Math Help!!!?

Question

Let R denote the region enclosed between the graph of y=x^2 and the graph of y=2x.
- Find the area of region R.
- Find the volume of the solid obtained by revolving the region R about the y- axis.

Answer 1

Two curves intersect at x^2 = 2x => (x = 2, y = 4).
The area between curves is:

integral[from 0 to 2](2x - x^2)*dx = [from 0 to 2](x^2 - x^3/3] = 4/3.

The volume is:

integral[from y=0 to 4](integral[from x=y/2 to sqrt(y)](2*pi*x*dx)*dy)
= integral[from y=0 to 4](pi*[from x=y/2 to sqrt(y)](x^2))
= pi*integral[from y=0 to 4](y - y^2/4)
= pi*[from y=0 to 4](y^2/2 - y^3/12]
= 8*pi/3.

Answer 2

this is an integration problem
the two curves intersect each other at two points
x^2=2x
x=2 or x=0
area=int((y1-y2)dx) |x=0 to 2
=int((2x-x^2)dx) |x=0 to 2
=x^2-(1/3)x^3 |x=0 to 2
=4-(1/3)*8-0+0
=4/3

concernig the volume it requires another integration
but we should change the coordinate system to be more suitable
but actually I need some more time to do it.

Answer 3

Volumes around the y axis are given by the formula
Vy = pi*int(x^2)dy. You therefore start by rewriting each as
x = f(y) and then putting it into this formula.