Answer
${I_h} = 4.75 \times {10^5}$ $g\cdot c{m^2}$
Work Step by Step
We have $R=10$ cm, $M=500$ g and $h=30$ cm. In this exercise, we consider the cylinder's axis of symmetry to be the $z$-axis.
Recall the Parallel-Axis Theorem in Exercise 63:
${I_h} = {I_z} + M{h^2}$
For ${I_z}$ we use the result in Exercise 45 for the cylinder:
${I_z} = \frac{1}{2}M{R^2}$
Thus,
${I_h} = \frac{1}{2}M{R^2} + M{h^2}$
$ = \frac{1}{2}\cdot500\cdot{10^2} + 500\cdot{30^2} = 4.75 \times {10^5}$
So, ${I_h} = 4.75 \times {10^5}$ $g\cdot c{m^2}$.
