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Math Help - Buoyancy Problem

  1. #1
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    Buoyancy Problem

    Hi

    I was just wondering if someone could help me with a buoyancy question. I'm trying to calculate the volume of water displaced by my river rapids boat depending on the depth it is submerged. The rubber section, which will be partly submerged, is bascially a dognut (torus) with the inside filled in. The distance from the centre of the tube to the centre of the torus, R is 3.6m and the tube radius, r is 0.4m. Its depth is 0.8m. I know that the volume of a torus = 2 (pie^2) x R x r^2, however my problem has the inside section included as well and I need to know how its volume increases with height. Any help would be much appreciated, thanks.
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  2. #2
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    Let x be the height of water inside the torus. We want to find the volume V(x) of water inside the torus.
    We know that 0< x < 2r.

    Now consider the cylinder of volume (pi*R^2)*x. The volume V(x) = (pi*R^2)*x - [(x/(2r))*(1/2)(pi*r^2)].

    I'm pretty sure that's right; let me know if you need more detail.
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  3. #3
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    Thanks a lot for the reply. Could you just explain the second part of the equation for me though please. I was thinking it would have been the volume of the cylinder + half the volume of the torus?
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  4. #4
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    Sorry!

    First off: Yikes, I left out a factor (2pi*R)! It should be V(x) = (pi*R^2)*x - [2pi*R*(x/(2r))*(1/2)(pi*r^2)], 0<x<2r.

    I was thinking it would have been the volume of the cylinder + half the volume of the torus?
    Yes, except with a minus - the total volume that the hole in the torus contains is V(2r) = (pi*R^2)*2r - [2pi*R*(2r/(2r))*(1/2)(pi*r^2)]=(volume of the cylinder) - (pi^2*r^2*R = half the volume of the torus.

    Explaining the subtracted part of V(x): think about the area (x/(2r)*(1/2)pi*r^2 that a cylinder of height x intersects with the torus. Then multiply this by 2pi*R, the circumference of circle part of the cylinder, to get a volume.

    Hope this helps.
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  5. #5
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    Thanks a lot mate.
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