Volume of Solid

  • Thread starter shards5
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  • #1
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Homework Statement



Find the volume of the solid bounded by the planes x = 0, y = 0, z = 0, and x + y + z = 9.

Homework Equations



. . . ?

The Attempt at a Solution


After drawing out the picture with z=0 I have a line going from 0,9 to 9,0 bounded by the x and y axis giving me a triangle.
Based on that I got the following domains.
0 <= x <= 9
0 <= y <= 9-x
Which I then use for the following double integral
[tex] \int^{9}_{0}\int^{9-x}_{0} 9 - x - y dy dx [/tex]
After the first integration I get.
9y-(y2)/2-x
After plugging in the limits and simplifying I get 81/2-x^2-x
After integrating the above I get: 81/2x-x3/3-x2/2
and plugging and chugging gives me 81 which is wrong. So . . . did I do my domain wrong or it is an integration mistake?
 

Answers and Replies

  • #2
LCKurtz
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Your setup is correct. Without seeing your steps in a readable form it's hard to tell you where you went wrong.
 
  • #3
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Your limits of integration look OK. I suspect you made an integration error or arithmetic mistake.

[tex] \int^{9}_{0}\int^{9-x}_{0} 9 - x - y~dy~dx [/tex]

After integrating with respect to y, I get
[tex]\int_0^9 81/2 - 9x + x^2/2~dx[/tex]

I get a value of 243.
 
  • #4
LCKurtz
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Your limits of integration look OK. I suspect you made an integration error or arithmetic mistake.

[tex] \int^{9}_{0}\int^{9-x}_{0} 9 - x - y~dy~dx [/tex]

After integrating with respect to y, I get
[tex]\int_0^9 81/2 - 9x + x^2/2~dx[/tex]

I get a value of 243.

Should get (1/6)*9*9*9 = 243/2.
 
  • #5
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Oops! I lost my denominator of 2 when I integrated x^2/2.
 

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