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nice

  • 5 years ago

how to rewrite a triple integral after changing limits ? ( from dxdzdy to dydxdz ) I have the question ...

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  1. nice
    • 5 years ago
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    second question in the attached file

  2. anonymous
    • 5 years ago
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    What is all that writing on the problem. Is that your attempt and you want us to check if it is right?

  3. nice
    • 5 years ago
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    no no it's the right answer, but I don't understand how !

  4. nice
    • 5 years ago
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    I need explanation ,, my exam is tomorrow :S :S :S

  5. amistre64
    • 5 years ago
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    1 or 2? or both?

  6. nice
    • 5 years ago
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    2 only ,,

  7. amistre64
    • 5 years ago
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    its all about translating what you got into a new direction to match the switch

  8. nice
    • 5 years ago
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    is there any strategy to get the right answer ?

  9. amistre64
    • 5 years ago
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    well, the outer one is a constant; moves from zero to it high point for starters; z = [0,4] \[\int_{0}^{4}dz\]

  10. amistre64
    • 5 years ago
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    the middle should be in terms of the outer... so it gets a z spot right?

  11. nice
    • 5 years ago
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    I think that I have always to draw ,, I think it's the only way !! but I was hopping to find simpler way ..

  12. amistre64
    • 5 years ago
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    simpler? maybe, but i tend to only now the hard way lol

  13. nice
    • 5 years ago
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    what are you studying ?

  14. amistre64
    • 5 years ago
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    whatever I can get my hands on :)

  15. nice
    • 5 years ago
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    yeah I mean simpler than drawing the graph

  16. amistre64
    • 5 years ago
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    ive taught myself all this stuff; and as i go thru the college courses I learn ways that i was to stupid to pick up on me own

  17. nice
    • 5 years ago
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    aha!! I'm suffering from my doctor in calculus this course, and I'm lost! So NOW I learned that I have to be independent specially on college,, right ?

  18. nice
    • 5 years ago
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    in college *

  19. amistre64
    • 5 years ago
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    dxdzdy dydxdz -------- -------- x: 4-2y-z y: (4-x-z)/2 x: 0 y: 0 z: 4-2y x: 4-z z: 0 x: 0 y: 2 z: 4 y: 0 z: 0 the first is a translation from point to plane x = 4-2y-z <=> y = (4-x-z)/2

  20. amistre64
    • 5 years ago
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    the second is a translation from the seems to keep the inner one ignored x = 4-2y-z (ignore the -2y from the inner) x=4-z

  21. amistre64
    • 5 years ago
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    same with the last? ignore the inners as 0? z = 4 -x translates to z=4

  22. amistre64
    • 5 years ago
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    but thats just a cursory view and i aint got nuthin to prove its a general rule

  23. amistre64
    • 5 years ago
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    maybe if we get some smarter than mes to verify or deny it :)

  24. amistre64
    • 5 years ago
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    see if it works on double integrals maybe....

  25. nice
    • 5 years ago
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    Thanks allot! I will see ...

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