A linear approximation to the error in volume can be written as
... ∆V = (∂V/∂d)·∆d + (∂V/∂h)·∆h
For V=(π/4)·d²·h, this is
... ∆V = 2·(π/4)·d·h·∆d + (π/4)·d²·∆h
Using ∆d = 0.05d and ∆h = 0.05h, this becomes
... ∆V = (π/4)·d²·h·(2·0.05 + 0.05) = 0.15·V
The nominal volume is
... V = (π/4)·d²·h = (π/4)·(2.2 m)²·(6.8 m) = 25.849 m³
Then the maximum error in volume is
... 0.15V = 0.15·25.849 m³ ≈ 3.877 m³
_____
Essentially, the error percentage is multiplied by the exponent of the associated variable. Then these products are added to get the maximum error percentage.
