Find the volume of the material to make a cup rotating the area between y=x+1 and y=2x^2 around x axis. x positive.

We have to find the volume of the cup created by rotating
the area between y = x + 1 and y = 2x^2 around the positive x-
axis.

The point of intersection of y = x + 1 and y = 2x^2
is

2x^2 = x + 1

=> 2x^2
- x - 1 = 0

=> 2x^2 - 2x + x - 1 =
0

=> 2x(x - 1) + 1(x - 1) =
0

=> (2x + 1)(x - 1) =
0

=> x = -1/2 and x =
1.

As we only have to consider the positive x values we
take it from x = 0 to x = 1.

The volume that we are trying
to obtain is given by the difference of the volumes enveloped by the two
curves.

=>pi*Int[(x+1)^2 - (2x^2)^2 dx] , x = 0 to x
= 1

=> pi*Int [ x^2 + 2x + 1 - 4x^4 dx] , x = 0 to x
= 1

=> pi*(x^3/3 + 2*x^2/2 + x - 4x^5/5), x = 0 to x
= 1

=> pi*(1/3 + 1 + 1 -
4/5)

=> pi*(
23/15)

The required volume is (23/15)*pi cube
units.