Chapter 16 - Multiple Integration - 16.5 Applications of Multiple Integrals - Exercises - Page 891: 16

We show that the $y$-coordinate of the centroid is $\bar y = \left( {\frac{{2R}}{3}} \right)\left( {\frac{{\sin \alpha }}{\alpha }} \right)$

Referring to Figure 13, we see that $r$ ranges from $0$ to $R$, whereas $\theta$ varies from $\theta = \frac{\pi }{2} - \alpha $ to $\theta = \frac{\pi }{2} + \alpha $. Thus, the description of the region ${\cal D}$ in polar coordinates: ${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le R,\frac{\pi }{2} - \alpha \le \theta \le \frac{\pi }{2} + \alpha } \right\}$ Evaluate the area of ${\cal D}$: $A = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}A = \mathop \smallint \limits_{\theta = \frac{\pi }{2} - \alpha }^{\frac{\pi }{2} + \alpha } \mathop \smallint \limits_{r = 0}^R r{\rm{d}}r{\rm{d}}\theta $ $ = \frac{1}{2}\mathop \smallint \limits_{\theta = \frac{\pi }{2} - \alpha }^{\frac{\pi }{2} + \alpha } \left( {{r^2}|_0^R} \right){\rm{d}}\theta $ $ = \frac{1}{2}{R^2}\mathop \smallint \limits_{\theta = \frac{\pi }{2} - \alpha }^{\frac{\pi }{2} + \alpha } {\rm{d}}\theta = \frac{1}{2}{R^2}\left( {\frac{\pi }{2} + \alpha - \frac{\pi }{2} + \alpha } \right) = \alpha {R^2}$ So, $A = \alpha {R^2}$. Evaluate the average of $x$-coordinate: $\bar x = \frac{1}{A}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x{\rm{d}}A = \frac{1}{{\alpha {R^2}}}\mathop \smallint \limits_{\theta = \frac{\pi }{2} - \alpha }^{\frac{\pi }{2} + \alpha } \mathop \smallint \limits_{r = 0}^R {r^2}\cos \theta {\rm{d}}r{\rm{d}}\theta $ $ = \frac{1}{{3\alpha {R^2}}}\mathop \smallint \limits_{\theta = \frac{\pi }{2} - \alpha }^{\frac{\pi }{2} + \alpha } \left( {{r^3}|_0^R} \right)\cos \theta {\rm{d}}\theta $ $ = \frac{R}{{3\alpha }}\mathop \smallint \limits_{\theta = \frac{\pi }{2} - \alpha }^{\frac{\pi }{2} + \alpha } \cos \theta {\rm{d}}\theta $ $ = \frac{R}{{3\alpha }}\left( {\sin \theta |_{\frac{\pi }{2} - \alpha }^{\frac{\pi }{2} + \alpha }} \right)$ $ = \frac{R}{{3\alpha }}\left( {\sin \left( {\frac{\pi }{2} + \alpha } \right) - \sin \left( {\frac{\pi }{2} - \alpha } \right)} \right) = \frac{R}{{3\alpha }}\left( {\cos \alpha - \cos \alpha } \right) = 0$ So, $\bar x = 0$. Evaluate the average of $y$-coordinate: $\bar y = \frac{1}{A}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y{\rm{d}}A = \frac{1}{{\alpha {R^2}}}\mathop \smallint \limits_{\theta = \frac{\pi }{2} - \alpha }^{\frac{\pi }{2} + \alpha } \mathop \smallint \limits_{r = 0}^R {r^2}\sin \theta {\rm{d}}r{\rm{d}}\theta $ $ = \frac{1}{{3\alpha {R^2}}}\mathop \smallint \limits_{\theta = \frac{\pi }{2} - \alpha }^{\frac{\pi }{2} + \alpha } \left( {{r^3}|_0^R} \right)\sin \theta {\rm{d}}\theta $ $ = \frac{R}{{3\alpha }}\mathop \smallint \limits_{\theta = \frac{\pi }{2} - \alpha }^{\frac{\pi }{2} + \alpha } \sin \theta {\rm{d}}\theta $ $ = - \frac{R}{{3\alpha }}\left( {\cos \theta |_{\frac{\pi }{2} - \alpha }^{\frac{\pi }{2} + \alpha }} \right)$ $ = - \frac{R}{{3\alpha }}\left( {\cos \left( {\frac{\pi }{2} + \alpha } \right) - \cos \left( {\frac{\pi }{2} - \alpha } \right)} \right)$ $ = - \frac{R}{{3\alpha }}\left( { - \sin \alpha - \sin \alpha } \right) = \frac{{2R\sin \alpha }}{{3\alpha }}.$ So, $\bar y = \left( {\frac{{2R}}{3}} \right)\left( {\frac{{\sin \alpha }}{\alpha }} \right)$. So, the coordinates of the centroid: $\left( {\bar x,\bar y} \right) = \left( {0,\left( {\frac{{2R}}{3}} \right)\left( {\frac{{\sin \alpha }}{\alpha }} \right)} \right)$.