Re: 3D plot -Reply

• To: mathgroup at smc.vnet.net
• Subject: [mg36389] Re: [mg36364] 3D plot -Reply
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Wed, 4 Sep 2002 02:56:48 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```The problem is that I do not think there is any bounded 3d body
described by your conditions. Anyway, this is how you can use
Mathematica (in general)  to solve this sort of problem.

You need two packages:

In[1]:=
<<Graphics`InequalityGraphics`

In[2]:=
<<Calculus`Integration`

To see your object use:

InequalityPlot3D[y > 3x && y < 4 - x^2 && z < x^2 + 4, {x}, {y}, {z}]

However, you will just get some error messages and a picture that looks
two dimensional. If you wanted the volume, evaluate:

In[3]:=
Integrate[Boole[y>3x&&y<4-x^2&&z<x^2+4],{x},{y},{z}]

Out[3]=
-Infinity

If you impose some limits on z you will get a finite positive answer,
but one that clearly is unbounded:

In[20]:=
NIntegrate[Boole[y>3x&&y<4-x^2&&z<x^2+4],{x},{y},{z,-100,100}]

Out[20]=
2239.58

In[21]:=
NIntegrate[Boole[y>3x&&y<4-x^2&&z<x^2+4],{x},{y},{z,-1000,1000}]

Out[21]=
20989.6

Andrzej

On Wednesday, September 4, 2002, at 03:08  am, Shz Shz Oon wrote:

> Hi Andrzej Kozlowski,
>
> Thanks for trying to help.
> Sorry, I must have something missing in my previous description.
>
> I need to find out the volumn of a 3D object which form by the
> equation :
> z=x^2 +4 (as bottom surface)
> and on the xy plane which bounded by a parabola y=4-x^2 and y=3x line.
>
> How would I use Mathematica to plot out this 3D object or find out its
> volumn with only the equation given?
>
> Thank you!
> Shz Shz
>
>
>>>> Andrzej Kozlowski <akoz at mimuw.edu.pl> 3/September/2002 04:33pm >>>
> 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
>>
>>
>>
> <DISCLAIM.TXT>

```

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