Solid Mensuration

A closed cylindrical container 10 feet in height and 4 feet in diameter contains water with depth of 3 feet and 5 inches. What would be the level of the water when the tank is lying in horizontal position?

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  1. Am I overthinking this ????

    volume of water in cylinder being upright
    = π(2^2)(41/12) inches^3 , (arrhhhg -- non-metric units)
    = (41/3)π in^3

    Now lay it on its side, and make a diagram of the cross section
    Shade in the water which is a segment of the circle, label the endpoints as A and B. Mark the centre as C
    we know the volume = segment x 10
    (41/3)π = 10segment
    segment area = (41/30)π in^2

    We can see that
    segment = area of sector - area of triangle.
    (this is becoming harder than I anticipated)
    let the central angle of all this mess be θ radians
    area of triangle = (1/2)(2)(2)sinθ
    = 2sinθ

    area of whole circle = 4π
    our sector = θ/2π of that
    or (θ/2π)(2π) = θ in^2

    so the area of our triangle is θ - (41/30)π
    But we know area of triangle is 2sinθ

    This is getting really messy!!!!

    2sinθ = θ - (41/30)π
    I used Wolfram to find θ=3.53213 radians, but that makes no sense!!!!

    If this had been correct, I could now find the height of the triangle and then easily find the depth of the water.

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