81 693
Assignments Done
99,2%
Successfully Done
In November 2019

# Answer to Question #11207 in Mechanics | Relativity for ajay

Question #11207
the radius of sphere is measured to be 5.3+- 0.1cm. calculate the percentage error in surface area and volume.
1
2012-06-22T10:49:31-0400
Recall that the surface area of the sphere of radius r is equal to
S = 4 pi
r^2,
and the volume is
V = 4/3 pi r^3

for r=5.3 we get that

S = 4 * 3.1416 * 5.3^2 = 352.9894 cm^2
V = 4/3 * 3.1416 * 5.3^3 =
623.6145 cm^3

for r= 5.3 - 0.1 = 5.2
S0 = 4 * 3.1416 * 5.2^2 =
339.7947 cm^2
V0 = 4/3 * 3.1416 * 5.2^3 = 588.9774 cm^3

and for r=
5.3 + 0.1 = 5.4
S1 = 4 * 3.1416 * 5.4^2 = 366.4354 cm^2
V1 = 4/3 *
3.1416 * 5.4^3 = 659.5837 cm^3

Then the percentage error for surface area
is

(S-S0)/S = (352.9894-339.7947)/352.9894 * 100 = 3.7380 %
(S1-S)/S
= (352.9894-366.4354)/352.9894 * 100 = 3.8092 %

(V-V0)/V =
(623.6145-588.9774)/623.6145 * 100 = 5.5542 %
(V1-V)/V =
(659.5837-623.6145)/623.6145 * 100 = 5.7679 %

Thus percentage error
in surface area is in the interval [-3.7380%, 3.8092%],
while percentage
error in volume is in the interval [-5.5542%, 5.7679%]

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!