Daniel B. answered 02/23/23
A retired computer professional to teach math, physics
Let
c = 78.000 cm be the nominal circumference of the sphere,
Δc = 0.500000 cm be the possible error in the circumference,
C = c ± Δc represent the interval (c - Δc, c + Δc),
R = C/2π be the (interval containing) the radius of the sphere,
S = 4πR² be the (interval containing) the surface area of the sphere.
Rewrite by substituting:
S = 4π((c ± Δc)/2π)² = (c ± Δc)²/π = (c² ± 2cΔc + Δc²)/π
Applying linear approximation
S ≈ (c² ± 2cΔc)/π = c²/π ± (2c/π)Δc
The maximum error in area is
(2c/π)Δc = (2×78/π)0.5 = 24.828 cm²
The nominal surfase area is c²/π, and the relative error is the ratio
(2c/π)Δc / (c²/π) = 2Δc/c = 2×0.5/78 = 0.01282
Notice that the relative error in the surface area, 2Δc/c, is
twice the relative error in the circumference, Δc/c.