Calculus: Early Transcendentals 9th Edition

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

Chapter 10 - Section 10.4 - Areas and Lengths in Polar Coordinates - 10.4 Exercises - Page 699: 2

Answer

$\frac{e^{3\pi}-\sqrt{e^{3\pi}}}{4}$

Work Step by Step

The area of the region is determined by the integral $A=\int_{3\pi/4}^{3\pi/2}\frac{1}{2}r^2d\theta$, Evaluate $A$: $A=\int_{3\pi/4}^{3\pi/2}\frac{1}{2}(e^{\theta})^2d\theta$ $A=\int_{3\pi/4}^{3\pi/2}\frac{1}{2}e^{2\theta}d\theta$ $A=[\frac{e^{2\theta}}{4}]_{3\pi/4}^{3\pi/2}$ $A=\frac{e^{3\pi/2\cdot 2}}{4}-\frac{e^{3\pi/4\cdot 2}}{4}$ $A=\frac{e^{3\pi}}{4}-\frac{e^{3\pi/2}}{4}$ $A=\frac{e^{3\pi}-\sqrt{e^{3\pi}}}{4}$ Thus, the area is $\frac{e^{3\pi}-\sqrt{e^{3\pi}}}{4}$.
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