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M

  • 5 years ago

double integral

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  1. anonymous
    • 5 years ago
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    What do you need answered??

  2. M
    • 5 years ago
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    trying to use equation to write question but not working

  3. amistre64
    • 5 years ago
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    yeah, they blame it on user error lol

  4. M
    • 5 years ago
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    hmm i click and nothing

  5. amistre64
    • 5 years ago
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    {SS} dx.dy ; [a,b] [c,d] works

  6. anonymous
    • 5 years ago
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    hey amister!!!

  7. amistre64
    • 5 years ago
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    hi ;)

  8. amistre64
    • 5 years ago
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    i saw a cute pic and figured... why not help lol

  9. M
    • 5 years ago
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    when you switch from 1 to 2 for dx and 0 to lnx for dy you end up with 0 to ln2 and e^y to 2

  10. M
    • 5 years ago
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    it's impossible to do this without equation lol

  11. amistre64
    • 5 years ago
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    is it dx.dy order ; or dy.dx order?

  12. M
    • 5 years ago
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    but shouldn't dy switch to 0 to 2?

  13. M
    • 5 years ago
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    going from dydx to dxdy

  14. amistre64
    • 5 years ago
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    so your re adjusting the intervals right?

  15. M
    • 5 years ago
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    answer book has it 0 to ln2

  16. M
    • 5 years ago
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    switching the order

  17. amistre64
    • 5 years ago
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    {S{Sdy}dx} ; x interval = [1,2] ; y interval = [0, ln(x)] originally right?

  18. M
    • 5 years ago
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    yeah

  19. amistre64
    • 5 years ago
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    then teh easiest thing to do is graph it; with the lines y=0, y=ln(x), x=1, x=2

  20. M
    • 5 years ago
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    when you switch shouldn't it be [0, 2] and [e^y, 0]

  21. M
    • 5 years ago
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    i mean [0, 2] and [e^y, 2]

  22. amistre64
    • 5 years ago
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  23. M
    • 5 years ago
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    yeah i drew the pic but can't figure out the limits

  24. amistre64
    • 5 years ago
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    this is the area we want, you agree?

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  25. M
    • 5 years ago
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    very nice yes

  26. amistre64
    • 5 years ago
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    then we determine the switch this way right?

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  27. M
    • 5 years ago
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    yeah

  28. amistre64
    • 5 years ago
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    {S{S} dx} dy} ; new y = [0,e^2] then, we want it a straigh line ...

  29. amistre64
    • 5 years ago
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    ln(2) = y and new x = [e^y, 2]

  30. amistre64
    • 5 years ago
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    new y = [0,ln(2)] new x = [e^y,2] ; you said the book has an answer?

  31. M
    • 5 years ago
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    yeah

  32. amistre64
    • 5 years ago
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    y = ln(2) at the top right? so new y = [0,ln(2)]

  33. M
    • 5 years ago
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    that makes sense but when i look at another problem [0, 1] [4x, 4] dydx switches to [0, 4] [0, y/4] dxdy

  34. amistre64
    • 5 years ago
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  35. M
    • 5 years ago
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    whoops ignore that it's [4, 0 ][0, sqrt x]dydx to [0, 2][4, y^2]dxdy

  36. M
    • 5 years ago
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    [0, 4][0, sqrt x]dydx to [0, 2][y^2, 4]]dxdy

  37. M
    • 5 years ago
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    ^^ that's correct one lol

  38. amistre64
    • 5 years ago
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    gonna make me draw another pic aintcha lol

  39. M
    • 5 years ago
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    if i do it like the way you did it then should it new become [0, sqrt x][y^2, 4]dxdy?

  40. M
    • 5 years ago
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    lol you dont have to

  41. amistre64
    • 5 years ago
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    but I did lol

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  42. M
    • 5 years ago
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    yeah

  43. M
    • 5 years ago
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    this is so confusing

  44. amistre64
    • 5 years ago
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  45. amistre64
    • 5 years ago
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    your just switching out the new for the old; it takes practice, but clearly define the old and new

  46. M
    • 5 years ago
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    but the new limits would be [0, sqrt x][y^2, 4]dxdy

  47. M
    • 5 years ago
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    but answer says [0, 2][y^2, 4]dxdy

  48. M
    • 5 years ago
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    can you tell where i'm getting confused?

  49. amistre64
    • 5 years ago
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    look at the picture; the new limit goes to 0,2 becasue its no longer a part of the curve...

  50. amistre64
    • 5 years ago
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    sqrt(x) is an old limit for an old y; that curve is no longer a part of the new y

  51. amistre64
    • 5 years ago
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    just like old x = 0 is no longer a part of new x interval;

  52. amistre64
    • 5 years ago
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    new y matches old x in that it is a limit that is defined from a point to point.... new x matches new y in that it is a limit from a curve to a point.....

  53. amistre64
    • 5 years ago
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    its like reading a map backwards to get back home .... you have to reorient it to make sense

  54. M
    • 5 years ago
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    i think im going to have to look at all the pictures again very slowly by myself lol

  55. M
    • 5 years ago
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    thanks for the help! need to start from top again

  56. amistre64
    • 5 years ago
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    good luck ;)

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