# 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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