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Math Help - More volume questions, integration

  1. #1
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    More volume questions, integration

    Question 26-3-15
    Statement
    Using Shell method, find the volume generated by revolving the region bounded by the given curve about the x-axis.
    x = 4y - y^2 - 3, x = 0


    Finding y intecept
    0 = 4y - y^2 - 3
    Y intercept: 1

    Shell method: 2pixy
    = 2pi(4y-y^2-3)y
    = 2pi(4y^2-y^3-3y)

    When I calculate this now with y = 1 the answer is zero, which is wrong.


    Question 26-3-21
    Statement
    Using shell method, find the volume generated by revolving the region bounded by the given curve about the y axis.
    x^2 - 4y^2 = 4, x = 3


    Attempt
    2pixy principal formula
    Isolate y
    -4y^2 = 4 - x^2
    -y^2 = (4-x^2)/4

    -y^2 = 1 - (x^2/4)
    y = 1 - x/2

    Now to integrate
    = 2pix(1 - x/2)
    = 2pix - (x^2/2)

    not sure how to proceed.

    Thanks for looking
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  2. #2
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    Re: More volume questions, integration

    I will add a few attempts:

    26-3-8 Rotate about x-axis, use shell method, determine volume:
    y = x^(1/2), x = 0, y = 2
    dV = 2pixy
    Isolate x
    x = y^2
    dV = 2pi(y^2)y
    dV = 2piy^3
    dV = 1/4y^4
    dV = 2pi(1/4(2^4)) = 25.132

    Is this correct? Thanks.

    26-3-10 Rotate about x-axis, use disk method, determine volume:
    y = 4x - x^2
    y = 0
    dV = piy^2
    y = (4x-x^2)(4x-x^2)
    y = 16x^2 - 4x^3 - 4x^3 + x^4
    y = 16x^2 - 8x^3 + x^4

    0 = 4x-x^2

    I'm stuck trying to find a limit here.

    26-3-12 Rotate about x-axis, use shell method, determine volume:
    y = x^2, y = x
    dV = 2piy^2
    y^2 = (x^2)^2
    y^2 = x^4
    1/4x^4

    is this the correct path to take? Thanks.
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  3. #3
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    Re: More volume questions, integration

    1. To rotate a region about the x axis, use \displaystyle \begin{align*} V = \int_a^b{\pi \left[ f(x) \right] ^2 \, dx} \end{align*}

    By a simple sketch, it's relatively easy to see that you will need to evaluate a large volume and subtract a smaller one. So your volume is evaluated by

    \displaystyle \begin{align*} V = \int_0^1{ \pi \left( 2 + \sqrt{1 - x} \right)^2 \, dx } - \int_0^1{ \pi \left( 2 - \sqrt{1 - x} \right)^2 \, dx}  \end{align*}.


    Use this formula for the other problems too.
    Last edited by Prove It; February 28th 2013 at 06:17 PM.
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  4. #4
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    Re: More volume questions, integration

    uh

    ok, hate to say it but I'm still pretty lost, probably find a tutor next... i spent 2 hrs on this stuff this morning and its getting harder and harder. I got thru the first few questions easily enough though.

    or maybe I am really tired and need to look at it with fresh brain. I kinda see where you're going by subtracting volumes.
    Last edited by togo; February 28th 2013 at 10:40 PM.
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  5. #5
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    Re: More volume questions, integration

    Togo,

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    It's a relatively new service but we have some very skilled online math tutors that can help you. I would encourage you to check it out. Let me know if you need anything. You can contact me via email at thomas [ at ] tutoruniverse [ dot ] com.

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    Hope this helps!
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