Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 10 - Section 10.2 - Calculus with Parametric Curves - 10.2 Exercises - Page 655: 32


The area is $\frac{8~\sqrt{2}}{15}$

Work Step by Step

$y = \sqrt{t}$ $dy = \frac{1}{2\sqrt{t}}~dt$ Since the area is enclosed by the y-axis, we need to find the values of $t$ when $x=0$: $x = t^2-2t = 0$ $t(t-2) = 0$ $t = 0, 2$ It would be be helpful to know if the area is on the left or the right of the y-axis. When $t = 1$, $x = (1)^2-2(1) = -1 \lt 0$ Therefore, the area is to the left of the y-axis. Thus, we should integrate from $2$ to $0$ to find the area. We can find the area: $A = \int_{2}^{0} x~dy$ $A = \int_{2}^{0} (t^2-2t)~\frac{1}{2\sqrt{t}}~dt$ $A = \frac{1}{2}\int_{2}^{0} (t^{3/2}-2\sqrt{t})~dt$ $A = \frac{1}{2}(\frac{2}{5}t^{5/2}-\frac{4}{3}t^{3/2})\Big\vert_{2}^{0}$ $A = \frac{1}{2}[\frac{2}{5}(0)^{5/2}-\frac{4}{3}(0)^{3/2}]-\frac{1}{2}[\frac{2}{5}(2)^{5/2}-\frac{4}{3}(2)^{3/2}]$ $A = \frac{1}{2}[(0)-(\frac{8}{5}\sqrt{2}-\frac{8}{3}\sqrt{2})]$ $A = -\frac{1}{2}(\frac{24}{15}\sqrt{2}-\frac{40}{15}\sqrt{2})$ $A = -\frac{1}{2}(-\frac{16}{15}\sqrt{2})$ $A = \frac{8~\sqrt{2}}{15}$ The area is $\frac{8~\sqrt{2}}{15}$
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