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Notes

空间解析几何

距离公式

两点之间

A(x1,y1,z1),B(x2,y2,z2)A(x_{1},y_{1},z_{1}), B(x_{2},y_{2},z_{2}) 两点之间的距离

d=(x1−x2)2+(y1−y2)2+(z1−z2)2d = \sqrt{(x_{1}-x_{2})^{2} + (y_{1}-y_{2})^{2} + (z_{1}-z_{2})^{2} }

点到平面

平面 Σ:Ax+By+Cz+D=0\Sigma: Ax+By+Cz+D=0，点 M(x0,y0,z0)M(x_{0},y_{0},z_{0}) 之间的距离

d=∣Ax0+By0+Cz0+D∣A2+B2+C2d = { |Ax_{0}+By_{0}+Cz_{0}+D| \over \sqrt{A^{2} + B^{2} + C^{2} } }

点到直线

直线 L:x−x0m=y−y0n=z−z0pL: \displaystyle{x-x_{0} \over m} = {y - y_{0} \over n} = {z - z_{0} \over p}，点 M1(x1,y1,z1)∉LM_{1}(x_{1},y_{1},z_{1}) \not \in L

令 M0(x0,y0,z0)M_{0}(x_{0},y_{0},z_{0})，s⃗={m,n,p}\vec s=\{m, n, p\}，则点 M1M_{1} 到 L\displaystyle{ L } 的距离

d=∣M0M1→×s⃗∣∣s⃗∣d = { \left|\overrightarrow{M_{0}M_{1} } \times \vec s \right| \over |\vec s| }

球极坐标

变量如何转换

{x=rsin⁡φcos⁡θy=rsin⁡φsin⁡θz=rcos⁡φ\begin{cases} x = r \sin \varphi \cos \theta \\ y = r \sin \varphi \sin \theta \\ z = r \cos \varphi \end{cases}
  • φ\varphi 为球上一点与原点连线和 z\displaystyle{ z } 轴正方向的夹角，φ∈[0,π]\varphi \in [0, \pi]
  • θ\theta 为球上一点所在的横截面上，半径与 x\displaystyle{ x } 轴正方向的旋转角，θ∈[0,2π]\theta \in [0, 2 \pi]
dV=r2sin⁡φdrdθdφ\mathrm{d} V = r^{2} \sin \varphi \mathrm{d} r \mathrm{d} \theta \mathrm{d} \varphi

球极坐标积分

I=∭Ωf(x,y,z)dxdydz=∫02πdθ∫0πdφ∫0rf(x,y,z)r2sin⁡φdr\begin{aligned} I & = \underset{\Omega} {\iiint} f(x,y,z) \mathrm {d} x \mathrm{d} y \mathrm{d} z \\ & = \int _{0}^{2\pi} \mathrm{d} \theta \int_{0}^{\pi} \mathrm{d} \varphi \int _{0}^{r} f(x,y,z) r^{2} \sin \varphi \mathrm{d} r \end{aligned}

单叶双曲面和双叶双曲面

单叶双曲面

x2a2+y2b2−z2c2=1{x^{2} \over a^{2} } + {y^{2} \over b^{2} } - {z^{2} \over c^{2} } = 1

双叶双曲面

x2a2+y2b2−z2c2=−1{x^{2} \over a^{2} } + {y^{2} \over b^{2} } - {z^{2} \over c^{2} } = -1

旋转曲面表面积

参数方程形式

光滑参数曲线 C\displaystyle{ C } 由参数方程

{x=x(t)y=y(t)\displaystyle {\left\lbrace\begin{matrix*}[l] x= x{\left( t\right)}\\ y= y{\left( t\right)}\\\end{matrix*}\right.}

其中，t∈[α,β]\displaystyle t\in{\left[\alpha,\beta\right]}，且 y(t)≥0\displaystyle y{\left( t\right)}\ge 0. 则曲线 C\displaystyle{ C } 绕 x\displaystyle{ x } 轴旋转所得的旋转曲面表面积为

S=2π∫αβy(t)x′2(t)+y′2(t)dt\displaystyle S= 2\pi\int_{\alpha}^{\beta} y{\left( t\right)}\sqrt{ {\left. x^\prime\right.}^{2}{\left( t\right)}+{\left. y^\prime\right.}^{2}{\left( t\right)} }\text{d} t

直角坐标形式

{x=xy=y(x)\displaystyle {\left\lbrace\begin{matrix*}[l] x= x\\ y= y{\left( x\right)}\\\end{matrix*}\right.}

曲线 C\displaystyle{ C } 绕 x\displaystyle{ x } 轴旋转所得旋转曲面的表面积为

S=2π∫αβy(x)dS=2π∫αβy(x)1+y′2(x)dx\displaystyle S= 2\pi\int_{\alpha}^{\beta} y{\left( x\right)}\text{d} S= 2\pi\int_{\alpha}^{\beta} y{\left( x\right)}\sqrt{1+{\left. y^\prime\right.}^{2}{\left( x\right)} }{\left.\text{d} x\right.}

绕 y\displaystyle{ y } 轴旋转所得旋转曲面的表面积为

S=2π∫αβx(y)dS=2π∫αβx(y)1+x′2(y)dx\displaystyle S= 2\pi\int_{\alpha}^{\beta} x{\left( y\right)}\text{d} S= 2\pi\int_{\alpha}^{\beta} x{\left( y\right)}\sqrt{1+{\left. x^\prime\right.}^{2}{\left( y\right)} }{\left.\text{d} x\right.}

极坐标形式

由于用极坐标表示有

{x=r(θ)cos⁡θy=r(θ)sin⁡θ\displaystyle {\left\lbrace\begin{matrix*}[l] x= r{\left(\theta\right)} \cos{\theta}\\ y= r{\left(\theta\right)} \sin{\theta}\\\end{matrix*}\right.}

求导有

{x′=r′(θ)cos⁡θ−r(θ)sin⁡θy′=r′(θ)sin⁡θ+r(θ)cos⁡θ\displaystyle {\left\lbrace\begin{matrix*}[l] x^\prime= r^\prime{\left(\theta\right)} \cos{\theta}- r{\left(\theta\right)} \sin{\theta}\\ y^\prime= r^\prime{\left(\theta\right)} \sin{\theta}+ r{\left(\theta\right)} \cos{\theta}\\\end{matrix*}\right.}

则

S=2π∫θ1θ2r(θ)sin⁡θr2(θ)+r′2(θ)dθ\displaystyle S= 2\pi\int_{\theta_{1} }^{\theta_{2} } r{\left(\theta\right)} \sin{\theta}\sqrt{r^{2}{\left(\theta\right)}+{\left. r^\prime\right.}^{2}{\left(\theta\right)} }\text{d}\theta

弧长公式

参数方程形式

光滑参数曲线 C\displaystyle{ C } 由参数方程

{x=x(t)y=y(t)\displaystyle {\left\lbrace\begin{matrix*}[l] x= x{\left( t\right)}\\ y= y{\left( t\right)}\\\end{matrix*}\right.}

其中 t∈[α,β]\displaystyle t\in{\left[\alpha,\beta\right]}，其弧长为

s=∫αβx′2(t)+y′2(t)dt\displaystyle s=\int_{\alpha}^{\beta}\sqrt{ {\left. x^\prime\right.}^{2}{\left( t\right)}+{\left. y^\prime\right.}^{2}{\left( t\right)} }\text{d} t

直角坐标形式

{x=xy=y(x)\displaystyle {\left\lbrace\begin{matrix*}[l] x= x\\ y= y{\left( x\right)}\\\end{matrix*}\right.} s=∫αβ1+y′2(t)dx\displaystyle s=\int_{\alpha}^{\beta}\sqrt{1+{\left. y^\prime\right.}^{2}{\left( t\right)} }{\left.\text{d} x\right.}

极坐标形式

由于用极坐标表示有

{x=r(θ)cos⁡θy=r(θ)sin⁡θ\displaystyle {\left\lbrace\begin{matrix*}[l] x= r{\left(\theta\right)} \cos{\theta}\\ y= r{\left(\theta\right)} \sin{\theta}\\\end{matrix*}\right.}

求导有

{x′=r′(θ)cos⁡θ−r(θ)sin⁡θy′=r′(θ)sin⁡θ+r(θ)cos⁡θ\displaystyle {\left\lbrace\begin{matrix*}[l] x^\prime= r^\prime{\left(\theta\right)} \cos{\theta}- r{\left(\theta\right)} \sin{\theta}\\ y^\prime= r^\prime{\left(\theta\right)} \sin{\theta}+ r{\left(\theta\right)} \cos{\theta}\\\end{matrix*}\right.}

则

s=∫θ1θ2r2(θ)+r′2(θ)dθ\displaystyle s=\int_{\theta_{1} }^{\theta_{2} }\sqrt{r^{2}{\left(\theta\right)}+{\left. r^\prime\right.}^{2}{\left(\theta\right)} }\text{d}\theta