Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 6 - Review - Exercises: 11

Answer

$ V=\frac{4}{3} \pi \left[ h^{2}+2ah \right] ^{\frac{3}{2}} $

Work Step by Step

{Step 1 of 5}: First, find the points of intersection of the curves $ x^{2}-y^{2}=a^{2} $ or $ x=\sqrt[]{a^{2}-y^{2}} $ and $ x=a+h $ (where a>0,h>0) $ \sqrt[]{a^{2}-y^{2}}= \left( a+h \right) $ $ a^{2}-y^{2}= \left( a+h \right) ^{2} $ $ y^{2}= \left( a+h \right) ^{2}-a^{2} $ $ y^{2}=a^{2}+h^{2}+2ah-a^{2} $ $ y^{2}=h^{2}+2ah $ $ y^{2}=h \left( h+2a \right) $ $ y= \pm \sqrt[]{h \left( h+2a \right) } $ {Step 2 of 5}: As an example, take $ a=1 $ and $ h=1 $ and sketch the curve {Step 3 of 5}: Rotate this shaded region about the y-axis. Rotate this to produce a solid. If we cut a Horizontal slice from this solid, we get a washer. The outer radius of the washer is $ a+h $ The inner radius of the washer is $ \sqrt[]{a^{2}-y^{2}} $ The cross-sectional area of the washer $ A \left( y \right) = \pi \left[ \left( a+h \right) ^{2}- \left( \sqrt[]{a^{2}-y^{2}} \right) ^{2} \right] $ $ = \pi \left[ a^{2}+h^{2}+2ah-a^{2}-y^{2} \right] $ $ = \pi \left[ \left( h^{2}+2ah \right) -y^{2} \right] $ {Step 4 of 5}: The volume of the solid $ V= \int _{-\sqrt[]{k \left( k+2a \right) }}^{+\sqrt[]{k \left( k+2a \right) }}A \left( y \right) dy $ Let $ h \left( h+2a \right) =h^{2}+2ah=k $ {Step 5 of 5}: Then $ V= \pi \int _{-\sqrt[]{k}}^{\sqrt[]{k}} \left( k-y^{2} \right) dy $ $ = \pi \left[ ky-\frac{y^{3}}{3} \right] _{-\sqrt[]{k}}^{\sqrt[]{k}} $ [By the fundamental theorem of calculus, part 2] $ = \pi \left[ \left( k\sqrt[]{k}-\frac{ \left( \sqrt[]{k} \right) ^{3}}{3} \right) - \left( -k\sqrt[]{k}+\frac{ \left( \sqrt[]{k} \right) ^{3}}{3} \right) \right] $ $ = \pi \left[ 2k\sqrt[]{k}-\frac{2}{3} \left( \sqrt[]{k} \right) ^{3} \right] $ $ = \pi \left[ 2 \left( \sqrt[]{k} \right) ^{3}-\frac{2}{3} \left( \sqrt[]{k} \right) ^{3} \right] $ $ = \pi \left[ \frac{4}{3} \left( \sqrt[]{k} \right) ^{3} \right] $ $ = \pi \left[ \frac{4}{3}k^{\frac{3}{2}} \right] $ $ V=\frac{4}{3} \pi \left[ h^{2}+2ah \right] ^{\frac{3}{2}} $
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