空间解析几何

距离公式

两点之间

$\displaystyle{ A \left( x _{ \left\lbrace 1 \right\rbrace } , y _{ \left\lbrace 1 \right\rbrace } , z _{ \left\lbrace 1 \right\rbrace } \right) , B \left( x _{ \left\lbrace 2 \right\rbrace } , y _{ \left\lbrace 2 \right\rbrace } , z _{ \left\lbrace 2 \right\rbrace } \right) }$ 两点之间的距离

$$d = \sqrt{(x_{1}-x_{2})^{2} + (y_{1}-y_{2})^{2} + (z_{1}-z_{2})^{2} }$$

点到平面

平面 $\Sigma: Ax+By+Cz+D=0$，点 $\displaystyle{ M \left( x _{ \left\lbrace 0 \right\rbrace } , y _{ \left\lbrace 0 \right\rbrace } , z _{ \left\lbrace 0 \right\rbrace } \right) }$ 之间的距离

$$d = { |Ax_{0}+By_{0}+Cz_{0}+D| \over \sqrt{A^{2} + B^{2} + C^{2} } }$$

点到直线

直线 $L: \displaystyle{x-x_{0} \over m} = {y - y_{0} \over n} = {z - z_{0} \over p}$，点 $M_{1}(x_{1},y_{1},z_{1}) \not \in L$

令 $\displaystyle{ M _{ \left\lbrace 0 \right\rbrace } \left( x _{ \left\lbrace 0 \right\rbrace } , y _{ \left\lbrace 0 \right\rbrace } , z _{ \left\lbrace 0 \right\rbrace } \right) }$，$\vec s=\{m, n, p\}$，则点 $\displaystyle{ M _{ \left\lbrace 1 \right\rbrace } }$ 到 $\displaystyle{ L }$ 的距离

$$d = { \left|\overrightarrow{M_{0}M_{1} } \times \vec s \right| \over |\vec s| }$$

球极坐标

变量如何转换

$$\begin{cases} x = r \sin \varphi \cos \theta \\ y = r \sin \varphi \sin \theta \\ z = r \cos \varphi \end{cases}$$

• $\varphi$ 为球上一点与原点连线和 $\displaystyle{ z }$ 轴正方向的夹角，$\varphi \in [0, \pi]$
• $\theta$ 为球上一点所在的横截面上，半径与 $\displaystyle{ x }$ 轴正方向的旋转角，$\theta \in [0, 2 \pi]$

$$\mathrm{d} V = r^{2} \sin \varphi \mathrm{d} r \mathrm{d} \theta \mathrm{d} \varphi$$

球极坐标积分

$$\begin{aligned} I & = \underset{\Omega} {\iiint} f(x,y,z) \mathrm {d} x \mathrm{d} y \mathrm{d} z \\ & = \int _{0}^{2\pi} \mathrm{d} \theta \int_{0}^{\pi} \mathrm{d} \varphi \int _{0}^{r} f(x,y,z) r^{2} \sin \varphi \mathrm{d} r \end{aligned}$$

单叶双曲面和双叶双曲面

单叶双曲面

$${x^{2} \over a^{2} } + {y^{2} \over b^{2} } - {z^{2} \over c^{2} } = 1$$

双叶双曲面

$${x^{2} \over a^{2} } + {y^{2} \over b^{2} } - {z^{2} \over c^{2} } = -1$$

旋转曲面表面积

参数方程形式

光滑参数曲线 $\displaystyle{ C }$ 由参数方程

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

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

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

直角坐标形式

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

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

$$\displaystyle S= 2\pi\int_{\alpha}^{\beta} y{\left( x\right)}\text{d} S= 2\pi\int_{\alpha}^{\beta} y{\left( x\right)}\sqrt{1+{\left. y^\prime\right.}^{2}{\left( x\right)} }{\left.\text{d} x\right.}$$

绕 $\displaystyle{ y }$ 轴旋转所得旋转曲面的表面积为

$$\displaystyle S= 2\pi\int_{\alpha}^{\beta} x{\left( y\right)}\text{d} S= 2\pi\int_{\alpha}^{\beta} x{\left( y\right)}\sqrt{1+{\left. x^\prime\right.}^{2}{\left( y\right)} }{\left.\text{d} x\right.}$$

极坐标形式

由于用极坐标表示有

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

求导有

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

则

$$\displaystyle S= 2\pi\int_{\theta_{1} }^{\theta_{2} } r{\left(\theta\right)} \sin{\theta}\sqrt{r^{2}{\left(\theta\right)}+{\left. r^\prime\right.}^{2}{\left(\theta\right)} }\text{d}\theta$$

弧长公式

参数方程形式

光滑参数曲线 $\displaystyle{ C }$ 由参数方程

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

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

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

直角坐标形式

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

极坐标形式

由于用极坐标表示有

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

求导有

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

则

$$\displaystyle s=\int_{\theta_{1} }^{\theta_{2} }\sqrt{r^{2}{\left(\theta\right)}+{\left. r^\prime\right.}^{2}{\left(\theta\right)} }\text{d}\theta$$