Find the total surface area of a hollow cylinder open at both ends, if its length is 12 cm, external diameter is 8 cm and the thickness is 2 cm.

Given, external diameter = 8 cm and h = 12 cm

External radius (R) = $\dfrac{\text{diameter}}{2} = \dfrac{8}{2} = 4$ cm.

Let internal radius be r cm.

Thickness = (R - r)

⇒ 2 = 4 - r

⇒ r = 4 - 2

⇒ r = 2 cm.

By formula,

Total surface area of hollow cylinder = 2π(Rh + rh + R² - r²)

$= 2 \times \dfrac{22}{7} (4 \times 12 + 2 \times 12 + 4^2 - 2^2) \\[1em] = \dfrac{44}{7} (48 + 24 + 16 - 4) \\[1em] = \dfrac{44}{7} \times 84 \\[1em] = \dfrac{3696}{7} \\[1em] = 528 \text{ cm}^2.$

Hence, the total surface area of the given hollow cylinder open at both ends is 528 cm².