Mathematica基础——三重积分

电脑

Mathematica

先来画图：RegionPlot3D[x^2+y^2<=z<=1,{x,-1,1},{y,-1,1},{z,-0.5,1.5}

求zhege图形的体积：Integrate[Boole[x^2+y^2<=z<=1],{x,-Infinity,Infinity},{y,-Infinity,Infinity},{z,-Infinity,Infinity}]

如果每一个点的密度都是该点的竖坐标，求这个“物体”的质量：Integrate[z Boole[x^2+y^2<=z<=1],    {x,-Infinity,Infinity},     {y,-Infinity,Infinity},      {z,-Infinity,Infinity}]

计算三重积分，题目如下图。Integrate[(x^2+y^2) Boole[x^2+y^2<=4&&Sqrt[x^2+y^2]<=z<=2],  {x,-Infinity,Infinity},   {y,-Infinity,Infinity},    {z,-Infinity,Infinity}]

对Sqrt[x^2+y^2+z^2]求积分，积分区域是x^2+y^2+z^2<=z。直接积分：Integrate[Sqrt[x^2+y^2+z^2] Boole[x^2+y^2+z^2<=z ] ,   {x,-Infinity,Infinity},    {y,-Infinity,Infinity},     {z,-Infinity,Infinity}]

转而尝试球坐标变换：令：{x->r Sin[u] Cos[v],y->r Sin[u] Sin[v],z->r Cos[u]}于是有：dx dy dz->(r^2)Sin[u] dr du dv原题可以化为：Integrate[r^2 Sin[u] Sqrt[x^2+y^2+z^2]/.{x->r Sin[u] Cos[v],y->r Sin[u] Sin[v],z->r Cos[u]},{v,0,2 Pi},{u,0,Pi/2},{r,0,Cos[u]}]