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1. Diameter of Sphere and Computation of its Area and Volume
The area of a sphere of radius,
r
, is given by A = 4 π
r
2
. The volume, by (4/3) π
r
3
. Since
the diameter,
d
, is twice the radius, (
d
= 2
r
)we can rewrite the above as: A = π
d
2
and V = (1/6) π
d
3
Suppose our sphere is a basketball and we measure its diameter with a ruler whose smallest
division is 1 inch. (Such a ruler is shown below.) Using our rule of thumb, we can read the ruler
to the nearest 0.2 in.). We measure the diameter to be 9.4 in. Our understanding is that our
diameter may have an error of ±0.2 in. due to our visual estimation which corresponds to a
percent error
of 100 X 0.2 / 9.4 = 2.1%. We seek to find out what the impact of such an error
will be on the area and volume of the ball.
A simple way is to calculate the area and volume of the sphere from the value of our measured
diameter, and then to add and subtract the uncertainty and recalculate the area and volume.
Diameter
(in)
Area (in
2
)
Volume
(in
3
)
We write the Area
and Volume in
scientific notation
because of the
number of signficant
figures justified in
their values
9.4
2.8 X 10
2
4.4 X 10
2
9.4  0.2 =
9.2
2.7 X 10
2
4.1X 10
2
9.4 + 0.2 =
9.6
2.9 X 10
2
4.6 X 10
2
The
Range
of the volume (the difference between the maximum and minimum values) is 4.6 X
10
2
 4.1 X 10
2
= 5 X 10
1
(I.e., ~50 in
3
separate the minimum and maximum values!). Note that
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 Fall '09
 sdsd

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