Answer
${y_{CM}} = \frac{1}{{12}}$
Work Step by Step
Let ${\cal W}$ denote the solid (B) in Figure 4. The description of ${\cal W}$ in cylindrical coordinates:
${\cal W} = \left\{ {\left( {r,\theta ,z} \right)|0 \le r \le 1,0 \le \theta \le 2\pi ,0 \le z \le \frac{1}{2}r\sin \theta + \frac{3}{2}} \right\}$
We assume in this exercise that the mass density is uniform. So, the center of mass coincides with the centroid.
Evaluate the volume of ${\cal W}$:
$V = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} {\rm{d}}V = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 \mathop \smallint \limits_{z = 0}^{\left( {r\sin \theta } \right)/2 + 3/2} r{\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 r\left( {z|_0^{\left( {r\sin \theta } \right)/2 + 3/2}} \right){\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 \left( {\frac{1}{2}{r^2}\sin \theta + \frac{3}{2}r} \right){\rm{d}}r{\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\left( {\frac{1}{6}{r^3}\sin \theta + \frac{3}{4}{r^2}} \right)|_0^1} \right){\rm{d}}\theta $
$ = \mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\frac{1}{6}\sin \theta + \frac{3}{4}} \right){\rm{d}}\theta $
$ = \left( { - \frac{1}{6}\cos \theta + \frac{3}{4}\theta } \right)|_0^{2\pi } = \frac{3}{2}\pi $
Evaluate ${y_{CM}}$:
${y_{CM}} = \frac{1}{V}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal W}^{} y{\rm{d}}V = \frac{2}{{3\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 \mathop \smallint \limits_{z = 0}^{\left( {r\sin \theta } \right)/2 + 3/2} {r^2}\sin \theta {\rm{d}}z{\rm{d}}r{\rm{d}}\theta $
$ = \frac{2}{{3\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 {r^2}\sin \theta \left( {z|_0^{\left( {r\sin \theta } \right)/2 + 3/2}} \right){\rm{d}}r{\rm{d}}\theta $
$ = \frac{2}{{3\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } \mathop \smallint \limits_{r = 0}^1 \left( {\frac{1}{2}{r^3}{{\sin }^2}\theta + \frac{3}{2}{r^2}\sin \theta } \right){\rm{d}}r{\rm{d}}\theta $
$ = \frac{2}{{3\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\left( {\frac{1}{8}{r^4}{{\sin }^2}\theta + \frac{1}{2}{r^3}\sin \theta } \right)|_0^1} \right){\rm{d}}\theta $
$ = \frac{2}{{3\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {\frac{1}{8}{{\sin }^2}\theta + \frac{1}{2}\sin \theta } \right){\rm{d}}\theta $
Using the Double-angle formulas in Section 1.4:
${\sin ^2}x = \frac{1}{2}\left( {1 - \cos 2x} \right)$
we get
${y_{CM}} = \frac{1}{{24\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } \left( {1 - \cos 2\theta } \right){\rm{d}}\theta + \frac{1}{{3\pi }}\mathop \smallint \limits_{\theta = 0}^{2\pi } \sin \theta {\rm{d}}\theta $
$ = \frac{1}{{24\pi }}\left( {\left( {\theta - \frac{1}{2}\sin 2\theta } \right)|_0^{2\pi }} \right) - \frac{1}{{3\pi }}\left( {\cos \theta |_0^{2\pi }} \right)$
$ = \frac{1}{{24\pi }}\left( {2\pi } \right) - \frac{1}{2}\left( {1 - 1} \right) = \frac{1}{{12}}$
So, ${y_{CM}} = \frac{1}{{12}}$.
