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Math Help - Solids of Rotation.

  1. #1
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    Solids of Rotation.

    My current object with this problem is just to get to the integral. Once I have that I can solve for volume myself.

    The Question is:

    The volume of the solid obtained by rotating the region enclosed by

    \[x = 5 y, \quad y^3 = x \quad (\mbox{with } y\geq 0)\]

    about the y-axis can be computed using the method of disks or washers via an integral

    \(\displaystyle V = \int_a^b\) part I'm having trouble with, however it is with respect to y; so dy

    with limits of integration \(a = 0\) and   \(b =\sqrt{5}\)

    The volume is \(V =?\)
    Here's a graph of the function as well:


    My Attempt:
    I use the flowing formula:

    \int_{0}^{\sqrt{5}}(\pi(outer\, radius)^2-\pi(inner\, radius)^2)dy

    \int_{0}^{\sqrt{5}}(\pi(y^3)^2-\pi(5y)^2)dy
    This is what I think the integral should be, but it's not.

    \pi\int_{0}^{\sqrt{5}}((y^6)-(25y^2))dy

    25\pi\int_{0}^{\sqrt{5}}((y^6)-(y^2))dy

    \mid_{0}^{\sqrt{5}}25\pi(\frac{y^{7}}{7}-\frac{y^{3}}{3})
    Last edited by Zanderist; December 1st 2010 at 01:43 PM.
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  2. #2
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    As far as I can see, your only mistake is to have the inner and outer radii the wrong way round. For 0\leqslant y\leqslant\sqrt5, 5y is bigger than y^3, so 5y should be the outer radius. Remember that you are taking horizontal disks, so the larger radius is the one that is further from the y-axis.

    Edit. Just seen another mistake: you have multiplied the  y^6 by 25, as well as the y^2 . Obviously you shouldn't have done that.
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  3. #3
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    Quote Originally Posted by Opalg View Post
    Edit. Just seen another mistake: you have multiplied the  y^6 by 25, as well as the y^2 . Obviously you shouldn't have done that.
    I do not see this. Or at least do not understand what your trying to point out.

    However, I'm thinking you mean when I pulled out the 25 of the integral.

    Okay I got the volume now, that's this problem solved.
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