Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 15 - Topics In Vector Calculus - 15.2 Line Integrals - Exercises Set 15.2 - Page 1109: 29

Answer

Result: \[ \int_0^1 (yzdx - xzdy + xydz) = 1 - e^3. \]

Work Step by Step

The path of integration is given by the parametrized curve: \[ \begin{align*} x &= e^t, \\ y &= e^{3t}, \\ z &= e^{-t}, \end{align*} \] for \(0 \leq t \leq 1\). Therefore, the integral can be performed as: \[ \begin{align*} \int_0^1 (yzdx - xzdy + xydz) &= \int_0^1 \left(e^{2t} e^t dt - 1(3dt)e^{3t} + e^{4t} (-dt)e^{-t}\right) \\ &= \int_0^1 (e^{3t} - 3e^{3t} - e^{3t})dt \\ &= \int_0^1 (-3e^{3t})dt \\ &= -3\left[3e^{3t}\right]_0^1 \\ &= 1 - e^3. \end{align*} \] Result: \[ \int_0^1 (yzdx - xzdy + xydz) = 1 - e^3. \]
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