Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 10 - Parametric Equations and Polar Coordinates - 10.2 Exercises - Page 676: 61

Answer

$$S = \frac {2\pi}{1215}( {247\sqrt {13} + 64} )$$

Work Step by Step

1. Write the formula for the area of the surface obtained by rotating a curve about the $x$-axis. $$S = \int_{\alpha}^{\beta} 2\pi y \sqrt {(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \space dt$$ 2. Calculate: $\frac{dx}{dt}$ and $\frac{dy}{dt}$: $\frac{dx}{dt} = \frac{d(t^3)}{dt} = 3t^2$ $\frac{dy}{dt} = \frac{d(t^2)}{dt} = 2t $ 3. Substitute the given values into the formula: $S = \int_{0}^{1} 2\pi (t^2) \sqrt {(3t^2)^2 + (2t)^2} \space dt$ $S = \int_{0}^{1} 2\pi (t^2) \sqrt {9t^4 + 4t^2} \space dt$ $S = \int_{0}^{1} 2\pi (t^2) \sqrt {t^2 (9t^2 + 4)} \space dt$ $S = \int_{0}^{1} 2\pi (t^2)t \sqrt {(9t^2 + 4)} \space dt$ $u = 9t^2 + 4 \longrightarrow t^2 = \frac{u - 4} 9$ $\frac{du}{dt} = 18t \longrightarrow t = \frac{1}{18}\frac{du}{dt}$ $S = \int_{\alpha}^{\beta} 2\pi ( \frac{u - 4} 9)(\frac{1}{18}\frac{du}{dt}) \sqrt {u} \space dt$ $S = \frac {\pi}{81} \int_{\alpha}^{\beta} ({u - 4})(\frac{du}{dt}) \sqrt {u} \space dt$ $S = \frac {\pi}{81} \int_{\alpha}^{\beta} ({u - 4}) \sqrt {u} \space du$ $S = \frac {\pi}{81} \int_{\alpha}^{\beta} ({u^{3/2} - 4u^{1/2}}) \space du$ $S = \frac {\pi}{81} [\frac{2u^{5/2}}{5} - \frac{8u^{3/2}}{3}]_{\alpha}^{\beta}$ $S = \frac {\pi}{81} [\frac{2(9t^2 + 4)^{5/2}}{5} - \frac{8(9t^2 + 4)^{3/2}}{3}]_{0}^{1}$ $S = \frac {\pi}{81}( \frac{2(9(1)^2 + 4)^{5/2}}{5} - \frac{8(9(1)^2 + 4)^{3/2}}{3} - (\frac{2(9(0)^2 + 4)^{5/2}}{5} - \frac{8(9(0)^2 + 4)^{3/2}}{3}))$ $S = \frac {\pi}{81}( \frac{2(13)^{5/2}}{5} - \frac{8(13)^{3/2}}{3} - (\frac{2( 4)^{5/2}}{5} - \frac{8( 4)^{3/2}}{3}))$ $S = \frac {\pi}{81}( \frac{2(13)^{2}\sqrt {13}}{5} - \frac{8(13)\sqrt {13}}{3} - ( \frac{2(4^2) \sqrt 4}{5} - \frac{8(4) \sqrt 4}{3}))$ $S = \frac {\pi}{81}( \frac{338\sqrt {13}}{5} - \frac{104\sqrt {13}}{3} - \frac{64}{5} + \frac{64}{3})$ $S = \frac {\pi}{81}( \frac{494\sqrt {13}}{15} + \frac{128}{15} )$ $S = \frac {\pi}{81}( \frac{494\sqrt {13} + 128}{15} )$ $S = \frac {\pi}{1215}( {494\sqrt {13} + 128} )$ $S = \frac {2\pi}{1215}( {247\sqrt {13} + 64} )$
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