Calculus: Early Transcendentals 8th Edition

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

Chapter 6 - Review - Exercises: 14

Answer

$ \pi \int _{0}^{1} \left( 2-x^{2} \right) ^{2}- \left( 2-\sqrt[]{x} \right) ^{2} $dx

Work Step by Step

{Step 1 of 3} Consider the region enclosed by the graphs of $ y=\sqrt[]{x} $ and $ y=x^{2} $.The region is displayed below. {Step 2 of 3} Clearly the graphs intersect at the origin and at (1,1). We use the method of cylindrical shells. A typical shell is shown above. The radius of the shell is $ r \left( y \right) =2-y $.To express the height of the shell, we need to rewrite the functions as functions of y. For $ y=x^{2} $ we have $ x=\sqrt[]{y} $ . For $ y=\sqrt[]{x} $ we have $ x=y^{2} $ .Thus the height of the shell is $ h \left( y \right) =\sqrt[]{y}-y^{2} $ .The volume V of the resulting solid of revolution is given by $ V=2 \pi \int _{0}^{1}r \left( y \right) h \left( y \right) dy $ $ =2 \pi \int _{0}^{1} \left( 2-y \right) \left( \sqrt[]{y}-y^{2} \right) dy $ {Step 3 of 3} This can also be done as an integral along the x-axis using washers. In that case the outer radius is $ R \left( x \right) =2-x^{2} $ and the inner radius is $ 2-\sqrt[]{x} $ .Then the volume is given by $ \pi \int _{0}^{1}R \left( x \right) ^{2}-r \left( x \right) ^{2}dx= \pi \int _{0}^{1} \left( 2-x^{2} \right) ^{2}- \left( 2-\sqrt[]{x} \right) ^{2} $dx.
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