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Math Help - Solid Volume rotation around y-axis

  1. #1
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    Solid Volume rotation around y-axis

    I need to find the volume of the solid formed when rotating f(x) = 4x-x^2 and y=4 and x=0 about the y-axis.

    Using the disc/washer method.

    I already did it using the shell method and got 128/3 pi.

    I am getting confused on how to do this, I believe it will need to be setup in 2 parts? Refer to my image sketch: Where I think the 2 parts are split by x=2. From x=0 to x=2 I see a disc method and then from x =2 to x=4 I see the washer method. However because the disc/washer method is perpendicular to the axis of rotation (y) that means that the thicknes of the disc/washers will be dy, which means the limits of integration would be in terms of y. Therefore I don't see how I can set it up with 2 equations with different integrals.
    Last edited by mr fantastic; October 12th 2011 at 01:02 AM. Reason: Bolded the edit made by the OP.
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  2. #2
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    Re: Solid Volume rotation around y-axis

    You're reading it as the volume of revolution of the shaded area, yes? In that case there should be mention at least of the line x = 4.

    Edit: sorry, I do agree with your 128/3 pi.

    Your disc radius, i.e. the length of the strip in your pic, should be

     x = 2 - \sqrt{4 - y}

    and then the washer distance will be the same again. So you don't need to split it. Double  2 - \sqrt{4 - y} is what you need to square before integrating with respect to y. Edit: no, pardon me. Doubling won't do. Didn't think it through. Sorry. More below.
    Last edited by tom@ballooncalculus; October 12th 2011 at 07:26 AM.
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  3. #3
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    Re: Solid Volume rotation around y-axis

    Quote Originally Posted by tom@ballooncalculus View Post
    You're reading it as the volume of revolution of the shaded area, yes? In that case there should be mention at least of the line x = 4.

    I get 32/3 pi from washers/discs as well as shells.

    Your disc radius, i.e. the length of the strip in your pic, should be

     x = 2 - \sqrt{4 - y}

    and then the washer distance will be the same again. So you don't need to split it. Double  2 - \sqrt{4 - y} is what you need to square before integrating with respect to y.
    I figured it that y=4 was the upper and 4x-x^2 was the lower so I did the area under y=4 and cutout the bottom function. I added x=4 myself because thats where the lower function ends but now I am debating to myself if the shaded area should really just be 4x-x^2 given they never tell me x=4.
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  4. #4
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    Re: Solid Volume rotation around y-axis

    I take it you mean under 4x - x^2. But then why mention y = 4? What's the precise wording?

    Perhaps they mean the disc without the washer.
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    Re: Solid Volume rotation around y-axis

    "Use the shell method and then the disc/washer method to find the volume of the solid formed by revolving the region bounded by the graph of: f(x)=4x-x^2, y=4 and x=0 about the y-axis."
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  6. #6
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    Re: Solid Volume rotation around y-axis

    Sounds like the disc without the washer. Perhaps just as well. I was wrong about doubling it. Pardon me.
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    Re: Solid Volume rotation around y-axis

    Quote Originally Posted by tom@ballooncalculus View Post
    Sounds like the disc without the washer. Perhaps just as well. I was wrong about doubling it. Pardon me.
    So the y=4 is redundant information or is my shaded region from my picture correct?

    I used the following for my shell method:

    thickness = dx
    r = x
    h = 4x-x^2

    Integral from 0 to 4 [ (2*pi*x) (4x-x^2) ] dx
    2pi * Integral from 0 to 4 [ (4x^2-x^3) ] dx
    2pi * [ (4/3)x^3 - (1/4)x^4 ] -> 2pi*[(4/3)*4^3-(1/4)*4^4] = 2Pi*[64/3] = 128/3 pi

    I guess that's incorrect?
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  8. #8
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    Re: Solid Volume rotation around y-axis

    But what you've got there is the area under the graph, right? If so, fine.

    If you want the shaded region, then with shells it's

    2 \pi \int _0^4 x [ 4 - (4x - x^2)]\ dx = 2 \pi \int _0^4 4x - 4x^2 + x^3\ dx

    for both parts, and obviously then just

    2 \pi \int _0^2 4x - 4x^2 + x^3\ dx

    for the inside part. With discs, the inside part is

    \pi \int _0^4 (2 - \sqrt{4 - y})^2\ dx = \pi \int _0^4 8 - y - 4\sqrt{4 - y}\ dx

    and the other, which is a washer, is

    \pi \int _0^4 4^2 - (2 + \sqrt{4 - y})^2\ dx = \pi \int _0^4 8 + y - 4\sqrt{4 - y}\ dx
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