Calculus 2
posted by Amy on .
An underground tank full of water has the following shape:
Hemisphere  5 m radius. at the bottom
Cylinder  radius 5 m and height 10m in the middle
Circular cone radius 5 m and height 4 m at the top
The top of the tank is 2 m below the ground surface and is connected to the surface by a spout. find the work required to empty the tank by pumping all of the water out of the tank up to the surface.
density of water = 1000 kg/m^3
Gravity = 10 m/s^2
I am doing to where I have three parts to this question. I find the work of all of them then add the work done of all 3 together.. However, I cannot figure out how to find the work done for the hemisphere OR the circular cone. Please help me solve this out I have no idea where to start!

OK, Here is the hemisphere.
we have a hemisphere with base 16 feet below ground and bottom 21 feet below ground.
We need its volume and the distance of the cg below ground.
The volume is easy, half a sphere
(1/2) (4/3) pi r^3 = (2/3) pi 125 = 250 pi/3
the centroid of a sphere is 3/8 r from the base as derived here:
http://mathworld.wolfram.com/Hemisphere.html
Therefore the center of mass of the hemisphere is
21 +(3/8)5
below earth
therefore we must lift a weight of water of
rho g (250 pi/3) a distance of (21+15/8) meters
that is in Joules
use rho = 10^3 kg/m^3 and g = 10 m/s^2 
rho g (250 pi/3) a distance of (16+15/8) meters

Now do the cone the same way
base is at 6 meters
volume = (1/3) pi r^2(4)
cg is at [6  (1/4)4] meters below ground 
Can you do the rest now?

Why would you do the distance of hemisphere from 0 to 16, when we are doing just the hemipsher alone then adding it to the rest after.. wouldnt distance by 5dy

You are lifting the water from the cg of the hemisphere all the way to the surface.
that is 16 meters to the top of the hemisphere plus another 15/8 to the cg