8 The continuous random variable \(X\) has probability density function f given by $$f ( x ) = \begin{cases} \frac { 1 } { 6 } x & 2 \leqslant x \leqslant 4
0 & \text { otherwise } \end{cases}$$ The random variable \(Y\) is defined by \(Y = X ^ { 3 }\). Show that \(Y\) has probability density function g given by $$g ( y ) = \begin{cases} \frac { 1 } { 18 } y ^ { - \frac { 1 } { 3 } } & 8 \leqslant y \leqslant 64
0 & \text { otherwise } \end{cases}$$ Find \(\mathrm { E } ( Y )\).