The radius of a sphere is measured as 7 centimeters, with a possible error of 0.025 centimeter. Required:
a. Use differentials to approximate the possible propagated error, in cm3, in computing the volume of the sphere.
b. Use differentials to approximate the possible propagated error in computing the surface area of the sphere.
c. Approximate the percent errors in parts (a) and (b).

a) dV(s)  =  15,386 cm³

b) dS(s) = 4,396 cm²

c) dV(s)/V(s) = 1,07 %    and   dS(s)/ S(s)  =  0,71 %

Step-by-step explanation:

a) The volume of the sphere is

V(s) = (4/3)*π*x³        where x is the radius

Taking derivatives on both sides of the equation we get:

dV(s)/ dr  =  4*π*x²    or

dV(s)  =  4*π*x² *dr

the possible propagated error in cm³ in computing the volume of the sphere is:

dV(s)  = 4*3,14*(7)²*(0,025)

dV(s)  =  15,386 cm³

b) Surface area of the sphere is:

V(s) = (4/3)*π*x³  

dV(s) /dx  =  S(s) = 4*π*x³

And

dS(s) /dx  = 8*π*x

dS(s) = 8*π*x*dx

dS(s) = 8*3,14*7*(0,025)

dS(s) = 4,396 cm²

c) The approximates errors in a and b are:

V(s) =  (4/3)*π*x³     then

V(s) = (4/3)*3,14*(7)³

V(s) = 1436,03 cm³

And  the possible propagated error in volume is from a)  is

dV(s)  =  15,386 cm³

dV(s)/V(s)  = [15,386 cm³/1436,03 cm³]* 100

dV(s)/V(s) = 1,07 %

And for case b)

dS(s) = 4,396 cm²

And the surface area of the sphere is:

S(s) =  4*π*x³        ⇒   S(s) =  4*3,14*(7)²    ⇒ S(s) = 615,44 cm²

dS(s) = 4,396 cm²

dS(s)/ S(s)  =  [ 4,396 cm²/615,44 cm² ] * 100

dS(s)/ S(s)  =  0,71

the answer is 11.2 (i know i’m late)