Radius of a garden roller (r) = 1.4 / 2 = 0.7 m and length of the garden roller (h) = 2 m ∴ Curved surface area of garden roller = 2πrh . So, the area levelled of a garden roller in one revolution is 8.8 m2. Now, area levelled of a garden in 5 revolutions = 8.8 x 5= 44m2