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A cylinder water tank with a radius 2 feet and length 6 feet…

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A cylinder water tank with...
A cylinder water tank with a radius 2 feet and length 6 feet is filled with water to a depth of 3 feet when in a horizontal position. If the tank is turned upright, what is the depth of the water? Give answer in terms of pi.
Submitted: 8 years ago.Category: Math Homework
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Answered in 47 minutes by:
6/18/2010
Math Tutor or Teacher: Emanuel, Bachelor's Degree replied 8 years ago
Emanuel
Emanuel, Bachelor's Degree
Category: Math Homework
Satisfied Customers: 614
Experience: Aerospace engineer, expert in Maths and Physics
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In the beginning, the cylind is horizontal, so its height is 4 (the diameter) and it's 6 feet long.
It gets filled with water up to 3 meters: imagine the cylinder to be transparent, if you look at one of its basis, you'll see a circle (4 feet tall) full of water up to 3 meters.
The first thing we want to do is finding the area of the part of circle where we see the water:graphic

AHC and BHC are right triangles with hypotenuse double than one of the sides, hence β=60°.
The area of the triangle ABC is AB*HC/2
AB=2*(AC*sin(β))=2√(3)
HC=1
Area ABC=(2√(3))*1/2=√(3)
The area of the whole circle is πr²=4π
The area of the circle filled with water, without the triangle ABC, is:
A_circle*(360°-2β)/(360°)=A_circle*(2/3)=4π*(2/3)=8π/3
Area filled with water: A'=(8π/3)+√(3)
The cylinder is long 6, so the volume of the water inside the cylinder is:
V'=h*A'=6*[(8π/3)+√(3)]=16π+6√(3)

When the cylinder is upright, the volume of water is the same, but this time the area of the basis is the whole circle:
V'=A*h'=4πh'

We can now find the depth of the water:
16π+6√(3)=4πh'
h'=[16π+6√(3)]/(4π)
h'=4 + [3√(3)]/(2π)

EDIT: sorry, just noticed I typed a 4 in the last line of the answer, where it should have been 2 - fixed now

Emanuel

Edited by Emanuel on 6/18/2010 at 12:20 AM EST
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