a=2% error and b=5% error. Area=?% error

Question

mass 3% error and length of cube 2% error in measurement . How much percentage error in volume

Answer 1

PART 1:

a=2% error, b=5% error, area =ab. a(1±0.02)b(1±0.05)=ab(1±0.07) approx, so the area error is about 7%.

PART 2:

If the side of the cube has length a(1±0.02) the volume is a³(1±0.02)³. 

Error in volume is a³(1±0.02)³/a³=(1±0.02)³=1±0.06 approx, 6% error.

Density of cube is constant and density=mass/volume.

mass is m(1±0.03), density=m(1±0.03)/[a³(1±0.02)³].

If there were no error margin, density, d=m/a³.

Error in density=(1±0.03)/(1±0.02)³=(1±0.03)(1±0.02)^-3=(1±0.03)(1∓0.06) approx.

Error=1±0.09 approx, 9% error in density.

CHECK 

Suppose cube has side length 100mm with an error of 2%, then maximum length=102mm and minimum length=98mm.

Suppose mass is 100mg with an error of 3%, then maximum mass is 103mg and minimum is 97mg. Density is minimum when we have (minimum mass)/(maximum volume) and is maximum when we have (maximum mass)/(minimum volume). 

Therefore 97/102³≤d≤103/98³ mg/mm³, 0.0000914≤d≤0.00010944 mg/mm³. Expected density=0.0001 mg/mm³.

Therefore d ranges from 91.4% and 109.4% of its expected value. We can see 91.4% is 8.6% below expectation and 109.4 is 9.4% above.

The estimate of ±9% is reasonably accurate for the error in the density.