Let R be the plane of region encircled by the curves y=2x-x^2 and y=X.

Let us find the intersection points :

2x- x^2  = x

x = x^2    

x^2  - x  = 0

x( x - 1)  = 0      so   the intersection points are   x = 0   and x  = 1

a.  The area of the region is given by

1                                     1                                                   1

∫   2x - x ^2  - x  dx    =    ∫  x  - x^2   dx   =   [ x^2/2  - x^3/3]   =  1/2  - 1/3  =  1/6  units^2

0                                     0                                                   0

b) calculate the volume of the solid obtained by rotating R about X-axis

Using the "washer" method, this is given by  :

    1

pi ∫  [2x - x^2] ^2   -  [x]^2  dx    =

   0

   1

pi ∫  4x^2  - 4x^3 + x^4  - x^2  dx   =

   0

    1

pi  ∫  3x^2  - 4x^3 +  x^4   dx    =

    0

                                1

[ x^3   -  x^4 + x^5/5 ]      =   1 - 1  + 1/5   =     1/5  units ^3

                               0

c) calculate the volume of the solid obtained by rotating R about y-axis

The integration is made easier if we use the "shell" method

The volume is given  by :

      1

2pi ∫   x  *   (2x - x^2  - x) ] dx   =

     0

     1

2pi ∫ x  [x - x^2]  dx     =

     0

     1

2pi ∫  x^2  - x^3   dx   =

     0

                              1

2pi  [ x^3/3  - x^4/4]     =    2pi  [ 1/3  - 1/4]  =  2pi ( 1/12)  =  pi / 6    units ^3

                              0