第一型曲线积分

• 参数方程表达$x=x(t),y=y(t),z=z(t)$

$$dl=\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2+z^{\prime }(t{)}^2}dt$$

• 特殊参数方程：

  • 极坐标

  $$\begin{gathered} x=r(t)\cos \theta (t),y=r(t)\sin \theta (t) \\ dl=\sqrt{(dr{)}^2+(rd\theta {)}^2} \\ =\sqrt{r^{\prime 2}+r^2{\theta }^{\prime 2}}dt \end{gathered}$$

  • 柱坐标

  $$\begin{gathered} x=r(t)\cos \theta (t) \\ y=r(t)\sin \theta (t) \\ z=z(t) \\ dl=\sqrt{(dr{)}^2+(rd\theta {)}^2+(dz{)}^2} \end{gathered}$$

  • 球坐标

  $$\begin{gathered} x=r(t)\sin \theta (t)\cos \phi (t) \\ y=r(t)\sin \theta (t)\sin \phi (t) \\ z=r(t)\cos \theta (t) \\ dl=\sqrt{(dr{)}^2+(r\sin \theta d\phi {)}^2+(rd\theta {)}^2} \end{gathered}$$

  • 上式的计算方法：

  $$\begin{gathered} \left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}r\sin \theta \cos \phi \\ r\sin \theta \sin \phi \\ r\cos \theta \end{array}\right)分别对r,\phi ,\theta 求导得 \\ \left(\begin{array}{c}\sin \theta \cos \phi \\ \sin \theta \sin \phi \\ \cos \theta \end{array}\right),\left(\begin{array}{c}r\sin \theta \cdot (-\sin \phi )\\ r\sin \theta \cos \phi \\ 0\end{array}\right),\left(\begin{array}{c}r\cos \theta \cos \phi \\ r\cos \theta \sin \phi \\ -r\sin \theta \end{array}\right) \\ 将上面三个向量单位化后的向量记作e_r,e_{\phi },e_{\theta }，看作dr,d\phi ,d\theta ， \\ 那么求导得到的三个向量就是e_r,r\sin \theta e_{\phi },r{e}_{\theta }，对应了三个正交的坐标方向的增量 \\ dl=\sqrt{(dr{)}^2+(r\sin \theta d\phi {)}^2+(d\theta {)}^2} \end{gathered}$$

第一型曲面积分

• 参考重积分应用中已经涉及到的面积微元

• 在参数方程$(x,y,z)=f(u,v)$下

$$\begin{gathered} r_u=(\frac{\partial x}{\partial u},\frac{\partial y}{\partial u},\frac{\partial z}{\partial u}) \\ {r}_v=(\frac{\partial x}{\partial v},\frac{\partial y}{\partial v},\frac{\partial z}{\partial v}) \\ dS=\sqrt{det\left|\begin{array}{cc}{r}_u\cdot r_u& r_u\cdot r_v\\ r_v\cdot r_u& r_v\cdot r_v\end{array}\right|}dudv \end{gathered}$$

• $du,dv$前的系数恰好是$r_u,r_v$张成的平行四边形的面积

$${\iint }_{\Omega }f(x,y,z)dS={\iint }_{{\Omega }^{\ast }}f(u,v)\sqrt{det\left|\begin{array}{cc}{r}_u\cdot r_u& r_u\cdot r_v\\ r_v\cdot r_u& r_v\cdot r_v\end{array}\right|}dudv$$

• 特别的在$(x,y,z)=(x,y,z(x,y))$下

$$dS=\sqrt{1+(\frac{\partial z}{\partial x}{)}^2+(\frac{\partial z}{\partial y}{)}^2}dxdy$$

• 另有在$n$维空间中的$n-1$维超平面可以用$n-1$个基的叉积的模长计算，泛用性不如上面