Answer
a) A=$ \frac{5}{12} $
b)V = $ \frac{41 \pi }{105} $
c) V= $ \frac{13 \pi }{30} $
Work Step by Step
{Step 1 of 9}
A) First, find the points of intersection of the curves y=x3 and y=2x-x2
$x^3=2x-x^2$
$x^3+x^2-2x=0$
$x(x^2+x-2)=0$
$x=0$ or $ x2+x-2=0$
$x=0$ or $x2+2x-x-2=0$
$x=0$ or $(x+2)(x-1)=0$
$x=0$ or $x=-2 $ or $x=1$
The points of intersection are (0,0), (-2,-8), and (1,1).
Consider the two points (0,0) and (1,1), which are in the first quadrant.
{Step 2 of 9}
Sketch the curves.
{Step 3 of 9}
Find the area of the shaded region, which is labelled R.
A=$ \int _{0}^{1} \left[ \left( 2x-x^{2} \right) -x^{3} \right] dx $
=$ \left[ x^{2}-\frac{1}{3}x^{3}-\frac{1}{4}x^{4} \right] _{0}^{1} $
=1-$ \frac{1}{3} $
-$ \frac{1}{4} $
A=$ \frac{5}{12} $
{Step 4 of 9}
B) Rotate this region about the x-axis. Use the slicing method. Consider a strip in this region parallel to the y-axis. Rotate this region about the x-axis to produce a washer.
{Step 5 of 9}
The outer radius of the washer is $2x-x^2$
The inner radius of the washer is $x^3$
The cross-sectional area of the washer
A(x) =$ \pi \left[ \left( 2x-x^{2} \right) ^{2}- \left( x^{3} \right) ^{2} \right] $
=$ \pi \left[ 4x^{2}+x^{4}-4x^{3}-x^{6} \right] $
{Step 6 of 9}
Then the volume of the solid
V=$ \int _{0}^{1}A \left( x \right) dx $
=$ \pi \int _{0}^{1} \left[ 4x^{2}+x^{4}-4x^{3}-x^{6} \right] $dx
=$ \pi \left[ \frac{4}{3}x^{3}+\frac{1}{5}x^{5}-x^{4}-\frac{1}{7}x^{7} \right] _{0}^{1} $
=$ \pi \left[ \frac{4}{3}+\frac{1}{5}-1-\frac{1}{7} \right] $
V = $ \frac{41 \pi }{105} $
{Step 7 of 9}
( C ) rotate region R about the y-axis and use the cylindrical shell method. Consider a vertical strip in region R. rotate the region about the y-axis to produce a cylindrical shell with a radius of x and a height of $(2x-x^2-x^3)$.
{Step 8 of 9}
Draw a Figure
{Step 9 of 9}
The circumference of the shell is 2$ \pi x $
The volume of the solid
V = $ \int _{0}^{1} \left( 2 \pi x \right) \left( 2x-x^{2}-x^{3} \right) dx $
V= 2$ \pi \int _{0}^{1} \left( 2x^{2}-x^{3}-x^{4} \right) dx $
=2$ \pi \left[ \frac{2}{3}x^{3}-\frac{1}{4}x^{4}-\frac{1}{5}x^{5} \right] _{0}^{1} $
=2$ \pi \left[ \frac{2}{3}-\frac{1}{4}-\frac{1}{5} \right] $
V= $ \frac{13 \pi }{30} $
