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Volumes of Revolutions

  1. Jul 28, 2006 #1
    There is a glass chamber that is to be filled with water. The chamber is divided into two parts, the outer being filled with water and the inner being empty. The chamber is such that a person can stand inside without getting wet.
    It is modelled by the equations [tex]f(x) = \frac{243}{1820x^2} - \frac{243}{1820} \mbox{and} g(x) = \frac{2}{3}f(x) = \frac{486}{5460x^2} - \frac{486}{5460}[/tex]

    The chamber has a height of 0.9m and a width of 1m, as shown in the attached image.

    1. Find the total volume to fill the flask in cubic cm.

    Alright, so what I did was rearrange each equation into terms of y as the solid is revolved around the y-axis.
    [tex]f(y) = \sqrt{\frac{243}{1820\left(y + \frac{243}{1820}\right)}}[/tex]
    [tex]g(y) = \sqrt{\frac{486}{5460 \left(y + \frac{486}{5460}\right)}}[/tex]

    To find the volume, it is upper bound - lower bound, so:
    [tex]V = \pi\int_{0}^{0.9} \left(\frac{243}{1820\left(y + \frac{243}{1820}\right)}\right) - \left(\frac{486}{5460\left(y + \frac{486}{5460}\right)}\right) dy[/tex]

    It is at this point that I am stuck. I have tried to rearrange into one fraction, but then I am unable to integrate it. Maybe somebody could shed some light as to what I could do, or a simpler way of going about this? Thanks.

    Attached Files:

  2. jcsd
  3. Jul 28, 2006 #2
    Remember that [tex]\int\frac{f'(x)}{f(x)}dx=\ln\midf(x)\mid+c[/tex], provided f(x) is of order of 1.
  4. Jul 28, 2006 #3
    Can I just split that into two separate integrals and then mulitply the answer by [tex]\pi[/tex]?

    If so, after simplifying I get:
    [tex]V = \pi\left[\left(\frac{243}{1820}ln\left(\frac{1881}{243}\right)\right) - \left(\frac{486}{5460}ln\left(\frac{5400}{486}\right)\right)\right] = 185,064.4144 cm^3[/tex]
  5. Jul 28, 2006 #4
    Yes, you can do that but make sure you change the name of the function to other than [tex]V(y)[/tex] before multiplying the answer by [tex]\pi[/tex].

    Yes, the answer is correct. [tex]V=1.85\times10^5\ cm^3[/tex]
  6. Jul 28, 2006 #5
    Excellent. I now to have find the constant rate of flow if the chamber is to fill in 20 min. Would that just be [tex]\frac{dV}{dt} = \frac{185,064.4144}{20}[/tex]?
  7. Jul 31, 2006 #6
    Is my last post correct?
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