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lovekblue

  • 3 years ago

Need help in integral area involve 3d objects

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  1. lovekblue
    • 3 years ago
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  2. Algebraic!
    • 3 years ago
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    what's your question?

  3. Algebraic!
    • 3 years ago
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    ah I see those are multiple choices...

  4. Algebraic!
    • 3 years ago
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    did you get the 'sketch the region' one?

  5. Algebraic!
    • 3 years ago
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    basically just asking you to pick which of those areas is bounded by y=5, x=4 and the function...

  6. Algebraic!
    • 3 years ago
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    you there?

  7. Algebraic!
    • 3 years ago
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    the first choice is bounded by y=10, x=4 and the function... the second choice is bounded by y=10, x=0, x=4 and the function...

  8. lovekblue
    • 3 years ago
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    yes i am here..

  9. lovekblue
    • 3 years ago
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    I know the answer for the first picture already, not sure about the 2nd one @Algebraic!

  10. lovekblue
    • 3 years ago
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    For the first picture, i think it is the last one for the 2nd picture, i am thinking either 1st or 4th one , what do u think ?

  11. Algebraic!
    • 3 years ago
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    naw, the last picture is bounded by x=0 y=0 and the function...

  12. Algebraic!
    • 3 years ago
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    (and x=4)

  13. lovekblue
    • 3 years ago
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    so for the first picture, u also think is the last one?

  14. lovekblue
    • 3 years ago
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    for my function, x is not 0 u mean?

  15. Algebraic!
    • 3 years ago
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    just trace around each of those red regions and see what functions or lines your tracing...

  16. Algebraic!
    • 3 years ago
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    what lines do you trace for the last choice?

  17. Algebraic!
    • 3 years ago
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    |dw:1352777609189:dw|

  18. lovekblue
    • 3 years ago
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    yea, that's what i have too. dont really know how i got that..but yeah.. Do you possibly know how to volume the volume of this?

  19. Algebraic!
    • 3 years ago
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    |dw:1352777630991:dw|

  20. Algebraic!
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    |dw:1352777716676:dw|

  21. lovekblue
    • 3 years ago
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    dont u need to integrate the area function to volume , and then something to do with the height..?

  22. Algebraic!
    • 3 years ago
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    so... is that the region described in your problem statement??

  23. lovekblue
    • 3 years ago
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    yes

  24. Algebraic!
    • 3 years ago
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    really? your problem statement makes no mention of x=0 or y=0 ...

  25. Algebraic!
    • 3 years ago
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    it does mention y=5 however...

  26. lovekblue
    • 3 years ago
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    well, but it says rotate about y= 10, and y= 5

  27. Algebraic!
    • 3 years ago
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    naw..

  28. lovekblue
    • 3 years ago
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    then u can kinda know there is a hole in between

  29. lovekblue
    • 3 years ago
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    umm..wait..so which picture do you think it is for the first set of picture? i thought we have the same answer

  30. Algebraic!
    • 3 years ago
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    the region is bounded by x=4 y=5 and y=5e^-x

  31. Algebraic!
    • 3 years ago
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    like I said trace the boundary of the regions you have been given to find which one works with that description..

  32. Algebraic!
    • 3 years ago
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    there's no point in guessing here... even though this is multiple guess

  33. lovekblue
    • 3 years ago
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    i still think is the last one..

  34. lovekblue
    • 3 years ago
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    that's why i want to know which one u think it is

  35. Algebraic!
    • 3 years ago
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    why is it the last one..

  36. Algebraic!
    • 3 years ago
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    are you talking about the first part or the second part?

  37. lovekblue
    • 3 years ago
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    first part ..

  38. Algebraic!
    • 3 years ago
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    because if you do the first part correctly, the answer to the second part should be obvious

  39. Algebraic!
    • 3 years ago
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    ok..

  40. lovekblue
    • 3 years ago
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    so do you agree that it is either the bottom left or bottom right picture is correct due to the fact that given y=5

  41. Algebraic!
    • 3 years ago
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    then why are you saying it's the last region?

  42. lovekblue
    • 3 years ago
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    why is it not the last picture ..lol |dw:1352778179844:dw|

  43. Algebraic!
    • 3 years ago
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    did you see the sketches where I traced the bounding functions??

  44. Algebraic!
    • 3 years ago
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    what are they??

  45. lovekblue
    • 3 years ago
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    the curve is y= 5e^-x , y =5 , x=4

  46. Algebraic!
    • 3 years ago
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    what functions bound the region in the last picture?

  47. Algebraic!
    • 3 years ago
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    hint: I already did it for you.

  48. lovekblue
    • 3 years ago
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    x=0, y=0..which i dont really get u ..o_o

  49. Algebraic!
    • 3 years ago
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    trace the boundary

  50. Algebraic!
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    |dw:1352778560923:dw|

  51. Algebraic!
    • 3 years ago
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    what's that portion?

  52. Algebraic!
    • 3 years ago
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    say "y=0"

  53. Algebraic!
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    |dw:1352778627593:dw|

  54. Algebraic!
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    what's that portion?

  55. lovekblue
    • 3 years ago
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    are you trying to say that radius doesnt bound to x=0?

  56. lovekblue
    • 3 years ago
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    which it does indeed make sense when i look in the 2nd part

  57. Algebraic!
    • 3 years ago
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    I'd focus on the first part, if I were you..

  58. Algebraic!
    • 3 years ago
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    picking which region is bounded by y=5, x=4 and y= 5e^-x

  59. lovekblue
    • 3 years ago
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    i'm not sure what u mean by bounded..do you mean by starting from there?

  60. Algebraic!
    • 3 years ago
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    trace the boundary of each region

  61. Algebraic!
    • 3 years ago
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    like I did... for the last region pictured in your choices...

  62. lovekblue
    • 3 years ago
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    if i do what u did, then the answer will be the bottom left

  63. Algebraic!
    • 3 years ago
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    yes

  64. lovekblue
    • 3 years ago
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    but my question with that picture is.. don't u need to do 5- 5e^-t in order to get tht area?

  65. Algebraic!
    • 3 years ago
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    not necessarily..

  66. Algebraic!
    • 3 years ago
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    you'll be using washers probably, since you haven't learned about shells yet (I assume)...

  67. Algebraic!
    • 3 years ago
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    so the inner radius is 5 and the outer radius is 10-5e^-x

  68. lovekblue
    • 3 years ago
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    yea i didnt yet.. i didnt even learn these stuff much when my assignment is due 2 days after..

  69. lovekblue
    • 3 years ago
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    ya, i do get that part

  70. Algebraic!
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    |dw:1352779152955:dw|

  71. lovekblue
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    is that region not what the last picture look like?

  72. Algebraic!
    • 3 years ago
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    for the second part... yep.

  73. lovekblue
    • 3 years ago
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    can i ask what does "region " mean..

  74. Algebraic!
    • 3 years ago
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    the red areas in the first part.

  75. lovekblue
    • 3 years ago
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    yea..what does that red area mean

  76. lovekblue
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    is it like the path of rotation or something?

  77. Algebraic!
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    no.

  78. Algebraic!
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    it's the thing you're sweeping around the specified axis in order to generate a solid

  79. Algebraic!
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    |dw:1352779636199:dw|

  80. Algebraic!
    • 3 years ago
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    take that region as an example... it's a rectangle...

  81. Algebraic!
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    rotate about the x axis and you sweep out a cylinder..

  82. lovekblue
    • 3 years ago
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    so it's kinda you are sweeping that region, and that sweeping motion generate the solid figure?

  83. Algebraic!
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    |dw:1352779702320:dw|

  84. Algebraic!
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    yes, exactly.

  85. lovekblue
    • 3 years ago
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    okay thank you so so much!!!!!!

  86. lovekblue
    • 3 years ago
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    Do you still have time to guide me on filding the volume?

  87. Algebraic!
    • 3 years ago
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    the second multiple choice question? or a new question?

  88. lovekblue
    • 3 years ago
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    it's the same question. but it asks to find the volume

  89. lovekblue
    • 3 years ago
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    cuz finding the volume is actually the first part..but the graphing part seems easier so i skipped to that first ..

  90. Algebraic!
    • 3 years ago
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    sure... we are using the washers in the sketch (and in the last choice to the second question)

  91. Algebraic!
    • 3 years ago
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    inner radius is 5, outer radius is 10 - 5e^-x

  92. Algebraic!
    • 3 years ago
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    area of a washer will be pi( (10 - 5e^-x)^2 - 5^2 )

  93. lovekblue
    • 3 years ago
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    yea..that's all i have too.. not sure how to integrate that..

  94. Algebraic!
    • 3 years ago
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    \[\pi \int\limits_{ }^{} (10-5e^{-x})^2 - 5^2 dx\]

  95. Algebraic!
    • 3 years ago
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    \[\pi \int\limits_{0}^{4} 75 -100e^{-x} +25e^{-2x} dx\]

  96. Algebraic!
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    that should be pretty easy to integrate...

  97. lovekblue
    • 3 years ago
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    oh ok..i didnt combine them.. ok thanks, i will try that out :)

  98. lovekblue
    • 3 years ago
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    @Algebraic! umm.. how can i leave it as exact numbers..?

  99. lovekblue
    • 3 years ago
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    i got pi ((300+10 e^-4 - 25/2 e^-8)- (10-25/2))

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