# 多重积分

# 柱面坐标与球面坐标

# 柱面坐标

(r,θ,z)(r,\theta,z) 中 ,r,θr,\theta表示在xOy\textit{xOy}平面上的投影位置,zz表示与xOy\textit{xOy}平面的距离

x=rcosθy=rsinθr2=x2+y2z=z x = rcos\theta \\ y = rsin\theta \\ r^2 = x^2 + y^2 \\ z = z

# 柱面坐标上的积分

dV=rdrθdzdV = rdr\theta dz

函数ff在立体区域VV上的积分公式就是

Vf(r,θ,z)rdrdθdz \iiint\limits_V f(r,\theta,z)rdrd\theta dz

# 球面坐标

x=rsinφcosθy=rsinφsinθz=rcosφ x = r\sin\varphi \cos\theta \\ y = r\sin\varphi \sin\theta \\ z = r\cos\varphi

# 球面坐标上的积分

ΔVΔrrΔθrsinθΔφ=r2sinθΔφΔrΔθVf(x,y,z)dV=Vf(r,θ,φ)r2sinθdrdθdφ \begin{align} & \begin{align} \Delta V &\approx \Delta r\cdot r\Delta\theta \cdot r\sin\theta\Delta\varphi \\ &= r^2\sin\theta\Delta\varphi \Delta r\Delta\theta \end{align}\\\\ &\iiint\limits_V f(x,y,z)dV = \iiint\limits_V f(r,\theta,\varphi )r^2\sin\theta drd\theta d\varphi \end{align}

# 质量、质心、矩

# 质量

σ(x,y,z)为密度函数质量=Vσ(x,y,z)dxdydz当密度σ(x,y,z)=1时,质量等于体积 \begin{align} & \sigma(x,y,z) 为密度函数 \\\\ & 质量 = \iiint\limits_V \sigma(x,y,z)dxdydz \\\\ & 当密度 \sigma(x,y,z) = 1 时,质量等于体积 \end{align}

# 质心

x=Vσ(x,y,z)xdxdydzVσ(x,y,z)dxdydzy=Vσ(x,y,z)ydxdydzVσ(x,y,z)dxdydzz=Vσ(x,y,z)zdxdydzVσ(x,y,z)dxdydz \begin{align} \overline{x} = \frac{ \iiint\limits_V\sigma(x,y,z)x dxdydz }{ \iiint\limits_V\sigma(x,y,z) dxdydz } \\\\ \overline{y} = \frac{ \iiint\limits_V\sigma(x,y,z)y dxdydz }{ \iiint\limits_V\sigma(x,y,z) dxdydz } \\\\ \overline{z} = \frac{ \iiint\limits_V\sigma(x,y,z)z dxdydz }{ \iiint\limits_V\sigma(x,y,z) dxdydz } \\\\ \end{align}

# 惯性矩

对于质点,它的惯性矩等于mr2mr^2,其中mm表示质量,而rr则是该质点与转动中心之间的距离

质量密度为σ(x,y,z)的物体惯性矩等于Vσ(x,y,z)(x2+y2)dxdydz柱面坐标:Vσ(r,θ,z)(r2)rdrdθdz \begin{align} & 质量密度为\sigma(x,y,z)的物体惯性矩等于 \\\\ & \iiint\limits_V\sigma(x,y,z)(x^2+y^2)dxdydz \\\\ & 柱面坐标: \\\\ & \iiint\limits_V\sigma(r,\theta,z)(r^2)rdrd\theta dz \\\\ \end{align}

# 坐标变换

在切换坐标系的时候,应该用什么取代dV=dxdydzdV = dxdydz?

假设在某一个新的坐标系里,描述空间中一个点的坐标变成了u,v,wu,v,w,而不是x,y,zx,y,z。假设这套坐标与x,y,zx,y,z的关系为x=fx(u,v,w),y=fy(u,v,w),z=fz(u,v,w)x=f_x(u,v,w),y=f_y(u,v,w),z=f_z(u,v,w)

此时,函数f(x,y,z)f(x,y,z)的三重积分公式Vf(x,y,z)dV\iiint\limits_Vf(x,y,z)dV就会变换成Vf(u,v,w)dV\iiint\limits_Vf(u,v,w)dV,但是后边这个dVdV得用新的坐标系来表示,也就是要找出在uvwuvw坐标系下,一个小盒子形状区域的体积。

设向量r=xi+yj+zk\bold{r} = x\bold{i} + y\bold{j} + z\bold{k},则盒子状区域的体积差不多与下边三个向量的三重积相等

ru,rv,rw, \frac{ \partial \bold{r} }{\partial u}, \frac{ \partial \bold{r} }{\partial v}, \frac{ \partial \bold{r} }{\partial w},

即,dVdV等于这三个向量的三重内积

dV=xuyuzuxvyvzvxwywzw雅克比行列式dudvdw=(x,y,z)(u,v,w)雅克比行列式缩写dudvdw dV = \underbrace{ \begin{vmatrix} \frac{ \partial x }{\partial u} & \frac{ \partial y }{\partial u} & \frac{ \partial z }{\partial u} \\\\ \frac{ \partial x }{\partial v} & \frac{ \partial y }{\partial v} & \frac{ \partial z }{\partial v} \\\\ \frac{ \partial x }{\partial w} & \frac{ \partial y }{\partial w} & \frac{ \partial z }{\partial w} \\\\ \end{vmatrix}}_{ 雅克比行列式 } du\,dv\,dw = \underbrace{ \left|\frac{\partial(x,y,z)}{\partial(u,v,w)}\right| }_{雅克比行列式缩写}du\,dv\,dw

雅克比行列式,代表了u,v,wu,v,w坐标和x,y,zx,y,z坐标之间的体积伸缩因子

由此,u,v,wu,v,w坐标系下的三重积分可以写成

Vf(u,v,w)(x,y,z)(u,v,w)dudvdw \iiint\limits_Vf(u,v,w)\left|\frac{\partial(x,y,z)}{\partial(u,v,w)}\right|du\,dv\,dw

例:从直角坐标(x,y,z)(x,y,z)到球面坐标(ρ,φ,θ)(\rho,\varphi,\theta)的转换

x=ρsinφcosθy=ρsinφsinθz=ρcosφr=ρsinφcosθi+ρsinφsinθj+ρcosφkrρ=sinφcosθi+sinφsinθj+cosφkrφ=ρcosφcosθi+ρcosφsinθjρsinφkrθ=ρsinφsinθi+ρsinφcosθjsinφcosθsinφsinθcosφρcosφcosθρcosφsinθρsinφρsinφsinθρsinφcosθ0=ρ2sinφVf(x,y,z)dV=Vf(ρ,φ,θ)ρ2sinθdρdφdθ \begin{align} & x = \rho\sin\varphi\cos\theta \\ & y = \rho\sin\varphi\sin\theta \\ & z = \rho\cos\varphi \\\\ & \bold{r} = \rho\sin\varphi\cos\theta\bold{i} + \rho\sin\varphi\sin\theta\bold{j} + \rho\cos\varphi\bold{k} \\ & \frac{\partial{\bold{r}}}{\partial{\rho}} =\sin\varphi\cos\theta \bold{i} + \sin\varphi\sin\theta\bold{j} + \cos\varphi\bold{k} \\ & \frac{\partial{\bold{r}}}{\partial{\varphi}} = \rho\cos\varphi\cos\theta\bold{i} + \rho\cos\varphi\sin\theta\bold{j} - \rho\sin\varphi\bold{k} \\ & \frac{\partial{\bold{r}}}{\partial{\theta}} = -\rho\sin\varphi\sin\theta \bold{i} + \rho\sin\varphi\cos\theta\bold{j} \\\\ & \begin{vmatrix} \sin\varphi\cos\theta & \sin\varphi\sin\theta & \cos\varphi \\ \rho\cos\varphi\cos\theta & \rho\cos\varphi\sin\theta & -\rho\sin\varphi \\ -\rho\sin\varphi\sin\theta & \rho\sin\varphi\cos\theta & 0 \end{vmatrix} = \rho^2\sin\varphi \\\\ &\therefore \iiint\limits_V f(x,y,z)dV = \iiint\limits_V f(\rho,\varphi,\theta )\rho^2\sin\theta d\rho\,d\varphi\,d\theta \end{align}