# geometry math

posted by
**CrazyFeet** on
.

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.