解析几何 周维祺
一条曲线绕一条定直线旋转一周所得的曲面
上文中的曲线称为该曲面的母线
上文中的直线称为该曲面的（旋转）轴
母线方程：{F(x0,y0,z0)=0G(x0,y0,z0)=0\begin{cases}F(x_0,y_0,z_0)=0 \\ G(x_0,y_0,z_0)=0\end{cases}{F(x0,y0,z0)=0G(x0,y0,z0)=0;
轴过点(a0,b0,c0)(a_0,b_0,c_0)(a0,b0,c0)，方向(a,b,c)(a,b,c)(a,b,c)
曲线上一点M(x0,y0,z0)M(x_0,y_0,z_0)M(x0,y0,z0)的运动轨迹所在平面与轴垂直
圆上每一点到轴上一个定点(a0,b0,c0)(a_0,b_0,c_0)(a0,b0,c0)的距离都相等
联立，得：{F(x0,y0,z0)=0G(x0,y0,z0)=0r2=(x0−a0)2+(y0−b0)2+(z0−c0)2(x−a0)2+(y−b0)2+(z−c0)2=r2a(x−x0)+b(y−y0)+c(z−z0)=0\begin{cases}F(x_0,y_0,z_0)=0 \\ G(x_0,y_0,z_0)=0 \\ r^2=(x_0-a_0)^2+(y_0-b_0)^2+(z_0-c_0)^2 \\ (x-a_0)^2+(y-b_0)^2+(z-c_0)^2=r^2 \\ a(x-x_0)+b(y-y_0)+c(z-z_0)=0 \end{cases}⎩⎨⎧F(x0,y0,z0)=0G(x0,y0,z0)=0r2=(x0−a0)2+(y0−b0)2+(z0−c0)2(x−a0)2+(y−b0)2+(z−c0)2=r2a(x−x0)+b(y−y0)+c(z−z0)=0;
消去x0,y0,z0x_0,y_0,z_0x0,y0,z0得关于x,y,zx,y,zx,y,z的方程即旋转曲面的方程
过点(0,0,1)(0,0,1)(0,0,1)，方向为(2,1,0)(2,1,0)(2,1,0)的直线，绕直线x=y=zx=y=zx=y=z旋转
设(x0,y0,z0)(x_0,y_0,z_0)(x0,y0,z0)是母线上一点,则(x0,y0,z0)=(2t,t,1)(x_0,y_0,z_0)=(2t,t,1)(x0,y0,z0)=(2t,t,1)
轴方向：(1,1,1)(1,1,1)(1,1,1)；轴上一点：(0,0,0)(0,0,0)(0,0,0)
{(x−x0)+(y−y0)+(z−z0)=0x2+y2+z2=x02+y02+z02\begin{cases}(x-x_0)+(y-y_0)+(z-z_0)=0 \\ x^2+y^2+z^2=x_0^2+y_0^2+z_0^2 \end{cases}{(x−x0)+(y−y0)+(z−z0)=0x2+y2+z2=x02+y02+z02
{(x0,y0,z0)=(2t,t,1)(x−x0)+(y−y0)+(z−z0)=0x2+y2+z2=x02+y02+z02\begin{cases} (x_0,y_0,z_0)=(2t,t,1) \\ (x-x_0)+(y-y_0)+(z-z_0)=0 \\ x^2+y^2+z^2=x_0^2+y_0^2+z_0^2 \end{cases}⎩⎨⎧(x0,y0,z0)=(2t,t,1)(x−x0)+(y−y0)+(z−z0)=0x2+y2+z2=x02+y02+z02
{x+y+z−1=3tx2+y2+z2=5t2+1\begin{cases} x+y+z-1=3t \\ x^2+y^2+z^2=5t^2+1 \end{cases}{x+y+z−1=3tx2+y2+z2=5t2+1
x2+y2+z2=59(x+y+z−1)2+1x^2+y^2+z^2=\frac{5}{9}(x+y+z-1)^2+1x2+y2+z2=95(x+y+z−1)2+1
求 {(y−b)2+z2=a2x=0\begin{cases} (y-b)^2+z^2=a^2 \\ x=0 \end{cases}{(y−b)2+z2=a2x=0绕zzz轴旋转所得的曲面
设(0,y0,z0)(0,y_0,z_0)(0,y0,z0)是母线上一点
轴方向：(0,0,1)(0,0,1)(0,0,1)；轴上一点：(0,0,0)(0,0,0)(0,0,0)
{x2+y2+z2=y02+z020⋅(x−0)+0⋅(y−y0)+1⋅(z−z0)=0\begin{cases} x^2+y^2+z^2=y_0^2+z_0^2 \\ 0\cdot(x-0)+0\cdot(y-y_0)+1\cdot(z-z_0)=0 \end{cases}{x2+y2+z2=y02+z020⋅(x−0)+0⋅(y−y0)+1⋅(z−z0)=0
z=z0,z=z_0,\quadz=z0, y0=±x2+y2y_0=\pm\sqrt{x^2+y^2}y0=±x2+y2, 代入(y0−b)2+z02=a2(y_0-b)^2+z_0^2=a^2(y0−b)2+z02=a2
{F(x,y)=0z=0\begin{cases} F(x,y)=0 \\ z=0 \end{cases}{F(x,y)=0z=0 绕xxx轴：F(x,±y2+z2)=0F(x,\pm\sqrt{y^2+z^2})=0F(x,±y2+z2)=0;
{F(x,y)=0z=0\begin{cases} F(x,y)=0 \\ z=0 \end{cases}{F(x,y)=0z=0 绕yyy轴：F(±x2+z2,y)=0F(\pm\sqrt{x^2+z^2},y)=0F(±x2+z2,y)=0;
绕zzz轴：仍在xyxyxy平面上
{F(y,z)=0x=0\begin{cases} F(y,z)=0 \\ x=0 \end{cases}{F(y,z)=0x=0 绕yyy轴：F(y,±x2+z2)=0F(y,\pm\sqrt{x^2+z^2})=0F(y,±x2+z2)=0;
{F(y,z)=0x=0\begin{cases} F(y,z)=0 \\ x=0 \end{cases}{F(y,z)=0x=0 绕zzz轴：F(±x2+y2,z)=0F(\pm\sqrt{x^2+y^2},z)=0F(±x2+y2,z)=0;
绕xxx轴：仍在yzyzyz平面上
{F(x,z)=0y=0\begin{cases} F(x,z)=0 \\ y=0 \end{cases}{F(x,z)=0y=0 绕xxx轴：F(x,±y2+z2)=0F(x,\pm\sqrt{y^2+z^2})=0F(x,±y2+z2)=0;
{F(x,z)=0y=0\begin{cases} F(x,z)=0 \\ y=0 \end{cases}{F(x,z)=0y=0 绕zzz轴：F(±x2+y2,z)=0F(\pm\sqrt{x^2+y^2},z)=0F(±x2+y2,z)=0;
绕yyy轴：仍在xzxzxz平面上
保留旋转轴坐标，用±其它两坐标平方和\pm\sqrt{其它两坐标平方和}±其它两坐标平方和代替另一坐标
求{b2x2+a2y2=a2b2z=0\begin{cases} b^2x^2+a^2y^2=a^2b^2 \\ z=0 \end{cases}{b2x2+a2y2=a2b2z=0 分别绕xxx轴和yyy轴旋转所得的曲面方程
旋转曲面的定义
母线和轴的概念
旋转曲面方程的计算
坐标平面上的曲线绕坐标轴旋转所得的方程