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Double Integral Test question

  1. Apr 6, 2005 #1
    *This was accidently posted in the 'Calculus & Analysis' section. Moderators can delete that one. Sorry.*

    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

    **Main questions:

    1) Are my limits of integration set correctly?
    2) Is my resulting answer correct?

    Thanks for reviewing.

    Last edited: Apr 6, 2005
  2. jcsd
  3. Apr 6, 2005 #2
    bump. i would really like some help on this.
  4. Apr 6, 2005 #3
    You have your y limits backward. The lowest possible value of y is zero and the value of y calculated from the line equation ranges from 0 to 3 for the allowed values of x.

    You made a small mistake in the y integration, the -18 should be -9. Although the answer is still 9.

    The answer with the correct y limit order is 6.
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