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  1. anonymous
    • one year ago
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    @King.Void.

  2. Astrophysics
    • one year ago
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    First thing you want to do is sketch it out

  3. anonymous
    • one year ago
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    yes u want to or should i?

  4. Astrophysics
    • one year ago
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    You should

  5. anonymous
    • one year ago
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    |dw:1440417292230:dw|

  6. anonymous
    • one year ago
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    srry i dont know how to do it properly its a bit rough

  7. Astrophysics
    • one year ago
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    |dw:1440417409578:dw|

  8. anonymous
    • one year ago
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    yes

  9. anonymous
    • one year ago
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    when it says label co-ordinates of 4 vertices does that mean what u did on the graph

  10. anonymous
    • one year ago
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    |dw:1440417708934:dw|

  11. Astrophysics
    • one year ago
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    So what is the area you should be integrating

  12. anonymous
    • one year ago
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    is it where the all curves are

  13. anonymous
    • one year ago
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    |dw:1440418416259:dw|

  14. Astrophysics
    • one year ago
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    I guess it wants it above y=8

  15. Astrophysics
    • one year ago
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    So we probably need a better drawing

  16. phi
    • one year ago
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    it wants this region

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  17. Astrophysics
    • one year ago
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    Yeah exactly

  18. phi
    • one year ago
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    the top point is at 9,9^3 or (9,729)

  19. Astrophysics
    • one year ago
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    |dw:1440418815806:dw|

  20. Astrophysics
    • one year ago
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    Just putting your pic up in draw phi :P

  21. anonymous
    • one year ago
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    \[\int\limits_{9}^{2}x ^{3} dx +\int\limits_{9}^{2} 8 dx -\int\limits_{9}^{8} x dx\]

  22. anonymous
    • one year ago
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    is it something like this

  23. phi
    • one year ago
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    part (b) wants the integral with respect to y

  24. anonymous
    • one year ago
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    yes

  25. Astrophysics
    • one year ago
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    You should have dy

  26. Astrophysics
    • one year ago
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    So you're doing horizontal strips not vertical, that's part c

  27. anonymous
    • one year ago
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    how?

  28. phi
    • one year ago
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    |dw:1440419331010:dw|

  29. phi
    • one year ago
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    the first integral has a right bound formed by y=x the right bound is the curve y= x^3 each rectangle will have width x1-x2 and height dy we need x1 and x2 in terms of y x1 (right bound) is x=y so y x2 (left bound is x^3= y so x= y^(1/3) the integral is \[ \int_8^9 y-y^\frac{1}{3} \ dy \] then we need a second integral for y=9 to 9^3

  30. phi
    • one year ago
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    follow?

  31. anonymous
    • one year ago
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    yes

  32. phi
    • one year ago
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    can you post the 2nd integral?

  33. anonymous
    • one year ago
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    \[\int\limits_{9}^{3} y \]

  34. anonymous
    • one year ago
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    \[\int\limits_{3}^{9} y dy\]

  35. anonymous
    • one year ago
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    srry the first one was a mistake

  36. phi
    • one year ago
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    then we need a second integral for y=9 to 9^3

  37. anonymous
    • one year ago
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    \[\int\limits_{3}^{9} 8\]

  38. phi
    • one year ago
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    First, identify the dimensions of the "thin rectangle" |dw:1440420324066:dw| we want the area of that "box" we want a positive number, so we would do Xright - Xleft that gives the width then multiply by the height dy (Xr - Xleft)* dy that is the first step

  39. anonymous
    • one year ago
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    ok

  40. phi
    • one year ago
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    next step: we don't want to use an x variable. but we know Xright is a constant = 9 so we have (8 - Xleft) dy now we need to get rid of Xleft (i.e. put it in terms of y) if we are given a y value, can you find Xleft?

  41. phi
    • one year ago
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    **(9 - Xleft) dy

  42. anonymous
    • one year ago
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    (9-x) dy

  43. phi
    • one year ago
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    ok, but that x you have in (9-x) is some number at a specific y say the "box" is at y=8. what is the value of that x? (remember that x is the x of the point on the curve y=x^3 when y is 8)

  44. anonymous
    • one year ago
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    \[\sqrt[3]{y}\]

  45. phi
    • one year ago
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    yes, in general, (and specifically, when y=8, x is 2) but we want the general answer, because as we change y we want the "correct x" x= y^(1/3) (I would use exponents, easier to integrate)

  46. phi
    • one year ago
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    so fix up this: (9-x) dy in other words, replace that x with a function of y

  47. anonymous
    • one year ago
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    (9-y^(1/3) ) dy

  48. anonymous
    • one year ago
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    \[\int\limits_{3}^{9} [ 9- y ^{\frac{ 1 }{ 3 }} ]\]

  49. anonymous
    • one year ago
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    dy

  50. phi
    • one year ago
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    that is the area of a single thin box we know "add them up" , which is to say, integrate starting from the lowest y (which is point B in the graph) up to the highest y (which is off the graph)

  51. anonymous
    • one year ago
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    ok

  52. anonymous
    • one year ago
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    \[\int\limits_{8}^{9} y-y ^{\frac{ 1 }{ 3}} dy\] +\[\int\limits_{9}^{3} [9-y ^{\frac{ 1 }{ 3}}] dy\]

  53. phi
    • one year ago
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    how are you getting 3 for the upper limit?

  54. anonymous
    • one year ago
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    srry mistake

  55. anonymous
    • one year ago
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    \int\limits_{3}^{9} [9-y ^{\frac{ 1 }{ 3}}] dy

  56. phi
    • one year ago
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    no, the upper limit is where x^3 meets x=9

  57. phi
    • one year ago
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    the left side curve is y= x^3 and the right boundary is x=9 the left side gets closer to x=9 as y gets bigger they intersect when y= 729

  58. anonymous
    • one year ago
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    how did u get Y= 729

  59. anonymous
    • one year ago
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    didnt u just use the two integrals and then integrate

  60. phi
    • one year ago
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    the very first part of the problem is to find the 4 points where the various curves meet the shape is like this: |dw:1440422110612:dw|

  61. phi
    • one year ago
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    we find where y=x^3 meets x=9 obviously they meet when x=9 in the first equation, and the y value is y=9^3 = 729

  62. phi
    • one year ago
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    notice we use two integrals. the first one has a different "right boundary" we use that one from y=8 to y=9 then we switch to the new right boundary (of x=9), and use that all the way up to y=729

  63. anonymous
    • one year ago
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    yea

  64. phi
    • one year ago
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    so the problem is \[ \int\limits_{8}^{9} y-y ^{\frac{ 1 }{ 3}}\ dy + \int\limits_{9}^{729} 9-y ^{\frac{ 1 }{ 3}} \ dy\]

  65. anonymous
    • one year ago
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    oh wow

  66. anonymous
    • one year ago
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    6.45+1573.29 =1579.74

  67. anonymous
    • one year ago
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    is this right?

  68. phi
    • one year ago
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    yes

  69. anonymous
    • one year ago
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    thank u so much u have alot of patience

  70. anonymous
    • one year ago
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    part c?

  71. anonymous
    • one year ago
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    i show u the solution i have done

  72. phi
    • one year ago
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    part c is more complicated. the first step is draw in a thin vertical rectangle, and figure out its height and width

  73. anonymous
    • one year ago
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    |dw:1440423107863:dw|

  74. anonymous
    • one year ago
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    so do we get the values respect to x

  75. anonymous
    • one year ago
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    \[\int\limits_{2}^{9} x ^{3} dx - \int\limits_{2}^{9} 8 dx - \int\limits_{8}^{9}\]

  76. anonymous
    • one year ago
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    \[\int\limits_{8}^{9} x dx *\]

  77. phi
    • one year ago
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    there are a few ideas we have to keep clear. I would begin with a picture and draw in the rectangle.

  78. phi
    • one year ago
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    |dw:1440423583733:dw|