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 - 12.4 Area and Arc Length in Polar - Exercises - Page 624: 11

Answer

$area = \frac{1}{{48}}{\pi ^3}$

Work Step by Step

Since $r=\theta$, the rectangular coordinates of a point on the curve corresponding to $\theta$ is given by $\left( {x,y} \right) = \left( {r\cos \theta ,r\sin \theta } \right)$, $\left( {x,y} \right) = \theta \left( {\cos \theta ,\sin \theta } \right)$. For the interval $0 \le \theta \le 2\pi $, we evaluate several points in rectangular coordinates corresponding to $\theta = 0,\frac{\pi }{4},\frac{\pi }{2},...,2\pi $ and list them on the following table: $\begin{array}{*{20}{c}} \theta \\ 0\\ {\frac{\pi }{4}}\\ {\frac{\pi }{2}}\\ {\frac{{3\pi }}{4}}\\ \pi \\ {\frac{{5\pi }}{4}}\\ {\frac{{3\pi }}{2}}\\ {\frac{{7\pi }}{4}}\\ {2\pi } \end{array}\begin{array}{*{20}{c}} {\left( {x,y} \right)}\\ {\left( {0,0} \right)}\\ {\left( {\frac{\pi }{8}\sqrt 2 ,\frac{\pi }{8}\sqrt 2 } \right)}\\ {\left( {0,\frac{\pi }{2}} \right)}\\ {\left( { - \frac{{3\pi }}{8}\sqrt 2 ,\frac{{3\pi }}{8}\sqrt 2 } \right)}\\ {\left( { - \pi ,0} \right)}\\ {\left( { - \frac{{5\pi }}{8}\sqrt 2 , - \frac{{5\pi }}{8}\sqrt 2 } \right)}\\ {\left( {0, - \frac{{3\pi }}{2}} \right)}\\ {\left( {\frac{{7\pi }}{8}\sqrt 2 , - \frac{{7\pi }}{8}\sqrt 2 } \right)}\\ {\left( {2\pi ,0} \right)} \end{array}$ Then we plot the points in rectangular coordinates and sketch the curve by connecting these points. From the figure we see that the region bounded by the curve and the first quadrant corresponds to the interval $0 \le \theta \le \frac{\pi }{2}$. So, the area of this region by Eq. (2) of Theorem 1 is $area = \frac{1}{2}\cdot\mathop \smallint \limits_0^{\pi /2} {\theta ^2}{\rm{d}}\theta = \frac{1}{6}{\theta ^3}|_0^{\pi /2} = \frac{1}{{48}}{\pi ^3}$
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