The total surface area of a hollow metal cylinder, open at both ends, of external radius $8 \; cm$ and height $10 \; cm$ is $338 \pi \; cm^2$. Taking $r$ to be the internal radius, write down an equation in $r$ and use it to find the thickness of the metal in the cylinder.
External radius of the hollow cylinder $= R = 8 \; cm$
Internal radius of the hollow cylinder $= r \; cm$
Height of the hollow cylinder $= h = 10 \; cm$
Total surface area of the hollow cylinder $= A = 2 \pi R h + 2 \pi r h + 2 \pi \left(R^2 - r^2\right)$
i.e. $\;$ $2 \pi \left(Rh + rh + R^2 - r^2\right) = 338 \pi$
i.e. $\;$ $Rh + rh + R^2 - r^2 = 169$
i.e. $\;$ $8 \times 10 + 10 r + 8^2 - r^2 = 169$
i.e. $\;$ $10 r - r^2 = 25$
i.e. $\;$ $r^2 - 10r + 25 = 0$
i.e. $\;$ $\left(r - 5\right)^2 = 0$ $\implies$ $r = 5 \; cm$
$\therefore \;$ Thickness of the metal in the cylinder $= R - r = 8 - 5 = 3 \; cm$