There is an upside-down cone that is 12m high and has a circular base with a radius of 4m that is being filled with liquid. Make V, the volume, a function of the height of the liquid, h. What is the volume of the cone when the water is 5m deep?
V = (1/3)pi*r^2*h
This formula doesn't help cuz the cone is upside-down. Is there another formula?
Actually,
V = (1/3)pi*r^2*h
is correct for the volume of liquid in an upside-down cone, if h is the liquid depth, measured from the tip at the bottom. In this case, r = (h/3) is the radius of the pool of liquid inside when the liquid is h deep. r is a function of h.
Therefore
V = (1/9)pi*h^3
Plug in h = 5 for the volume when the water is 5m deep.
V = Pir^2h/3
The volume of the entire cone is V = 3.14(4^2)12/3 = 201.06
Too find the volume when the water level is 5 ft., you must subtract the volume of the cone above the 5 ft. level or a cone of 7 ft. height and a base radius of 4 - 1.666 = 2.333 ft.
V(5) = 201.06 - 3.14(2.333^2)7/3 = 161.16 cub.ft.
To find the volume of the cone when the water is 5m deep, we need to subtract the volume of the cone above the 5m level from the total volume of the cone.
The total volume of the cone is given by V = (1/3)πr^2h, where r is the radius of the circular base and h is the height of the cone.
In this case, the radius of the cone at a height h is given by r = (h/3). Therefore, the volume of the cone when the water is h meters deep is V = (1/9)πh^3.
To find the volume when the water level is 5m, we substitute h = 5 into the formula:
V(5) = (1/9)π(5^3) = (1/9)π125 = 125π/9 ≈ 43.98 m^3.
So, the volume of the cone when the water is 5m deep is approximately 43.98 cubic meters.