Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 16 - Multiple Integration - 16.5 Applications of Multiple Integrals - Exercises - Page 891: 26

Answer

The center of mass: $\left( {{x_{CM}},{y_{CM}}} \right) = \left( {0,\frac{{3\pi R}}{{16}}} \right)$.

Work Step by Step

We have a region ${\cal D}$ defined by the semicircle ${x^2} + {y^2} \le {R^2}$, $y \ge 0$; and the mass density $\delta \left( {x,y} \right) = y$. The description of ${\cal D}$ in polar coordinates: ${\cal D} = \left\{ {\left( {r,\theta } \right)|0 \le r \le R,0 \le \theta \le \pi } \right\}$ Im polar coordinates: $\delta \left( {r\cos \theta ,r\sin \theta } \right) = r\sin \theta $. 1. Evaluate the mass of ${\cal D}$: $M = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \delta \left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{\theta = 0}^\pi \mathop \smallint \limits_{r = 0}^R {r^2}\sin \theta {\rm{d}}r{\rm{d}}\theta $ $ = \left( {\mathop \smallint \limits_{\theta = 0}^\pi \sin \theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^R {r^2}{\rm{d}}r} \right)$ $ = - \left( {\cos \theta |_0^\pi } \right)\left( {\frac{1}{3}{r^3}|_0^R} \right)$ $ = - \left( { - 1 - 1} \right)\left( {\frac{1}{3}{R^3}} \right) = \frac{2}{3}{R^3}$ 2. Evaluate the $x$-coordinate of the center of mass: ${x_{CM}} = \frac{1}{M}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} x\delta \left( {x,y} \right){\rm{d}}A = \frac{3}{{2{R^3}}}\mathop \smallint \limits_{\theta = 0}^\pi \mathop \smallint \limits_{r = 0}^R {r^3}\cos \theta \sin \theta {\rm{d}}r{\rm{d}}\theta $ $ = \frac{3}{{2{R^3}}}\left( {\mathop \smallint \limits_{\theta = 0}^\pi \cos \theta \sin \theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^R {r^3}{\rm{d}}r} \right)$ Since $\mathop \smallint \limits_{\theta = 0}^\pi \cos \theta \sin \theta {\rm{d}}\theta = \frac{1}{2}\mathop \smallint \limits_{\theta = 0}^\pi \sin 2\theta {\rm{d}}\theta = - \frac{1}{4}\left( {\cos 2\theta |_0^\pi } \right) = 0$ so, ${x_{CM}} = 0$. 3. Evaluate the $y$-coordinate of the center of mass: ${y_{CM}} = \frac{1}{M}\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} y\delta \left( {x,y} \right){\rm{d}}A = \frac{3}{{2{R^3}}}\mathop \smallint \limits_{\theta = 0}^\pi \mathop \smallint \limits_{r = 0}^R {r^3}{\sin ^2}\theta {\rm{d}}r{\rm{d}}\theta $ $ = \frac{3}{{2{R^3}}}\left( {\mathop \smallint \limits_{\theta = 0}^\pi {{\sin }^2}\theta {\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^R {r^3}{\rm{d}}r} \right)$ Using the Double-angle formulas in Section 1.4: ${\sin ^2}x = \frac{1}{2}\left( {1 - \cos 2x} \right)$ we get ${y_{CM}} = \frac{3}{{4{R^3}}}\left( {\mathop \smallint \limits_{\theta = 0}^\pi \left( {1 - \cos 2\theta } \right){\rm{d}}\theta } \right)\left( {\mathop \smallint \limits_{r = 0}^R {r^3}{\rm{d}}r} \right)$ ${y_{CM}} = \frac{3}{{16{R^3}}}\left( {\left( {\theta - \frac{1}{2}\sin 2\theta } \right)|_0^\pi } \right)\left( {{r^4}|_0^R} \right)$ ${y_{CM}} = \frac{{3R}}{{16}}\left( \pi \right)$ Thus, the center of mass: $\left( {{x_{CM}},{y_{CM}}} \right) = \left( {0,\frac{{3\pi R}}{{16}}} \right)$.
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