Suppose the ends of the cylindrical storage tank in the figure arc circles of radius 3 ft and the cylinder is 20 ft long. Determine the volume of the oil in the tank to the nearest cubic foot if the rod shows a depth of 2 ft.
assuming the rod lies along a diameter of the circle, you need to find the area of a circular segment, knowing its height and the circle's radius.
Draw a diagram. If the depth is d, then the central angle subtending the surface of the oil can be found using
cos(θ/2) = (r-d)/r
and the area of the segment is
a = 1/2 r^2 (θ-sinθ)
Now multiply by the length of the tank to get the volume.
In one line, that would be
v = 20 (1/2 r^2 (θ-sinθ))
= 10*9 (2 arccos((r-d)/r) - √(r^2-(r-d)^2)
= 90(2 arccos(1/3) - √8/3)
= 136.72
To determine the volume of the oil in the tank, we can break it down into two parts: the volume of the cylinder portion and the volume of the two semi-circular ends.
First, let's calculate the volume of the cylinder:
The formula for the volume of a cylinder is given by V = πr^2h, where r is the radius of the base and h is the height (or length) of the cylinder.
In this case, the radius (r) is given as 3 ft and the length (h) is 20 ft.
So, the volume of the cylinder is V_cylinder = π(3 ft)^2(20 ft) = 180π ft^3.
Next, let's determine the volume of the two semi-circular ends:
The volume of a sphere is given by V = 4/3πr^3, but in this case, we only need half of that volume since we are dealing with semi-circular ends.
So, the volume of each semi-circular end is V_end = (1/2)(4/3π(3 ft)^3) = 18π ft^3.
Now, to find the total volume of the oil in the tank, we need to subtract the volume of the two semi-circular ends from the volume of the cylinder:
V_total = V_cylinder - 2V_end = 180π ft^3 - 2(18π ft^3) = 144π ft^3.
Finally, we need to round the volume to the nearest cubic foot:
Using a calculator, we can approximate π as 3.14.
So, V_total ≈ 144(3.14) ft^3 ≈ 452.16 ft^3.
Therefore, the volume of the oil in the tank, to the nearest cubic foot, is approximately 452 ft^3.