[Date Index]
[Thread Index]
[Author Index]
RE: 3D plot
*To*: mathgroup at smc.vnet.net
*Subject*: [mg36395] RE: [mg36364] 3D plot
*From*: "DrBob" <drbob at bigfoot.com>
*Date*: Wed, 4 Sep 2002 21:22:06 -0400 (EDT)
*Reply-to*: <drbob at bigfoot.com>
*Sender*: owner-wri-mathgroup at wolfram.com
First define the functions
z[x_] := x^2 + 4;
y1[x_] := 4 - x^2;
y2[x_] := 3x;
Figure out where the y limits coincide:
Solve[y1[x] == y2[x], x]
{{x -> -4}, {x -> 1}}
Which is bigger in between?
y1[0] - y2[0]
4
y1 is. Confirm that with a plot. (y2 is linear):
Plot[{y1[x], y2[x]}, {x, -5, 5}]
Next figure out where z is zero:
Solve[z[x] == 0, x]
{{x -> -2*I}, {x -> 2*I}}
It's never zero for real x, but is it positive, or negative?
z[0]
4
Look at the plot, just for fun:
Plot[{z[x]}, {x, -4, 1}]
z is positive for all x (but only the interval [-4,1] MATTERS). The
volume you want, therefore, is
Integrate[z[x]*(y1[x] - y2[x]), {x, -4, 1}]
625/4
or:
g[x_] = Integrate[z[x]*(y1[x] - y2[x]), x]
g[1] - g[-4]
16*x - 6*x^2 - (3*x^4)/4 - x^5/5
625/4
Bobby Treat
-----Original Message-----
From: Shz Shz Oon [mailto:OONSSHZ at hitachi.com.my]
To: mathgroup at smc.vnet.net
Subject: [mg36395] [mg36364] 3D plot
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: Mathematica 4.2 Notebook warning with Mathematica 4.1**
Next by Date:
**Trouble with Export**
Previous by thread:
**Re: 3D plot -Reply**
Next by thread:
**RE: Re: 3D plot -Reply**
| |