3. A water tank is in the shape of a right circular cone
of altitude 10 feet and base radius
5 feet, with its base at the ground. If the tank is
full, find an integral but do not evaluate for the work
done in pumping the top 5 feet of the water out of the top.
Let h be the height off the ground of a horizontal
slice of water in the tank. Let r be the radius of that slice
(which is circular). Using similar triangles,
we get (10 - h)/r = (total height)/(base radius) = 10/5 = 2,
so r = (10 - h)/2:
___|___ radius = r, at height h
total height = 10 |
_________|_________ base radius = 5, at height h = 0
Each slice of water of thickness dh weighs 62.4(Pi)r2 dh pounds,
and must be lifted 10 - h feet, for 62.4(Pi)((10 - h)/2)2(10 - h)dh ft.lbs.
We must do this for h = 5 to h = 10 (i.e., for the upper 5 feet of water).
So the answer is the integral from h = 5 to h = 10 of
62.4(Pi)((10 - h)/2)2(10 - h)dh.