Calculus (3rd Edition)

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

Chapter 6 - Applications of the Integral - 6.3 Volumes of Revolution - Exercises - Page 305: 24

Answer

$$\frac{4802\pi }{5}$$

Work Step by Step

Given $$x=4-y,\ \ x=16-y^2 $$ We first find the intersection points \begin{align*} 4-y&=16-y^2\\ y^2-y -12&=0 \end{align*} Then $y=-3,\ 4$ Then the volume of the solid obtained by rotating the region enclosed by the graph of $x=4-y,\ \ x=16-y^2$ about the $y-$axis is given by \begin{align*} V&=\pi\int_{-3}^{4}[f(y)]^2-[g(y)]^2dy\\ &=\pi \int_{-3}^{4}\left[\left(16-y^{2}\right)^{2}-(4-y)^{2}\right] d y\\ &=\pi \int_{-3}^{4}\left[y^{4}-33 y^{2}+8 y+240\right] d y\\ &=\pi \left(\frac{y^{5}}{5}-\frac{33 y^{3}}{3}+\frac{8 y^{2}}{2}+240 y\right)\bigg|_{-3}^{4}\\ &=\frac{4802\pi }{5} \end{align*} We use the Mathematica program to plot the solid.
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