Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 6 - Section 6.2 - Volume - 6.2 Exercises - Page 456: 10

Answer

$V = \pi \int_0^2 {\left[ {{{\left( {6 - {y^2}} \right)}^2} - 4} \right]} dy$

Work Step by Step

$$\eqalign{ & y = \sqrt x ,{\text{ }}y = 0,{\text{ }}x = 4,{\text{ about }}x = 6 \cr & y = \sqrt x \Rightarrow {x^2} = y \cr & {\text{Graph of the region shown below }}\left( {{\text{Using Geogebra}}} \right) \cr & \cr & {\text{For this exercise the inner radius is given by:}} \cr & {r_i} = 6 - f\left( y \right) \cr & {r_i} = 6 - 4 \cr & {r_i} = 2 \cr & {\text{And the outer radius is given by:}} \cr & {r_o} = 6 - f\left( y \right) \cr & {r_o} = 6 - {y^2} \cr & {\text{The cross - sectional area is}} \cr & A\left( y \right) = \pi {\left( {{r_o}} \right)^2} - \pi {\left( {{r_i}} \right)^2} \cr & A\left( y \right) = \pi {\left( {6 - {y^2}} \right)^2} - \pi {\left( 2 \right)^2} \cr & {\text{The volume is given by}} \cr & V = \int_c^d {A\left( y \right)} dy \cr & V = \int_0^2 {\left[ {\pi {{\left( {6 - {y^2}} \right)}^2} - \pi {{\left( 2 \right)}^2}} \right]} dy \cr & V = \int_0^2 {\left[ {\pi {{\left( {6 - {y^2}} \right)}^2} - 4\pi } \right]} dy \cr & V = \pi \int_0^2 {\left[ {{{\left( {6 - {y^2}} \right)}^2} - 4} \right]} dy \cr} $$
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