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?

5 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
oops ... missed the horizontal part

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 sector :

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 segment = 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 ?