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MathGroup Archive 2002

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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





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