A Study Guide on Work
Produced by Anne Hawkins, Janine Pulley and Rachel McHenry
Revised by the Learning Center
Pumping Water
While quite different, this problem also involves a work integral.
When setting up this integral, it is instructional to consider the water in the
tanks as being composed of many stacked thin ‘slabs’ of thickness y . Total
work can be thought of as the sum of the work required to lift each
individual slab.
However, before we start with work, we must start with volume and area. In
our case, the tank is rectangular and the area of each slab is constant:
A  10 ft 12 ft   120 ft . From this we can get V , the change in volume of
one of these slabs.
V   area   thickness   120y
Force then equals the density (a constant which we have) times the volume,
V .
F  y    62.4 120y 
Since we are emptying this tank from the top, the variable quantity in this
problem is the vertical distance that the water is being pumped. As the water
level in the tank decreases, each ‘slab’ requires more work to pump out. This
is expressed symbolically as:
Top — y = 20— y
The integral for this particular problem will be
b
W    density F  y  dy
a
W    62.4 120  20  y  dy
20
0
The bounds are simply the heights of the top and the bottom of the tank
(defining the bottom as zero, even though the top is at ground level)
(Final answer: W = 1497600J)