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adnanchowdhury

  • 2 years ago

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

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  1. adnanchowdhury
    • 2 years ago
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    How do you solve it?

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

  3. bahrom7893
    • 2 years ago
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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

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

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

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

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

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

  9. bahrom7893
    • 2 years ago
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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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

  35. adnanchowdhury
    • 2 years ago
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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

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

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

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

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

  40. TuringTest
    • 2 years ago
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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

  41. adnanchowdhury
    • 2 years ago
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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.

  42. TuringTest
    • 2 years ago
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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

  43. myininaya
    • 2 years ago
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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

  44. TuringTest
    • 2 years ago
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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\[2\pi\int_{0}^{6}rhdr=2\pi\int_{0}^{6}y(\frac{y^2}{12}+1)=90\pi\]

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

  46. TuringTest
    • 2 years ago
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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

  47. adnanchowdhury
    • 2 years ago
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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

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

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

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

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

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

  53. TuringTest
    • 2 years ago
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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

  54. adnanchowdhury
    • 2 years ago
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    Oh I see.

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

  56. TuringTest
    • 2 years ago
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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

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

  58. adnanchowdhury
    • 2 years ago
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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).

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

  60. TuringTest
    • 2 years ago
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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.

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