Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2002
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2002

[Date Index] [Thread Index] [Author Index]

Search the Archive

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
>
>
>



  • Prev by Date: Re:Generating Two Unit Orthogonal Vectors to a 3D Vector
  • Next by Date: Re: 3D plot
  • Previous by thread: 3D plot
  • Next by thread: Re: 3D plot