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: 62

Answer

$$S =\frac {48 \pi} 5 $$

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 $\frac{dy}{dt}$: $\frac{dx}{dt} = \frac{d(3t - t^3)}{dt} = 3 - 3t^2$ $\frac{dy}{dt} = \frac{d(3t^2)}{dt} = 6t $ 3. Substitute the given values into the formula: $S = \int_{0}^{1} 2\pi (3t^2) \sqrt {(3-3t^2)^2 + (6t)^2} \space dt$ $S = \int_{0}^{1} 2\pi (3t^2) \sqrt {(9 - 18t^2 + 9t^4) + (36t^2)} \space dt$ $S = \int_{0}^{1} 2\pi (3t^2) \sqrt {(9 + 18t^2 + 9t^4) } \space dt$ $S = \int_{0}^{1} 2\pi (3t^2) \sqrt {(3 + 3t^2)^2 } \space dt$ $S = \int_{0}^{1} 2\pi (3t^2) (3 + 3t^2) \space dt$ $S = 2\pi \int_{0}^{1} (9t^2 + 9t^4) \space dt$ $S = 2\pi [3t^3 + \frac 9 5 t^5]^0_1$ $S = 2\pi (3(1)^3 + \frac 9 5 (1)^5) - 2\pi (3(0)^3 + \frac 9 5 (0)^5)$ $S = 2\pi (3 + \frac 9 5 ) $ $S = 2\pi (\frac {24} 5 ) $ $S =\frac {48 \pi} 5 $
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