Calculus (3rd Edition)

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

Chapter 6 - Applications of the Integral - 6.5 Work and Energy - Exercises - Page 317: 21

Answer

$W$ = $9800l\pi{r}^{3}$ $J$

Work Step by Step

at location $y$ length = $l$ width = $2\sqrt {r^{2}-y^{2}}$ thickness = $Δy$ $V$ = $(l)(2\sqrt {r^{2}-y^{2}})(Δy)$ $F$ = $(9.8)(1000)(2l\sqrt {r^{2}-y^{2}}Δy)$ $F$ = $19600l\sqrt {r^{2}-y^{2}}Δy$ distance = $r-y$ $V$ = $\int_{-r}^{r}19600l\sqrt {r^{2}-y^{2}}(r-y)dy$ $V$ = $\int_{-r}^{r}19600lr\sqrt {r^{2}-y^{2}}dy$ - $\int_{-r}^{r}19600ly\sqrt {r^{2}-y^{2}}dy$ $\int_{-r}^{r}19600lr\sqrt {r^{2}-y^{2}}dy$ = $\frac{1}{2}\pi{r}^{2}$ $\int_{-r}^{r}19600ly\sqrt {r^{2}-y^{2}}dy$ = $0$ $V$ = $\frac{1}{2}\pi{r}^{2}$ total work $W$ = $19600lr(\frac{1}{2}\pi{r}^{2})$-$19600l(0)$ $W$ = $9800l\pi{r}^{3}$ $J$
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