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Math Help - Problem based learning

  1. #1
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    I am not really sure what type of question is this. The question is below. Thank you.

    A hollow hemispherical bowl of radius 4cm has a smaller solid spherical marble of radius 1cm at the bottom. If the bowl is filled with water to a depth of 1.5cm, find the volume of water in the bowl.
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  2. #2
    dud
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    Calculate the volume of a hemispherical object with a radius of 1.5cm - the volume of the ball which is submerged in the water?
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  3. #3
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    Quote Originally Posted by poppy_wolf
    I am not really sure what type of question is this. The question is below. Thank you.

    A hollow hemispherical bowl of radius 4cm has a smaller solid spherical marble of radius 1cm at the bottom. If the bowl is filled with water to a depth of 1.5cm, find the volume of water in the bowl.
    I would solve this problem with Calculus. The most important step is to know that given a hemisphere filled with water to depth of a having radius r then its volume is \pi (\frac{2}{3}r^3-ar^2+\frac{1}{3}a^3). Then from that your subtract the volume of the marble inside which in this case has volume \frac{4}{3}\pi
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  4. #4
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    Quote Originally Posted by ThePerfectHacker
    I would solve this problem with Calculus. The most important step is to know that given a hemisphere filled with water to depth of a having radius r then its volume is \pi (\frac{2}{3}r^3-ar^2+\frac{1}{3}a^3). Then from that your subtract the volume of the marble inside which in this case has volume \frac{4}{3}\pi

    Will the solution be like this..

    Volume of bowl = 2/3 (4^3) - (1.5)(4^2) + 1/3 (1.5^3)
    = 17 13/24

    volume of water = 17 13/24 - <br />
\frac{4}{3}\pi<br />

    = 13.35287
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  5. #5
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    Did you forget to multiply by \pi in the first one?
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  6. #6
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    Hello,

    to make sure that I got your question I've attached a sketch of the situation as I understand it.

    The marble has a diametre of 2 cm, that means it will stuck out of the water if there is only a depth of 1.5 cm.
    Therefore you have to calculate two parts of spheres:

    V_{part\ of\ sphere}=\frac{\pi}{3}\cdot h^2 \cdot (3\cdot r-h)

    V_{1}=\frac{\pi}{3}\cdot 1.5^2 \cdot (3\cdot 4-1.5)

    V_{2}=\frac{\pi}{3}\cdot 1.5^2 \cdot (3\cdot 1-1.5)

    The difference of both volumes is the volume of water you're looking for.

    Bye
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  7. #7
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    I see what you are saying earboth. I did not take into account the fact that the sphere stuck out of the water!

    This is definetly a Calculus problem for volume. It is solved with the formula I posted here before.
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