积=\int_0^a2 \pi x^2\sqrt{1+4x^2}dx$dA=(2πy)ds表面积=∫0a​2πx21+4x2​dx
Ex5:球面面积:

$y=\sqrt{a^2-x^2},y'=\frac{-x}{\sqrt{a^2-x^2}}\\ 1+(\frac{-x}{\sqrt{a^2-x^2}})^2=\frac{a^2}{a^2-x^2}\\ area=\int_{x1}^{x2}2 \pi y ds\\ =\int_{x_1}^{x2}2 \pi \sqrt{a^2-x^2}\sqrt{\frac{a^2}{a^2-x^2}}dx\\ =\int_{x_1}^{x2}2\pi adx\\ =2\pi adx=2\pi a(x_2-x_1)$y=a2−x2​,y′=a2−x2​−x​1+(a2−x2​−x​)2=a2−x2a2​area=∫x1x2​2πyds=∫x1​x2​2πa2−x2​a2−x2a2​​dx=∫x1​x2​2πadx=2πadx=2πa(x2​−x1​)
Ex6:$\left\{\begin{matrix} x = 2 sint& \\ y = cost& \end{matrix}\right.${x=2sinty=cost​​,如图所示：

$\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$41​x2+y2=sin2t+cos2t=1dtds​=(2cost)2+(sint)2​弧长=∫02π​4cos2t+sin2t​dt椭球的表面积：绕y轴旋转dA=∫0π​2π(2sint)4cos2t+sin2t​dt