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
>
>
>