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Math Help - A spherical ball is completely immersed into an inverted right circular cone full of

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    A spherical ball is completely immersed into an inverted right circular cone full of

    A spherical ball is completely immersed into an inverted right circular cone full of water. After the ball is removed, it was found that, it was found that the water surface has dropped 6cm below the top of the cone. The radius of the cone is 12cm and its altitude is 36cm. Determine the area of contact with the water surface when it is full.
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    Re: A spherical ball is completely immersed into an inverted right circular cone full

    Hey alden35.

    For this problem you may have to use Archimedes principle and convert the volumes to surface areas.

    Just out of curiousity, have they talked about Archimedes principle on your class?
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    Re: A spherical ball is completely immersed into an inverted right circular cone full

    no
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    Re: A spherical ball is completely immersed into an inverted right circular cone full

    Archimedes principle is that (with regards to your problem) the volume of water lost is the volume of the thing that was submerged. In this case its because the water escaped from the cone but in other cases, the water will rise (since it can't escape anywhere).

    So you need to calculate the volumes of the water lost relative to the cone and sphere and then use this to get the surface area of the sphere for that particular region.

    Another hint is to remember that the region of the integral will be from r = 0 to r = h for some height h (due to symmetry of a sphere) so you will need to match the volume with its corresponding integral and then solve for h. You then use that to get the surface area of the sphere and you're done.
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    Re: A spherical ball is completely immersed into an inverted right circular cone full

    Quote Originally Posted by chiro View Post
    Archimedes principle is that (with regards to your problem) the volume of water lost is the volume of the thing that was submerged. In this case its because the water escaped from the cone but in other cases, the water will rise (since it can't escape anywhere).

    So you need to calculate the volumes of the water lost relative to the cone and sphere and then use this to get the surface area of the sphere for that particular region.

    Another hint is to remember that the region of the integral will be from r = 0 to r = h for some height h (due to symmetry of a sphere) so you will need to match the volume with its corresponding integral and then solve for h. You then use that to get the surface area of the sphere and you're done.
    I understand the question to calculate the surface area of the cone without base. pi*R*sqrt(h^2 + R^2)
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