Re: 3D plot

*To*: mathgroup at smc.vnet.net*Subject*: [mg36372] Re: [mg36364] 3D plot*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Wed, 4 Sep 2002 02:56:26 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

Mathematica could do this sort of thing if there were a three dimensional object described by your equations (as boundaries) but there isn't one. More precisely, the pair of equations {z=x^2 +4, y=4-x^2} describes a parabola in three space which you can plot with: g1=ParametricPlot3D[{x, 4 - x^2, x^2 + 4}, {x, -5, 5}] The equation y=3x describes the plane: g2 = ParametricPlot3D[{x, 3x, z}, {x, -5, 5}, {z, -25, 25}] You can see the two together in <<RealTime3D` Show[{g1,g2}] There are clearly two points of intersection. They can be found with: Solve[{z == x^2 + 4, y == 4 - x^2, y == 3*x}, {x, y, z}] {{z -> 5, y -> 3, x -> 1}, {z -> 20, y -> -12, x -> -4}} So where is the 3D object whose volume you want to find? Andrzej Kozlowski Toyama International University JAPAN On Tuesday, September 3, 2002, at 06:41 am, Shz Shz Oon wrote: > > Can I use Mathematica to find out the volumn of this 3 dimensional > object from > the equations : > > z=x^2 +4, y=4-x^2, y=3x > > > Thanks in advance! > Shz Shz > > >