MATH 271 Test #3T Solutions
(1) (20 points) A tank in the shape of a sphere with a radius of 5 meters is filled halfway
with water (density: 1000 kg/m
3
). The center of the tank is 20 meters underground.
How much work does it take to pump the water to the surface? Evaluate the integral.
(Note
g
= 9
.
8 m/s
2
.)
Solution:
The amount of work needed to pump the water to the surface is the
integral of the weight of the water (in a thin slice) times the distance it has to
be moved.
The weight of the water is the density of water times
g
, times the
crosssection area at a given depth, times
dx
. If the depth of the water is
x
, then
the crosssection area of the tank is
πr
2
, where
r
2
+ (
x

20)
2
= 5
2
(because the
tank is
20
meters below the surface and is spherical with radius
5
). The area is
only counted for the bottom half of the tank, so
20
≤
x
≤
25
. The integral which
represents the work needed to pump out all the water is
25
Z
20
(1000)(9
.
8)(
πr
2
)
x dx
= 9800
π
25
Z
20
(5
2

(
x

20)
2
)
x dx.
To evaluate the integral, expand
(
x

20)
2
=
x
2

40
x
+ 400
and multiply out the
polynomial to get
9800
π
25
Z
20
25
x

x
3
+ 40
x
2

400
x
= 9800
π

x
4
4
+
40
x
3

375
2
x
2
25
20
= 56
,
123
,
243
.
79 J
The substitution
u
=
x

20
could also be made.
Grading: This problem was graded on a
0
–
5
–
10
–
15
–
20
basis.
Grading for
common mistakes:

5
points if the integral wasn’t evaluated;
+7
points (total)
if the mass times a depth of
20 m
or
25 m
was used;
+10
points (or more) if the
expression involved an integra;
+15
points (or more) if the integral of an area
was found.
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 Spring '08
 Hurlbert
 Math, Calculus, ........., vescica piscis

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