Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 15 - Vector Analysis - 15.2 Exercises - Page 1063: 64

Answer

$\int_Cf(x,y) ds = 8\sqrt 2$

Work Step by Step

The path is given by , C: line from (0,0) to (4,4). This can be written as $\textbf r = x\textbf i + y \textbf j = 4 t\textbf i + 4 t\textbf j $ where $x(t) = 4t, y(t) = 4t, 0\le t \le 1$ $\textbf r' = 4 \textbf i + 4 \textbf j \Rightarrow \vert \textbf r' \vert = 4\sqrt 2$ $f(x,y) = y=4t$ Lateral surface area = $\int_C f(x,y) ds\\ =\int_C f(x,y).\vert \textbf r' \vert dt\\ = \int_{t=0}^{1}(4t.4\sqrt2)dt\\ = [8\sqrt2 t^2]_0^{1}=8\sqrt2$
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