The arc of the curve \(y ^ { 2 } = x ^ { 2 } + 8\) between the points where \(x = 0\) and \(x = 6\) is rotated through \(2 \pi\) radians about the \(x\)-axis. Show that the area \(S\) of the curved surface formed is given by
$$S = 2 \sqrt { 2 } \pi \int _ { 0 } ^ { 6 } \sqrt { x ^ { 2 } + 4 } \mathrm {~d} x$$
By means of the substitution \(x = 2 \sinh \theta\), show that
$$S = \pi \left( 24 \sqrt { 5 } + 4 \sqrt { 2 } \sinh ^ { - 1 } 3 \right)$$