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Math Help - Volume of a point inside a truncated cone

  1. #1
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    Exclamation Volume of a point inside a truncated cone

    Hey guys I want to use a 12oz cup to measure things with, but I can't figure out how I would go about doing this. The measurements are roughly 1.1 inch radius at base, 1.6 inch radius on top and 3.8 inch height.

    For example, if I wanted to fill it with 4oz of water, how would I calculate what height to fill it at to get this?

    Thanks!
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    Re: Volume of a point inside a truncated cone

    You want to use a 12 oz cup to measure things with. The measurements are 1.1 inch at base, 1.6 inch radius and 3.7 inch height.

    The formula you're searching for says
    Volume = pi/3*h(R^2+R*r+r^2)
    where R is the larger radius
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    Re: Volume of a point inside a truncated cone

    I know that, but let's say I want to fill it with 4.5oz of water or something. How would I know what height to fill it to? (without cutting it, etc.)
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    Re: Volume of a point inside a truncated cone

    Quote Originally Posted by nucci93 View Post
    I know that, but let's say I want to fill it with 4.5oz of water or something. How would I know what height to fill it to? (without cutting it, etc.)
    Since I'm from Europe I gotta love the metric system Use the given formula and solve for the unknown variables. I'll walk you through this:

    The formula says
    Volume = pi/3*h(R^2+R*r+r^2)

    We also know that the relationship between the height and the larger radius (top) is
    h/R = 3.7/1.6 <=> h = 3.7R/1.6

    In this case we then have
    Volume = pi/3*h(R^2+R*r+r^2) => 4.5 = pi/3*h(R^2+R*1.1+1.1^2)
    You can solve this equation by first using the above given relationship between R and h into this equation. This will give you only h as unknown to solve for.
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    Re: Volume of a point inside a truncated cone

    Quote Originally Posted by nucci93 View Post
    I know that, but let's say I want to fill it with 4.5oz of water or something. How would I know what height to fill it to? (without cutting it, etc.)
    in addition to converting ounces to cubic inches, you would need to use concepts from calculus (volumes of rotation) to solve this problem.

    R = upper radius of the glass

    r = lower radius

    H = glass height

    h = height to fill for desired volume

    V = desired volume in cubic inches

    V = \pi \left[\frac{(R-r)^2 \cdot h^3}{3H^2} + \frac{r(R-r) \cdot h^2}{H} + r^2 \cdot h\right]

    convert your desired volume to cubic inches, substitute in the radii and height measurements of the glass ... solve for h.
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