1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fubini's Theorem apply? Volume

  1. Mar 5, 2008 #1
    Find the volume of the region bounded by [tex] x = - 1, \ x = 1 [/tex], [tex] y = - 1, y = 1 [/tex] and [tex] z = x [/tex].

    Is this equaled to [tex] \int_{0}^{1} \int_{ - 1}^{1} \int_{ - 1}^{1} dx \ dy \ dz [/tex]?

    Can it also equal [tex] \int_{ - 1}^{1} \int_{ - 1}^{1} \int_{0}^{x} \ dz \ dx \ dy [/tex]?
     
  2. jcsd
  3. Mar 5, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Neither is right. Your limits don't enclose any bounded region.
     
  4. Mar 5, 2008 #3
    How would I fix it?
     
  5. Mar 5, 2008 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Add another limit. Are you sure you copied all of the limits from the problem you want to solve?
     
  6. Mar 5, 2008 #5
    It also has to be above the [tex] x-y [/tex] plane. Yeah I copied it correctly. So maybe the problem is wrong? You could also take the determinant right?
     
  7. Mar 5, 2008 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The first integral is completely wrong: it is the volume of the rectangular solid [itex]0\le z\le 1[/itex], [itex]-1\le y\le 1[/itex], [itex]-1\le x\le 1[/itex].

    For the second integral, where did you get the lower limit z= 0? That is what Dick is suggesting you may have missed.
     
  8. Mar 5, 2008 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It could be above the xy plane, if you say so. But nothing in the original boundaries forces that to be true. Take the determinant of what?
     
  9. Mar 5, 2008 #8
    If it is above the [itex] x-y [/itex] plane, then [itex] z [/itex] cannot be negative. Its bounded by the [itex] x-y [/itex] plane.
     
    Last edited: Mar 5, 2008
  10. Mar 5, 2008 #9

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    And if z can't be negative, then your lower x limit isn't -1 either.
     
  11. Mar 5, 2008 #10
    Then it would be [tex] \int_{-1}^{-1} \int_{0}^{1} \int_{0}^{x} \ dz \ dx \ dy [/tex]?
     
  12. Mar 5, 2008 #11

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If you add the restriction z>0 as well, yes.
     
  13. Mar 5, 2008 #12
    Thank you for your help Dick and HallsofIvy.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Fubini's Theorem apply? Volume
  1. Fubini's theorem help (Replies: 3)

Loading...