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Math Help - Volume of revolution problem

  1. #1
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    Volume of revolution problem

    This is the only problem on my homework that I can't seem to figure out.

    Find the volume of the solid formed by revolving the region bounded by the graphs of y=x^2+1, y=0, x=0, and x=1 about the y-axis using the washer method.

    Any help would be greatly appreciated.
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  2. #2
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    With these it is crucial to draw the picture.

    You have y=x^2+1 is a parabola at the top. See the attached image.

    So method by washers: the washers are perpendicular to the axis of rotation. So they are horizontal in this case, and integration is parallel to the axis of rotation (so we need x's as a function of y).

    With washers you need to figure you the two radii (inner and outer).
    Since the axis of rotation is the line x=0, we need the distances from this line.

    From y=0 to y=1, inner radius is 0, outer is 1 (note that the
    From y=1 to y=2, inner radius is ?, outer is 1

    Let's figure out the inner radius: we need to write the inner radius is a function of y. What we have is y = x^2+1
    y-1 = x^2
    \pm\sqrt{y-1} = x

    but this is the positive one (the curve is for positive x) so
    From y=1 to y=2, inner radius is \sqrt{y-1}, outer is 1

    Now (informally) the volume of the washer is \pi(R^2-r^2)dy where R is the outer and r is the inner radii.

    So

    \int_0^1\pi(1^2-0^2)dy + \int_1^2 \pi(1^2-(\sqrt{y-1})^2)dy

    Make sense?
    Attached Thumbnails Attached Thumbnails Volume of revolution problem-sage0.png  
    Last edited by meymathis; July 16th 2008 at 06:21 PM. Reason: make clearer
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  3. #3
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    Ok so after integrating I got:
    [\pi - 0] + [ 2\pi - 3\pi/2]<br />

    <br />
\pi + \pi/2 = 3\pi/2<br />

    I hope I did this right, I have been making little dumb mistakes after working on this stuff for hours.
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  4. #4
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    Quote Originally Posted by Zolthas View Post
    This is the only problem on my homework that I can't seem to figure out.

    Find the volume of the solid formed by revolving the region bounded by the graphs of y=x^2+1, y=0, x=0, and x=1 about the y-axis using the washer method.

    Any help would be greatly appreciated.
    ( I am late, as usual. My typing is just a little bit faster than the speed of a starfish walking. See if you can use my solution anyway.)

    By washer method?

    It is done easier by shell method, where dV = (2pi*x)[(x^2 +1)*dx] and the integration is from x=0 to x=1.

    If it should be by washer method, then divide the solid into two solids.

    One solid will be a disc that is below the y=1 horizontal line. It has a radius of 1 unit and a height of 1 unit also. Use disc method here.
    dV1 = pi[1^2]dy
    dV1 = pi dy

    The other solid is the one above the y=1 horizontal line. And here you can use the washer method.
    dV2 = pi[(1^2 -x^2)*dy]
    dV2 = pi[1 -(y-1)]dy
    dV2 = pi[2 -y]dy

    So for the original solid,
    V = (pi)INT.(0 to 1)[dy] +(pi)INT.(1 to 2)[2 -y]dy

    -------
    Either way, by shell method or by disc+washer method, you should get the volume to be 1.5pi cubic units.
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  5. #5
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    Looks good to me.
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  6. #6
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    Quote Originally Posted by ticbol View Post
    By washer method?

    It is done easier by shell method, where dV = (2pi*x)[(x^2 +1)*dx] and the integration is from x=0 to x=1.
    I agree that the shell method would be easier, but the problem makes you use the washer method.


    Either way, by shell method or by disc+washer method, you should get the volume to be 1.5pi cubic units.
    So I got the right answer then (3pi/2).

    Thanks a ton for the help guys, I really appreciate it.
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