Answer
The kinetic energy required to rotate the disk about the $x$-axis is $\frac{{25}}{2}M{R^2}$ joules.
Work Step by Step
We have the disk ${\cal D}$ defined by ${x^2} + {y^2} \le {R^2}$, with total mass $M$ kg. Assuming that the mass density is uniform, the mass density $\delta$ is given by $\delta = \frac{M}{A} = \frac{M}{{\pi {R^2}}}$.
Using polar coordinates, the description of ${\cal D}$:
${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le R,0 \le \theta \le 2\pi } \right\}$
Evaluate the moment of inertia ${I_x}$ in polar coordinates:
${I_x} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {y^2}\delta \left( {x,y} \right){\rm{d}}A = \frac{M}{{\pi {R^2}}}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^R {\left( {r\sin \theta } \right)^2}r{\rm{d}}r{\rm{d}}\theta $
$ = \frac{M}{{\pi {R^2}}}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } {{\sin }^2}\theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^R {r^3}{\rm{d}}r} \right)$
Consider the first integral on the right-hand side. Using the Double-angle formulas in Section 1.4:
${\sin ^2}x = \frac{1}{2}\left( {1 - \cos 2x} \right)$
we get
${I_x} = \frac{M}{{2\pi {R^2}}}\left( {\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {1 - \cos 2\theta } \right){\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^R {r^3}{\rm{d}}r} \right)$
${I_x} = \frac{M}{{8\pi {R^2}}}\left( {\left( {\theta - \frac{1}{2}\sin 2\theta } \right)|_0^{2\pi }} \right)\left( {{r^4}|_0^R} \right)$
$ = \frac{M}{{8\pi {R^2}}}\left( {2\pi } \right)\left( {{R^4}} \right) = \frac{M}{4}{R^2}$
With angular velocity $\omega = 10$ rad/s, the kinetic energy required to rotate the disk about the $x$-axis is
${\rm{RotationalKE}} = \frac{1}{2}{I_x}{\omega ^2} = \frac{1}{2}\cdot\frac{M}{4}{R^2}\cdot{10^2} = \frac{{25}}{2}M{R^2}$ joules
