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.
