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anonymous

  • 5 years ago

I need to setup the integral in three different ways.. solid bounded between the planes x=1, y=0, and 2x+y=6, bounded below by z=0 and bounded above by z=sqrt(x). Its just setting up the different orders of integrals, no solving.

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  1. watchmath
    • 5 years ago
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    |dw:1327285708927:dw|

  2. anonymous
    • 5 years ago
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    \[\int\limits_{1}^{3}\int\limits_{0}^{6-2x}\int\limits_{0}^{\sqrt{x} } f(x, y, z) dz dx dy\] Is this at all correct?

  3. watchmath
    • 5 years ago
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    you want triple or double integral?

  4. anonymous
    • 5 years ago
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    triple

  5. anonymous
    • 5 years ago
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    Sorry for the confusion.

  6. watchmath
    • 5 years ago
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    sorry I am in the middle of something. I can't help you right now. I'll be back if you patient :).

  7. anonymous
    • 5 years ago
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    Thanks!

  8. watchmath
    • 5 years ago
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    it is almost right but the order should be dz dy dx and also f(x,y,z)=1

  9. anonymous
    • 5 years ago
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    so would it be correct if it was dz dy dx? I need to find 2 more orders that's not the same, but will ultimately bring me to the same answer. I don't need to solve for f(x,y,z), just change the orders around..

  10. watchmath
    • 5 years ago
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    the next easiest is in the order of dz dx dy

  11. anonymous
    • 5 years ago
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    \[\int\limits_{0}^{6}\int\limits_{1}^{(6-y)/2}\int\limits_{0}^{\sqrt{x}} f(x,y,z) dz dx dy\] Would that be correct?

  12. watchmath
    • 5 years ago
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    what is the intersection between x=1 and 2x+y=6?

  13. anonymous
    • 5 years ago
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    3

  14. anonymous
    • 5 years ago
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    If only there is a one on one chat thing..

  15. watchmath
    • 5 years ago
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    2(1)+y=6 so y=4

  16. watchmath
    • 5 years ago
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    so the limit for dy is from 0 to 4

  17. anonymous
    • 5 years ago
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    OH. okay so it would be \[\int\limits_{0}^{4}\int\limits_{1}^{(6-y)/2}\int\limits_{0}^{\sqrt{x}} dz dx dy\]?

  18. watchmath
    • 5 years ago
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    yes :)

  19. anonymous
    • 5 years ago
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    now how would i do something like dx dz dy?

  20. anonymous
    • 5 years ago
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    Would this work out also?\[\int\limits_{1}^{\sqrt{3}}\int\limits_{0}^{z^2}\int\limits_{0}^{6-2x} dy dx dz\]?

  21. watchmath
    • 5 years ago
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    let see

  22. anonymous
    • 5 years ago
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    I'm so lost... I know that its not right..

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