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adnanchowdhury

Integration question: http://d.pr/RTc6

  • 2 years ago
  • 2 years ago

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  1. adnanchowdhury
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    How do you solve it?

    • 2 years ago
  2. adnanchowdhury
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    how do i integrate: \[π \int\limits_{0}^{6} (y ^{2}/12 +1)^{2} dy\]

    • 2 years ago
  3. bahrom7893
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    Rotate the rectangle first, then rotate the parabola itself, find individual volumes then subtract the volume of the solid generated by the parabola from the solid generated by the rectangle

    • 2 years ago
  4. bahrom7893
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    and for that integral, just multiply it out.

    • 2 years ago
  5. bahrom7893
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    And then integrate term by term that will make your life easy.

    • 2 years ago
  6. bahrom7893
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    actually finding the vol of the solid generated by the rectangle is easy.

    • 2 years ago
  7. adnanchowdhury
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    I've been taught to integrate like this when the limits are on the y axis: \[\int\limits_{a}^{b} π x ^{2} dy\]

    • 2 years ago
  8. bahrom7893
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    It generates a cylinder with radius 6 and height 4

    • 2 years ago
  9. bahrom7893
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    Yea but you don't really have to do that for the solid generated by the rectangle. If you rotate it around it generates a cylinder.

    • 2 years ago
  10. adnanchowdhury
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    What rectangle are you referring to?

    • 2 years ago
  11. bahrom7893
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    Imagine a line passing through (4,6) and (4,0)

    • 2 years ago
  12. bahrom7893
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    that will form a rectangle right? with the line y = 6, x = 4, and x and y axes

    • 2 years ago
  13. adnanchowdhury
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    yes, i see. But what do you after that?

    • 2 years ago
  14. TuringTest
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    you can do this with disk or shell method. Do you know both?|dw:1329324049040:dw|

    • 2 years ago
  15. bahrom7893
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    So find the volume of the solid generated by that rectangle - it will be a cylinder

    • 2 years ago
  16. adnanchowdhury
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    I haven't heard of the 'disk' or 'shell' method?

    • 2 years ago
  17. adnanchowdhury
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    What do i do after working out the vol. of solid rectangle?

    • 2 years ago
  18. bahrom7893
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    Find the volume generated by revolving the following area and then subtract it from the vol of the cylinder.

    • 2 years ago
  19. bahrom7893
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    |dw:1328460177913:dw|

    • 2 years ago
  20. adnanchowdhury
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    How would you integrate this? http://d.pr/2Eo2

    • 2 years ago
  21. bahrom7893
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    Just expand and then integrate term by term, it's not hard.

    • 2 years ago
  22. adnanchowdhury
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    without expanding? Like chain rule?

    • 2 years ago
  23. bahrom7893
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    I'm not sure if you can. Does the question force you to integrate without expanding?

    • 2 years ago
  24. adnanchowdhury
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    no but that is one technique i have to learn.

    • 2 years ago
  25. TuringTest
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    your formula is for a revolution about y, so it's not right

    • 2 years ago
  26. bahrom7893
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    no that's a diff question

    • 2 years ago
  27. myininaya
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    yeah bahrom is right i'm pretty sure the only way to integrate that is by expanding it

    • 2 years ago
  28. myininaya
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    that is the only way i see anyways

    • 2 years ago
  29. bahrom7893
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    I mean it's not u-sub, and I don't recognize any inverse formulas.. inverse tan won't work

    • 2 years ago
  30. bahrom7893
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    and integration by parts is gonna give u a horrible mess

    • 2 years ago
  31. TuringTest
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    ...but this integral is not supposed to represent the area in the above problem? separate Q you say?

    • 2 years ago
  32. myininaya
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    it is not so nasty to expand this it is the easiest method to expand

    • 2 years ago
  33. bahrom7893
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    Yea turing, it's separate

    • 2 years ago
  34. bahrom7893
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    I think expanding is the only way.

    • 2 years ago
  35. adnanchowdhury
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    Basicallly, I ended up with this integral, by rearranging the given equation to make x the subject. Then substituting into this: http://d.pr/PR8z

    • 2 years ago
  36. adnanchowdhury
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    The final answer is suppose to be 90π.

    • 2 years ago
  37. TuringTest
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    so it is supposed to be the same problem

    • 2 years ago
  38. TuringTest
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    ...which means the integral is wrong

    • 2 years ago
  39. bahrom7893
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    No, there are two diff problems here as far as i can tell, the integral is a completely diff question

    • 2 years ago
  40. TuringTest
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    No ,they just said they got the integral by subbing in for y, but they are integrating on the wrong axis with the wrong method

    • 2 years ago
  41. adnanchowdhury
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    That is the formula i am given in the book for integrating on the y-axis. So, i can't be integrating on the wrong axis.

    • 2 years ago
  42. TuringTest
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    \[2\pi\int_{0}^{6}y(\frac{y^2}{12}+1)dy\] is shell method, which is how I would do it

    • 2 years ago
  43. myininaya
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    \[\pi \int\limits_{0}^{6}(\frac{y^4}{144}+\frac{y^2}{6}+1) dy=\pi(\frac{y^5}{5 \cdot 144}+\frac{y^3}{3 \cdot 6}+y)|_0^6\] \[=\pi(\frac{6^5}{5 \cdot 144}+\frac{6^3}{3 \cdot 6}+6)=\pi(\frac{6^4 \cdot 6}{5 \cdot 144}+\frac{6^2}{3}+6)\] \[=\pi(\frac{6^4 }{5 \cdot 24}+\frac{ 6 \cdot 6}{3}+6)=\pi(\frac{6^4}{5 \cdot 24} +6(2)+6)\] \[=\pi(\frac{6 \cdot 6^3}{5 \cdot 24}+12+6)=\pi(\frac{6^3}{5 \cdot 4}+18)\] \[=\pi(\frac{6 \cdot 6^2}{5 \cdot 4}+18)=\pi(\frac{3 \cdot 6^2}{5 \cdot 2}+18)\] \[=\pi(\frac{3 \cdot 36}{10}+18)=\pi(\frac{12 \cdot 9}{10}+18)=\pi(\frac{108}{10}+18)\] so this is the same problem or different? lol i don't know anymore by the way getting the answer is hard without a calculator as you can see

    • 2 years ago
  44. TuringTest
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    shell method:|dw:1329325465495:dw|Look at the radius of each cylinder( y) and the height of each cylinder (x(y)) and we get\[2\pi\int_{0}^{6}rhdr=2\pi\int_{0}^{6}y(\frac{y^2}{12}+1)=90\pi\]

    • 2 years ago
  45. bahrom7893
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    oh wait so it was this question?

    • 2 years ago
  46. TuringTest
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    I am integrating along y because I used shell method, disk method would be a pain here because we would need two integrals

    • 2 years ago
  47. adnanchowdhury
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    @myininaya did it right. I think you used the same method as me, but you expanded x^2 in the beginning? @bahrom7893 yes

    • 2 years ago
  48. adnanchowdhury
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    I'm still quite confused about the shell method.

    • 2 years ago
  49. TuringTest
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    myin did the integral right, but it is not a solid about the x-axis

    • 2 years ago
  50. adnanchowdhury
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    Yes it is. ?

    • 2 years ago
  51. TuringTest
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    no, hers are disks about the y-axis

    • 2 years ago
  52. TuringTest
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    |dw:1329325925783:dw|like this but centered on the y-axis

    • 2 years ago
  53. TuringTest
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    area of each disk is pi*r^2, but if r is x(y) we are going around the y-axis

    • 2 years ago
  54. adnanchowdhury
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    Oh I see.

    • 2 years ago
  55. adnanchowdhury
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    This is rotation 360 degree about y-axis?

    • 2 years ago
  56. TuringTest
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    if you want shell method for your integral it can be done, but you will need to split the integral in two, or subtract the second area as bahrom did disks of radius y(x)=6 between 0<x<1 rings of inner radius y(x)=sqrt(12x-1) and outer radius y=6 between 1<x<6

    • 2 years ago
  57. adnanchowdhury
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    Ok, let me try.

    • 2 years ago
  58. adnanchowdhury
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    Yes! I got it. I did what bahrom7893 told me to do. Find volume of cylinder made by rectange on this graph. Then subtact it by the volume between where the line cross x-axis and where the rectangle ends (at x =4).

    • 2 years ago
  59. TuringTest
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    that'll work :D

    • 2 years ago
  60. TuringTest
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    Shell method\[2\pi\int_{0}^{6}y(\frac{y^2}{12}+1)dy\]Disk method (my way):\[\pi\int_{0}^{1}6^2dx+\pi\int_{1}^{6}6^2-(\sqrt{12(x-1)})^2dx\]Disk method bahrom's way\[\pi\int_{0}^{6}6^2dx-\pi\int_{1}^{6}(\sqrt{12(x-1)})^2dx\]all should give the same answer. Try to think about the difference in the concept of each formula if you can, that will broaden your understanding.

    • 2 years ago
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