Answer
$ V=\frac{64 \pi }{15} $
Work Step by Step
{Step 1 of 5}:
Find, find the points of intersection of $ y=2x $ and $ y=x^{2} $
$ x^{2}=2x $
$ x^{2}-2x=0 $
$ x \left( x-2 \right) =0 $
Or $ x=0 $
or $ x=2 $
The points of Intersection are (0, 0) and (2, 4).
{Step 2 of 5}:
Sketch the curves.
{Step 3 of 5}:
Rotate the shaded region about the x-axis. Sketch the solid obtained by rotating the shaded region about the x-axis and the washer.
{Step 4 of 5}:
The outer radius of the washer is 2x
The inner radius of the washer is x2
The cross-sectional area of the washer
$ A \left( x \right) = \pi \left[ \left( 2x \right) ^{2}- \left( x^{2} \right) ^{2} \right] $
$ = \pi \left[ 4x^{2}-x^{4} \right] $
{Step 5 of 5}:
The volume of the solid obtained by rotating the shaded region about the x-axis
$ V= \int _{0}^{2}A \left( x \right) dx $
$ = \int _{0}^{2} \pi \left[ 4x^{2}-x^{4} \right] dx $
$ = \pi \int _{0}^{2} \left( 4x^{2}-x^{4} \right) dx $
$ = \pi \left[ \frac{4}{3}x^{3}-\frac{1}{5}x^{5} \right] _{0}^{2} $
[By the fundamental theorem of calculus, part 2]
$ = \pi \left[ \frac{32}{3}-\frac{32}{5} \right] $
$ V=\frac{64 \pi }{15} $
