A hollow garden roller 63cm wide with a girth of 440cm

SOLUTION

Outer circumference of roller = 440 cm
$\therefore$ Outer radius (R) = $\frac{C}{2\pi }=\frac{440\times 7}{2\times 22}=70cm$
Thickness of iron = 4 cm
$\therefore$ Inner radius (r) = 70-4 = 66 cm
$\therefore$ Volume of iron used = $\pi h(R^2-r^2)$
= $\frac{22}{7}\times 63\left[\right(70{)}^{2}c{m}^3]$
= 198 (70+66)(70-66) $c{m}^3$
= $198\times 136\times 4=107712c{m}^3$