Calculus (3rd Edition)

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

Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - Chapter Review Exercises - Page 639: 35

Answer

total area $ = {\rm{e}} - {{\rm{e}}^{ - 1}}$

Work Step by Step

Notice from Figure 2, the right-hand side of the curve from the $y$-axis resides in the first and and the third quadrant. So, the limits of integration is $ - \frac{\pi }{2}$ and $\frac{\pi }{2}$. Let $A$ denote the region enclosed by this part of the curve. Using Eq. (2) of Theorem 1 of Section 12.4, the area of $A$ is $area = \frac{1}{2}\cdot\mathop \smallint \limits_{ - \pi /2}^{\pi /2} \cos \theta {{\rm{e}}^{\sin \theta }}{\rm{d}}\theta $ Let $t=\sin \theta$. So, $dt = \cos \theta d\theta $. The integral becomes $area = \frac{1}{2}\cdot\mathop \smallint \limits_{ - 1}^1 {{\rm{e}}^t}{\rm{d}}t$ $area = \frac{1}{2}{{\rm{e}}^t}|_{ - 1}^1 = \frac{1}{2}\left( {{\rm{e}} - {{\rm{e}}^{ - 1}}} \right)$ By symmetry, total area enclosed by the curve is twice the area of $A$. So, total area $ = {\rm{e}} - {{\rm{e}}^{ - 1}}$
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