Asked by Rebekah
Water is poured into a container in the sape of a right circlar cone with raiud 4t an heigh 16 feet. Express the volue (V) of the water in the cone as a funtio of the heigh (h) of the water.
thanks
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Answers
Answered by
Damon
I can not read your radius so call it R.
I assume the point of the one is down.
the radius of the water surface at height h is r = (16/R)h
the area of that water surface is pi r^2
The volume of a horizontal slice dh is
pi r^2 dh
which is
pi (16/R)^2 h^2 dh
so integrating the volume at height h is
V(h) = pi (16/R)^2 h^3 /3
if for example R were 4 feet then
V(h) = pi (16/3) h^3
I assume the point of the one is down.
the radius of the water surface at height h is r = (16/R)h
the area of that water surface is pi r^2
The volume of a horizontal slice dh is
pi r^2 dh
which is
pi (16/R)^2 h^2 dh
so integrating the volume at height h is
V(h) = pi (16/R)^2 h^3 /3
if for example R were 4 feet then
V(h) = pi (16/3) h^3
Answered by
Steve
or, just using similar cones, at height h, the surface of the water has a radius which can be found by
R/16 = r/h
r = hR/16
v = 1/3 pi r^2 * h
= pi/3 (hR/16)^2 * h
= pi/3 * R^2/256 * h
Hmmm. Better check both our maths - we disagree.
R/16 = r/h
r = hR/16
v = 1/3 pi r^2 * h
= pi/3 (hR/16)^2 * h
= pi/3 * R^2/256 * h
Hmmm. Better check both our maths - we disagree.
Answered by
Reiny
or , assuming you are saying the radius of the cone is 4 ft,
let the height of water be h
and the radius of the water level be r
r/h = 4/16 = 1/4
r = h/4
V = (1/3)πr^2 h
= (1/3)π(h^2/16)h
= (π/48)h^3
let the height of water be h
and the radius of the water level be r
r/h = 4/16 = 1/4
r = h/4
V = (1/3)πr^2 h
= (1/3)π(h^2/16)h
= (π/48)h^3
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