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Find the area bounded by the functions $y = 2x^3$ and $y = 2x^{1 \over 3}$.

1.)  Find points of intersection:

$2x^3 = 2x^{1 \over 3}$ implies $x^3 = x^{1 \over 3}$ implies  $x^9 = x$. \hfil \break
 Thus   $x^9 - x =  x(x^8 - 1)= 0$.  
Hence $x = 0$ and $x^8 - 1 = 0$ and $x^8 = 1$ and thus $x = 1, -1$


Hence the functions $y = 2x^3$ and $y = 2x^{1 \over 3}$ intersect when $x = -1, 0, 1$

2.)  Draw a rough graph:

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\centerline{\includegraphics[width=24ex]{6_2}}

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3.) Find area:

Method 1:  Use vertical rectangles:

$\int_{-1}^0 [2x^3 - 2x^{1 \over 3}]dx + \int_0^1 [2x^{1 \over 3} - 2x^{3}]dx$

\vfill

Method 2:  Use horizontal rectangles:

\centerline{\includegraphics[width=24ex]{6_2}}

If  $y = 2x^3$, then $x = ({y \over 2})^{1 \over 3}$. If  $y = 2x^{1 \over 3}$, then $x = ({y \over 2})^{3}$.

When $x = -1, y = -2$.  When $x = 0, y = 0$.  When $x = 1, y = 2$.  

$\int_{-2}^0 [({y \over 2})^{3}-  ({y \over 2})^{1 \over 3}]dy + \int_0^2 [ ({y \over 2})^{1 \over 3} - ({y \over 2})^{3}]dy$
\eject
Find the volume of the object obtained by rotating the area bounded by the functions $y = 2x^3$ and $y = 2x^{1 \over 3}$
about the $x-$axis ($y = 0$):


\centerline{\includegraphics[width=24ex]{6_2}}

$\pi \int_{-1}^0[(2x^{1 \over 3})^2 - (2x^{3})^2]dx + \pi \int_0^1 [(2x^{1 \over 3})^2 - (2x^{3})^2]dx$ 
{$ = \int_{-1}^1[(2x^{1 \over 3})^2 - (2x^{3})^2]dx$}

\vfill

Find the volume of the object obtained by rotating the area bounded by the functions $y = 2x^3$ and $y = 2x^{1 \over 3}$.
about $y = 4$:

\centerline{\includegraphics[width=24ex]{6_2}}

$\pi \int_{-1}^0[ (4 - 2x^{1 \over 3})^2 - (4 - 2x^{3})^2]dx +\pi \int_0^1 [(4 - 2x^{3})^2 - (4 - 2x^{1 \over 3})^2]dx$

\vfill

Find the volume of the object obtained by rotating the area bounded by the functions $y = 2x^3$ and $y = 2x^{1 \over 3}$.
about $y = -2$:

\centerline{\includegraphics[width=24ex]{6_2}}

$\pi \int_{-1}^0[ (2x^{ 3} - (-2) )^2 - (2x^{1 \over 3} - (-2) )^2]dx +\pi \int_0^1 [(2x^{1 \over 3} - (-2) )^2 - (2x^{3} - (-2) )^2]dx$

$ = \pi \int_{-1}^0[ (2x^{ 3} + 2 )^2 - (2x^{1 \over 3} + 2 )^2]dx +\pi \int_0^1 [(2x^{1 \over 3}+ 2 )^2 - (2x^{3} + 2 )^2]dx$


\eject

Find the volume of the object obtained by rotating the area bounded by the functions $y = 2x^3$  ($x = ({y \over 2})^{1 \over 3}$)and $y = 2x^{1 \over 3}$  ($x = ({y \over 2})^{3}$)
about the $y-$axis ($x = 0$):

\centerline{\includegraphics[width=24ex]{6_2}}

$\pi \int_{-2}^0 [( ({y \over 2})^{1 \over 3})^2 -  (({y \over 2})^{ 3})^2]dy + \pi \int_0^2 [ (({y \over 2})^{1 \over 3})^2 - (({y \over 2})^{3})^2]dy$
= $\pi \int_{-2}^2 [ (({y \over 2})^{1 \over 3})^2 -  (({y \over 2})^{ 3})^2 ]dy $

\vfill

Find the volume of the object obtained by rotating the area bounded by the functions $y = 2x^3$  ($x = ({y \over 2})^{1 \over 3}$)and $y = 2x^{1 \over 3}$  ($x = ({y \over 2})^{3}$)
about $x = 8$:


\centerline{\includegraphics[width=24ex]{6_2}}

$\pi \int_{-2}^0 [( 8 - ({y \over 2})^{3})^2 -  ( 8 - ({y \over 2})^{ 1 \over 3})^2]dy + \pi \int_0^2 [ ( 8 - ({y \over 2})^{1 \over 3})^2 - (  8 - ({y \over 2})^{3})^2]dy$

\vfill

Find the volume of the object obtained by rotating the area bounded by the functions $y = 2x^3$  ($x = ({y \over 2})^{1 \over 3}$)and $y = 2x^{1 \over 3}$  ($x = ({y \over 2})^{3}$)
about $x = -1$:

\centerline{\includegraphics[width=24ex]{6_2}}

$\pi \int_{-2}^0 [( ({y \over 2})^{1 \over 3} - 1 )^2 -  (({y \over 2})^{ 3} - 1 )^2]dy + \pi \int_0^2 [ (({y \over 2})^{3} - 1 )^2 - (({y \over 2})^{1 \over 3} - 1 )^2]dy$




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