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Double Integral Test question

  1. Apr 6, 2005 #1
    I took a test today. I wanted to know if I set my limits up correctly and got the right answer, because I've been having problems with that. Okay, here is the question:

    A space is bounded by x = 0, y = 0, xy-plane, and the plane: 3x + 2y + z = 6. Find the volume using a double integral.

    So, this is how I went about the problem...

    Since the space is bounded by the xy-plane, I set z = 0 for the plane. This gave me:

    [tex]3x + 2y = 6, y = \frac{-3x+6}{2}[/tex]

    That's just an equation of a line, so I plotted that on the xy-plane.

    At x = 0, y = 3 (0,3)
    At y = 0, x = 2 (2,0)

    This is the hard part for me...setting limits. I got:

    [tex]0\leq x\leq 2[/tex]


    [tex]\frac{-3x+6}{2}\leq y\leq 0[/tex]

    *sidenote: If it's wrong, would someone like to show me a simple strategy to setting limits? Also, if there's anything else you could do to help me set limits on integration, that would be really helpful.

    To go on with the problem, my resulting double integral was:

    [tex]\int_{x=0}^{x=2} \int_{y= \frac{-3x+6}{2}}^{y=0} (-3x-2y+6)dydx[/tex]

    integrating with respect to y first, I got:

    [tex]\int_{x=0}^{x=2} (-\frac{9}{4} x^2 + 9x-18)dx[/tex]

    resulting in answer = 9

    Thanks for reviewing.


    (yay, I learned how to LaTeX :) )
    Last edited: Apr 6, 2005
  2. jcsd
  3. Apr 6, 2005 #2
    oops, i think this belongs in the hw section... :(.
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