Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 17 - Line and Surface Integrals - 17.2 Line Integrals - Exercises - Page 934: 52

Answer

The work done by the field along the given path is $\frac{9}{2}{\pi ^2}$.

Work Step by Step

We obtain ${\bf{F}}\left( {{\bf{r}}\left( t \right)} \right) = \left( {\cos t,\sin t,t} \right)$ ${\bf{r}}'\left( t \right) = \left( { - \sin t,\cos t,1} \right)$ The work done by the field: $W = \mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = \mathop \smallint \limits_0^{3\pi } {\bf{F}}\left( {{\bf{r}}\left( t \right)} \right)\cdot{\bf{r}}'\left( t \right){\rm{d}}t$ $W = \mathop \smallint \limits_0^{3\pi } \left( {\cos t,\sin t,t} \right)\cdot\left( { - \sin t,\cos t,1} \right){\rm{d}}t$ $ = \mathop \smallint \limits_0^{3\pi } t{\rm{d}}t = \frac{1}{2}{t^2}|_0^{3\pi } = \frac{9}{2}{\pi ^2}$ So, the work done by the field along the given path is $\frac{9}{2}{\pi ^2}$.
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