Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 15 - Section 15.4 - Applications of Double Integrals - 15.4 Exercise - Page 1078: 16

Answer

The mass of the lamina is $$ \begin{aligned} m &= \pi k. \end{aligned} $$ The center of mass is located at the point $$ (\bar{x}, \bar{y})=\left(0, \frac{3}{ \pi}\right). $$

Work Step by Step

The boundary of a lamina consists of the semicircles $y=\sqrt {1-x^{2}}$ and $y=\sqrt {4-x^{2}}$ together with the portions of the x-axis that joins them, if the density at any point is inversely proportional to its distance from the origin. Then the distance from a point $(x,y)$ to the center of the origin is $\frac{1}{\sqrt {x^{2}+y^{2}}}$. Therefore the density function is $$ \rho(x, y)=\frac{k}{\sqrt {x^{2}+y^{2}}} $$ where $k$ is some constant. Both the density function and the shape of the lamina suggest that we convert to polar coordinates. Then $\sqrt {x^{2}+y^{2}}=r $ and the region $D$ is given by: $$ D=\left\{(r, \theta) | 1 \leq r \leq 2, \quad 0 \leq \theta \leq \pi \right\} .$$ The mass of the lamina is $$ \begin{aligned} m &=\iint_{D} \rho(x, y) d A \\ &=\iint_{D} \frac{k}{\sqrt {x^{2}+y^{2}}} d A \\ &=\int_{0}^{\pi} \int_{1}^{2} \frac{k}{r} \cdot r d r d \theta \\ &=k \int_{0}^{\pi} d \theta \int_{1}^{2} d r \\ &=k(\pi)\left[ r\right]_{1}^{2} \\ &= \pi k. \end{aligned} $$ Now, we find: $$ \begin{aligned} M_{y} &=\iint_{D} x \rho(x, y) d A \\ &=\int_{0}^{\pi} \int_{1}^{2}(r \cos \theta)(k / r) r d r d \theta \\ &=k \int_{0}^{\pi} \cos \theta d \theta \int_{1}^{2} r d r \\ &=k[\sin \theta]_{0}^{\pi}\left[\frac{1}{2} r^{2}\right]_{1}^{2} \\ &=k(0)\left(\frac{3}{2}\right)=0 \end{aligned} $$ and $$ \begin{aligned} \\ M_{x} &=\iint_{D} y \rho(x, y) d A \\ &=\int_{0}^{\pi} \int_{1}^{2}(r \sin \theta)(k / r) r d r d \theta \\ &=k \int_{0}^{\pi} \sin \theta d \theta \int_{1}^{2} r d r \\ &=k[-\cos \theta]_{0}^{\pi}\left[\frac{1}{2} r^{2}\right]_{1}^{2} \\ &=k(1+1)\left(\frac{3}{2}\right)\\ &=3 k \end{aligned} $$ Therefore, the center of mass is located at the point $$ (\bar{x}, \bar{y})=\left(0, \frac{3k}{ \pi k }\right)=\left(0, \frac{3}{ \pi}\right). $$
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