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MasterChief

  • one year ago

Volume of a graph (Washer)

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  1. MASTERCHIEF
    • one year ago
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    Find the volume of \[y=\sqrt{x}\] about x=4

  2. MASTERCHIEF
    • one year ago
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    I started with \[x=y^2\] so would that be the outer radius?

  3. SolomonZelman
    • one year ago
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    |dw:1441595115874:dw|roughly

  4. SolomonZelman
    • one year ago
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    x=y² is not the same as y=√x because it will have a twice larger volume

  5. MASTERCHIEF
    • one year ago
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    I'm sorry I forgot to mention it is bounded by y=0 so just quadrant 1

  6. SolomonZelman
    • one year ago
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    oh, and then if it is rotated about x=4, then I will assume that this is where the √x region ends at?

  7. MASTERCHIEF
    • one year ago
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    Yes

  8. SolomonZelman
    • one year ago
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    Oh, ok, so you know that your limits of integration are from 0 to 2. \(\large\color{black}{ \displaystyle \int_{0}^{2} \pi\left(y^2\right)dy }\)

  9. SolomonZelman
    • one year ago
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    So, you can tell that your radius is y², and it is from y=0 to y=2. THis is what I would do.

  10. SolomonZelman
    • one year ago
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    I missed it should e y^4

  11. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle \int_{0}^{2} \pi\left(y^2\right)^2dy }\) because radius squared. Sorry

  12. SolomonZelman
    • one year ago
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    So that is just an integral of \(\pi\)y\(^4\) from y=0 to y=2.

  13. MASTERCHIEF
    • one year ago
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    Wow I tried this the first time and I guess I integrated incorrectly so I was so confused!

  14. MASTERCHIEF
    • one year ago
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    Thank you

  15. SolomonZelman
    • one year ago
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    you can put integral into wolfram. what is important is to get a good practice of making a setup of the integral for volume. integration you know already....

  16. MASTERCHIEF
    • one year ago
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    Absolutely

  17. SolomonZelman
    • one year ago
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    32π/5 is what i got. (want to know how to make a π • ÷ × √ with no latex?)

  18. MASTERCHIEF
    • one year ago
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    Sure!

  19. DanJS
    • one year ago
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    here are some probs with the solns, pg3 has a nice little summary

  20. MASTERCHIEF
    • one year ago
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    That will definitely be useful, thank you

  21. SolomonZelman
    • one year ago
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    I am glitching a bit

  22. SolomonZelman
    • one year ago
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    Ok, so here is going to be a short guide. The algorithm is: \(1)\) Click ALT and hold it \(2)\) Click the "Number Code" (on the numberpad on the right of the keyboard- if you got one) \(3)\) Release the ALT ------------------------------ Number code Result 0, 2, 1, 5 × 2, 5, 1 √ 7 • 2, 4, 6 ÷ 2, 2, 7 π

  23. SolomonZelman
    • one year ago
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    there are some others too.... but these are useful examples

  24. MASTERCHIEF
    • one year ago
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    So if it was x=6 would the outer radius be 2+y^2 and the inner radius be 2?

  25. SolomonZelman
    • one year ago
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    if it was x=6, |dw:1441596092489:dw|

  26. MASTERCHIEF
    • one year ago
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    about x=6 I meant

  27. SolomonZelman
    • one year ago
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    Do you mean, if it was a region of y^2 bound by y=0 and x=4, but rotated about x=6?

  28. MASTERCHIEF
    • one year ago
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    Yes

  29. SolomonZelman
    • one year ago
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    Yes, then your radius is y²+2

  30. SolomonZelman
    • one year ago
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    and if you rotated in a same case about y=4+c then your radius is y²+c

  31. MASTERCHIEF
    • one year ago
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    I thought I had to subtract the inner area?

  32. SolomonZelman
    • one year ago
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    yeah, my bad, I am overheating let me think

  33. MASTERCHIEF
    • one year ago
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    |dw:1441596258832:dw|

  34. MASTERCHIEF
    • one year ago
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    Haha no problem

  35. SolomonZelman
    • one year ago
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    |dw:1441596426856:dw|

  36. MASTERCHIEF
    • one year ago
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    For this problem I can only use washer

  37. SolomonZelman
    • one year ago
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    Oh, you can do it with respect to x, and do f(x)-g(x)

  38. SolomonZelman
    • one year ago
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    But, shell is also good.

  39. SolomonZelman
    • one year ago
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    it is a matter of preference

  40. MASTERCHIEF
    • one year ago
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    Would f(x) be y^2+2 and g(x) be 2?

  41. MASTERCHIEF
    • one year ago
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    \[\int\limits_{0}^{2} (2+y^2)^2-(2)^2\] ?

  42. SolomonZelman
    • one year ago
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    lets rvw the washer with x's

  43. SolomonZelman
    • one year ago
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    |dw:1441596741353:dw|

  44. SolomonZelman
    • one year ago
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    of that integral should say π INTEGRAL f(x)²-g(x)² dx

  45. SolomonZelman
    • one year ago
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    from a to b

  46. MASTERCHIEF
    • one year ago
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    Yes I get that

  47. MASTERCHIEF
    • one year ago
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    I just don't understand what g(x) would be in this situation

  48. SolomonZelman
    • one year ago
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    and with y, you get the same thing|dw:1441596949137:dw|

  49. SolomonZelman
    • one year ago
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    the area of the cylinder with h=2, r=2 is 8π

  50. SolomonZelman
    • one year ago
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    Or you can say that g is 2.

  51. SolomonZelman
    • one year ago
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    if instead of 4, you had some z(x) boundary for the region, then the radius for g would be 6-z(x) (of course from y=0 to y=2)

  52. MASTERCHIEF
    • one year ago
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    Okay I think I understand this better now thank you

  53. SolomonZelman
    • one year ago
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    Anytime... thank you for refreshing me on these rotations:)

  54. SolomonZelman
    • one year ago
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    I mean, I really should tell you that Shell method rocks in so many cases, so try to use that as well. in any case, good luck!

  55. MASTERCHIEF
    • one year ago
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    Actually, did I plug this in incorrectly? http://www.wolframalpha.com/input/?i=integrate+from+0+to+2+pi%28%282%2By%5E2%29%5E2-4%29

  56. MASTERCHIEF
    • one year ago
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    The answer is 192π/5

  57. SolomonZelman
    • one year ago
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    for the problem where you want to know about the redion of y62 bound by y=0 and x=4, rotated about x=6....?

  58. SolomonZelman
    • one year ago
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    region*

  59. SolomonZelman
    • one year ago
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    y^2 (not y62)

  60. MASTERCHIEF
    • one year ago
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    yes

  61. SolomonZelman
    • one year ago
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    |dw:1441597714083:dw|

  62. SolomonZelman
    • one year ago
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    I actually do not get why the answer isn't what you got in wolfram.

  63. MASTERCHIEF
    • one year ago
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    hmmm

  64. SolomonZelman
    • one year ago
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    you can take the whole volume of radius y^2+2 with limits of y=0 to y=2, and subtract 8π cylinder in the middle, and you get precisely the same.

  65. SolomonZelman
    • one year ago
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    I have to go, it is almost 12am in my location. maybe I was answering a wrong question idk, but for what I asked, it should be 256π/15

  66. SolomonZelman
    • one year ago
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    Maybe my brain just shot down xD

  67. MASTERCHIEF
    • one year ago
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    Hahaha

  68. MASTERCHIEF
    • one year ago
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    But I really appreciated your time so thank you

  69. SolomonZelman
    • one year ago
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    I will look at it when I have time. Whatever I can do with my little knowledge:) gtg c(u)

  70. MASTERCHIEF
    • one year ago
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    Bye!

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