# Calculus 2

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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!

• Calculus 2 -

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

• whoops base at 16 meters -

rho g (250 pi/3) a distance of (16+15/8) meters

• Calculus 2 -

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

• Get this now? -

Can you do the rest now?

• Calculus 2 -

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 5-dy

• Calculus 2 -

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