Question

what is the percentage error in surface area of the sphere, when error in measuring its radius is 2%??

Answer 1

we have to find % error in surface area of the sphere .

where it is given that , % error in measuring its radius is 2%

we know, surface area of sphere , A = 4πr²
differentiate both sides ,
$dA=8\pi r.dr$

both sides dividing by A

$\frac{dA}{A}=\frac{8\pi rdr}{A}$

$\frac{dA}{A}=\frac{8\pi rdr}{4\pi r^2}$

$\frac{dA}{A}=2\frac{dr}{r}$

both sides multiply with 100 ,

$\frac{dA}{A}\times100=2\frac{dr}{r}\times100$

hence, we get % error in A = 2 × % error in r

so, % error in Area of sphere = 2 × 2 = 4%