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Math Help - Shell Method Problems

  1. #1
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    Question Shell Method Problems

    I'm having difficulties on finding volumes.

    Here's my problem: The area bounded by the curves y = x + 2 and y = x^2 revolves about y = 4. What is the resulting volume?

    How do I set this up? What's the radius function and height function? (I'm trying to use the shell method.)

    I thought the radius function would be (4 - y) and the height function (y^0.5 - y + 2). But this doesn't seem to work.

    And what if that area bounded revolved around the x-axis instead? How do I set that up?

    I know how to use the disk method here, but I'm at a loss to the set up with the shell method.

    Thanks for the help! I guess I'm not just "seeing" how these things work. That's why I don't know how to set them up.
    Last edited by IdentityProblem; August 26th 2014 at 06:16 PM.
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    Re: Shell Method Problems

    Well as a hint: If you are revolving about y = 4, then moving both functions down by 4 will rotate around the x-axis (y = 0).

    Thus you could try to find the area bounded by y = x - 2 and y = x^2 - 4 when rotated around the x axis (I'm sure you have formulas to do this).
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    Re: Shell Method Problems

    Well, I can't even figure this one out with the shell method when it's rotating about the x-axis itself.
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    Re: Shell Method Problems

    Surely you have a formula in your notes to find a volume by rotating around an axis. What is it?
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    Re: Shell Method Problems

    For the Shell Method, it would be 2pi*∫ from a to b of (the radius function)*(the height function).

    The radius function? I think(?) it would just be y in this case.
    The height function? I think it would be y0.5 - (y - 2).
    If I put this into the integral (from 0 to 4), I get 224pi/15.

    But I think the real answer should be 72pi/5.
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    Re: Shell Method Problems

    For the shell method, you need to split the integral into two:

    \int _0^1 + \int _1^4

    Answer:
    Spoiler:
    \frac{108 \pi }{5}
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    Unhappy Re: Shell Method Problems

    Quote Originally Posted by IdentityProblem View Post
    For the Shell Method, it would be 2pi*∫ from a to b of (the radius function)*(the height function).

    The radius function? I think(?) it would just be y in this case.
    The height function? I think it would be y0.5 - (y - 2).
    If I put this into the integral (from 0 to 4), I get 224pi/15.

    But I think the real answer should be 72pi/5.
    If I may, let me return to this. The region is just going around the x-axis in this case.

    With the disk and washer method, this makes sense to me.
    It would be pi times the integral from -1 to 2 of [ (x+2)^2 – (x^2)^2 ] with respect to x.
    This does give the answer 72pi/5.

    When I try the shell method, what's wrong with this set up…?
    2 pi times the integral from 0 to 4 of [ (y^0.5 – y + 2) (y) ] with respect to y.
    This gives me 224pi/15.

    I guess I'm not "getting" the concepts involved. I don't know what to do if I'm forced with the shell method in this problem.
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    Re: Shell Method Problems

    Whether you rotate around the x-axis or the line y=4, if you want to use the shell method, you need to split the integral into two:


    \int _0^1 + \int _1^4
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    Re: Shell Method Problems

    Quote Originally Posted by Idea View Post
    Whether you rotate around the x-axis or the line y=4, if you want to use the shell method, you need to split the integral into two:


    \int _0^1 + \int _1^4
    Thank you. But you see, my problem is that I don't understand why that is the case.

    So there must be something I'm missing conceptually.
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    Re: Shell Method Problems

    As you mentioned, rotating around the x-axis, using the disk method,

    \pi  \int_{-1}^2 \left((x+2)^2-x^4\right) \, dx=\frac{72 \pi }{5}

    Using the shell method the answer would be

    \int _0^12\pi  y \left(2\sqrt{y}\right)dy+\int _1^42\pi  y\left(\sqrt{y}-y+2\right)dy=\frac{72 \pi }{5}

    The 'radius function' is y but the 'height function' is (y^0.5 - y + 2) for y between 1 and 4, and it is 2(y^0.5) for y between 0 and 1

    For y between 1 and 4 the 'height function' is the length of a horizontal strip that extends from the straight line to the parabola
    For y between 0 and 1,the 'height function' is the length of a horizontal strip that extends from one branch
    of the parabola (x = -y^0.5) to another (x = y^0.5), so its length is 2(y^0.5)
    Last edited by Idea; August 31st 2014 at 03:43 AM.
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  11. #11
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    Re: Shell Method Problems

    Thank you, Idea!

    This is making a little more sense.
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