## Calculus: Early Transcendentals 8th Edition

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