Question
An oil tank in the shape of a right circular cylinder with horizontal axis is filled to a depth of 3 feet. If the circular ends of the tank have diameters of 10 feet and the length of the tank is 12 feet, how many cubic feet of oil are in the tank?
Answers
π * (d^2 / 4) * h = π * (100 / 4) * 3 = ? ft^3
The tank is lying on its side, so the above calculation will not work.
Draw a side view of the end of the tank. Since the radius of the tank is 5, the oil surface is a chord 2 feet from the center of the circle. That means that the width of the oil surface is 2√(5^2-2^2) = 2√21 ft across, and subtends an angle θ where sin(θ/2) = √21/5
The area of the oil is thus 1/2 r^2 (θ - sinθ)
Since sin(θ/2) = √21/5, sinθ = 2(√21/5)(2/5) = 4√21/25
thus the area is 1/2 * 25 (0.823 - 4√21/25) = 1.122 ft^2
That makes the volume of oil 1.122 * 12 = 13.464 ft^3
Draw a side view of the end of the tank. Since the radius of the tank is 5, the oil surface is a chord 2 feet from the center of the circle. That means that the width of the oil surface is 2√(5^2-2^2) = 2√21 ft across, and subtends an angle θ where sin(θ/2) = √21/5
The area of the oil is thus 1/2 r^2 (θ - sinθ)
Since sin(θ/2) = √21/5, sinθ = 2(√21/5)(2/5) = 4√21/25
thus the area is 1/2 * 25 (0.823 - 4√21/25) = 1.122 ft^2
That makes the volume of oil 1.122 * 12 = 13.464 ft^3
oops ... missed the horizontal part
should have realized it was to simple for calculus
should have realized it was to simple for calculus
The cylinder is lying on its side, so the question is a bit trickier.
Make a sketch of the circular cross-section
Draw the waterline at a depth of 3 ft, join that chord to the ends of the
diameter.
You should see 2 congruent right-angled triangles, each with
hypotenuse of 5 and a leg of 2 . (5-3 = 2)
Finding the base:
x^2 + 2^2 = 5^2
x = √21
So the length of the chord , (the water line) = 2√21
( I never used that , switched strategy to using angles. )
Central angle: half-central angle:
cosθ = 2/5, θ = 66.42°
so central angle = 132.84°
area of <b>sector </b>:
sector/(π(5^2)) = 132.84/360
sector = 28.982 ft^2
area of triangle sitting on chord:
= 1/2 (5)(5)sin 132.84
= 9.165 ft^2
area of <b>segment</b> = 28.982 - 9.165 = 19.8168
volume of water = surface area * length
= 19.8168*12 = 237.8 ft^3
check my calculations.
Make a sketch of the circular cross-section
Draw the waterline at a depth of 3 ft, join that chord to the ends of the
diameter.
You should see 2 congruent right-angled triangles, each with
hypotenuse of 5 and a leg of 2 . (5-3 = 2)
Finding the base:
x^2 + 2^2 = 5^2
x = √21
So the length of the chord , (the water line) = 2√21
( I never used that , switched strategy to using angles. )
Central angle: half-central angle:
cosθ = 2/5, θ = 66.42°
so central angle = 132.84°
area of <b>sector </b>:
sector/(π(5^2)) = 132.84/360
sector = 28.982 ft^2
area of triangle sitting on chord:
= 1/2 (5)(5)sin 132.84
= 9.165 ft^2
area of <b>segment</b> = 28.982 - 9.165 = 19.8168
volume of water = surface area * length
= 19.8168*12 = 237.8 ft^3
check my calculations.
@mathhelper when u were finding the base, where did x come from or what formula did u use ?