# 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.

Expert's answer

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%]

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%]

## Comments

## Leave a comment