Let R be the plane of region encircled by the curves y=2x-x^2 and y=X.
a) calculate the area of R
b) calculate the volume of the solid obtained by rotating R about X-axis
c) calculate the volume of the solid obtained by rotating R about y-axis
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