Answer
a) $ V=\frac{2 \pi }{15} $
b) $ V=\frac{ \pi }{6} $
c) $ V=\frac{8 \pi }{15} $
Work Step by Step
{Step 1 of 10}
First, find the points of intersection of the curves $ y=x $ and $ y=x^{2} $
For this, $ x=x^{2} $
$ x^{2}-x=0 $
$ x \left( x-1 \right) =0 or \left( x=0 or x=1 \right) $
The point of intersection are $ \left( 0,0 \right) $ and $ \left( 1,1 \right) $
.
{Step 2 of 10}
Sketch the curves.
{Step 3 of 10}
Rotate this shaded region about the x-axis. Consider a strip parallel to the y-axis in this shaded region. Rotation about the x-axis produces a washer.
The outer radius of the washer is $ x $
The inner radius of the washer is $ x^{2} $
{Step 4 of 10}
The cross-sectional area of the washer
$ A \left( x \right) = \pi \left[ x^{2}- \left( x^{2} \right) ^{2} \right] $
$ = \pi \left[ x^{2}-x^{4} \right] $
The volume of the solid
$ V= \int _{0}^{1} \pi \left( x^{2}-x^{4} \right) dx $
$ V= \pi \int _{0}^{1} \left( x^{2}-x^{4} \right) dx $
$ = \pi \left[ \frac{x^{3}}{3}-\frac{x^{5}}{5} \right] _{0}^{1} $
$ = \pi \left[ \frac{1}{3}-\frac{1}{5} \right] $
$ V=\frac{2 \pi }{15} $
{Step 5 of 10}
Consider the solid obtained by rotating this region about the y-axis. It is easier to use the cylindrical shell method to find the volume of this solid. Consider a strip at a distance of x from the origin. Rotation produces a cylindrical shell with radius x.
{Step 6 of 10}
Draw the Figure
{Step 7 of 10}
The circumference of the shell is $ 2 \pi x $, and the height of the shell is $ x-x^{2} $
The volume of the solid
$ V= \int _{0}^{1} \left( 2 \pi x \right) \left( x-x^{2} \right) dx $
$ =2 \pi \int _{0}^{1} \left( x^{2}-x^{3} \right) dx $
$ =2 \pi \left[ \frac{1}{3}x^{3}-\frac{1}{4}x^{4} \right] _{0}^{1} $
$ =2 \pi \left[ \frac{1}{3}-\frac{1}{4} \right] $
$ V=\frac{ \pi }{6} $
{Step 8 of 10}
Rotate this shaded region about $ y=2 $Use the slicing method
{Step 9 of 10}
The outer radius of the washer is $ 2-x^{2} $
The inner radius of the washer is $ 2-x $
The cross-sectional area of washer is
$ A \left( x \right) = \pi \left[ \left( 2-x^{2} \right) ^{2}- \left( 2-x \right) ^{2} \right] $
$ = \pi \left[ 4+x^{4}-4x^{2}-4-x^{2}+4x \right] $
$ = \pi \left[ x^{4}-5x^{2}+4x \right] $
{Step 10 of 10}
The volume of the solid obtained by rotating the bounded region about $ y=2 $
$ V= \int _{0}^{1}A \left( x \right) dx $
$ = \int _{0}^{1} \pi \left[ x^{4}-5x^{2}+4x \right] dx $
$ = \pi \int _{0}^{1} \left( x^{4}-5x^{2}+4x \right) dx $
$ = \pi \left[ \frac{1}{5}x^{5}-\frac{5}{3}x^{3}+4\frac{x^{2}}{2} \right] _{0}^{1} $
$ = \pi \left[ \frac{1}{5}-\frac{5}{3}+2 \right] $
$ V=\frac{8 \pi }{15} $
