Answer: To set up the integral, we divide the upper half of the aquarium into horizontal slices,
and for each slice, let x denote its distance from the top of the tank and ∆x denote
2
its thickness. (We choose horizontal slices because we want each drop of water in a
given slice to be the same distance from the top of the tank.) Using the formulae at
the beginning of this handout, we see that the work taken to pump such a slice out of
the tank is
work for a slice = W
= F · d
= (m · a) · d
= (ρ · V ) · a · d .
Since the length, width and thickness of the slice are given by 2 m, 1 m and ∆x m,
respectively, its volume is 2 · 1 · ∆x m3 = 2∆x m3
. Thus, the equation above becomes
work for a slice ≈
force
z }| {
mass
z }| {
(1000 kg/m
3
)
| {z }
density
(2∆x m
3
)
| {z }
volume
(9.8 m/s
2
)
| {z }
gravity
(x m)
| {z }
distance
= (1000)(9.8)(2)x · ∆x (kg · m/s
2
) · m
= (1000)(9.8)(2)x · ∆x N · m
= (1000)(9.8)(2)x · ∆x J .
Summing over our slices, this is
total work for top half of aquarium ≈
X(1000)(9.8)(2)x · ∆x J ,
where the sum is over the slices in the top half of the aquarium; that is, from distance
x = 0 to x = 1/2. As we refine our slices, this becomes the integral
total work = Z 1/2
0
(1000)(9.8)(2)x dx J
= (1000)(9.8)(2) Z 1/2
0
x dx J
= (1000)(9.8)(2)(1/8) J
= 2450 J .
