Question

If the radius of a sphere is measured as 9 cm with an error 0.03cm , then find the approximate error in calculating its volume.

Answer 1

$r=9,\delta r=0.03$
$v=\frac{4}{3}\pi r^3$
$\delta v=\frac{4}{3}\pi \left(3r^2\right)\delta r$
$=4\pi \left(9^2\right)\left(0.03\right)$
$=324\pi \left(0.03\right)$
$=9.72\pi c{m}^3$