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Double Integrals - interpreting what are my boundaries?

  1. Apr 21, 2008 #1
    Hmm ...

    The solid enclosed by the parabolic cylinders y=1-x^2, y=x^2-1 and the planes x+y+z=2, 2x+2y-z+10=0 by subtracting two volumes.

    Ok, so I know what my limits should be from the parabolic cylinders, but how was I supposed to know that the 2x+2y-z+10=0 plane doesn't play a role in my integral? When I was initially doing this problem, I thought I had to subtract the top plane from the bottom plane (lol I know, but I was just trying things).
     
    Last edited: Apr 21, 2008
  2. jcsd
  3. Apr 21, 2008 #2
    If you are doing this in 2-d i.e. using dx.dy
    the plane only define height (so nothing to do with limits)
    Remember: Areas and volumes
     
  4. Apr 22, 2008 #3

    HallsofIvy

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    If the planes form the "top" and "bottom" you can do this in either of two ways:
    The integral
    [tex]\int\int \int_{f(x,y)}^{g(x,y)} dzdydx= \int\int (g(x,y)- f(x,y))dydx[/tex]
    The integral of dzdxdy is just the integral of the limits for z integrated with respect to x and y.

    If, as I said, the planes for the top and bottom, solve their equations for z and ust those for g(x,y) and f(x,y). As for the integration in the xy plane, determine where the parabolas intersect, integrate between those x vaues and for the dy integral, integrate from y= x2-1 to y= 1- x2.
     
  5. Apr 22, 2008 #4
    Halls, this exercise is in the double-integral section, but I do love triple :)

    Question though ... so I have it set up as

    [tex]-2\int_0^1\int_0^{1-x^2}(x+y-2)dydy[/tex]

    Can I not take advantage of symmetry? B/c everywhere I look has it as ... [tex]\int_{-1}^1\int_{x^2-1}^{1-x^2}(2-x-y)dydx[/tex] or am I just messing up my algebra?
     
  6. Apr 22, 2008 #5
    I meant -4
     
  7. Apr 22, 2008 #6

    HallsofIvy

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    Have you calculated those two integrals to see if they are different?
     
  8. Apr 23, 2008 #7
    Uh huh ...

    [tex]-4\int_0^1\int_0^{1-x^2}(x+y-2)dydx[/tex]

    [tex]-4\int_0^1\left[x(1-x^2)+\frac{(1-x^2)^2}{2}-2(1-x^2)\right]dx[/tex]

    [tex]-4\left(\frac 1 2-\frac 1 4+\frac{1-\frac 2 3+\frac 1 5}{2}-2+\frac 2 3\right)=\frac{49}{15}[/tex]

    Answer = 64/3

    I guess I can't take advantage of symmetry on this problem.
     
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