Question

an error of 3.5% too large was made when measuring the radius of sphere. ignoring the products of small quantities, determine the approximate error in calculating i) the volume ii) the surface area, when they are calculated using the correct radius of measurements

Answer 1

let the correct volume be v
error in measurement of volume Δv
error in measurement of radius Δr=3.5%
v+Δv= 4/3(pi (r+Δr)^3)
Δv=v+Δv - v
Δv=4/3pi((r+Δr)^3-(r)^3)
=4/3*pi((Δr(r^2+2rΔr+Δr^2+r^2+r^2+rΔr)
=4/3pi((Δr(3r^2+3rΔr+Δr^2))
Δr^2 and Δr^3 can be neglected
=4/3pi((3r^2Δ r+0)
=4pi(r^2Δr)
we need r to find the error in volume,