Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 6 - Applications of Integration - 6.3 Volume by Slicing - 6.3 Exercises - Page 432: 46

Answer

$\dfrac{224\pi}{15}$

Work Step by Step

Our aim is to compute the volume of the revolution of the curve about the y-axis by using the Washer method. Washer method for computing the volume of the revolution of the curve: Let us consider two functions $f(x)$ and $g(x)$ (both are continuous functions) with $f(x) \geq g(x) \geq 0$ on the interval $[m, n]$ . Then the volume of the solid can be obtained by rotating the region under the graph about the y-axis and can be expressed as: $\ Volume, V=\pi \int_m^n [f(y)^2-g(y)^2] \ dy $ $\bf{Calculations:}$ $V= \pi \int_0^2 [(4)^2-(4-y^{2})^2] \ dy$ or, $= \pi \int_0^2 (8y^2-y^4) \ dy$ or, $= \pi \times ( \dfrac{8y^{3}}{3}-\dfrac{y^5}{5})_0^2$ or, $=\pi \times ( \dfrac{8(2)^{3}}{3}-\dfrac{(2)^5}{5})$ Hence, $V=\dfrac{224\pi}{15}$
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