Find the volume of the solid generated by revolving the region bounded by y = 4x and y = 4x^2 about the y-axis?

Since we are rotating two regions of the form y = f(x) about the y-axis, use the shell method.

Note that y = 4x and y = 4x^2 intersect when:

4x = 4x^2 ==> x = 0 and x = 1.

So, we will be integrating from 0 to 1.

Next, since 4x > 4x^2 for 0 < x < 1, we see that the height of each shell is:

upper function - lower function = 4x - 4x^2.

Then, since the radius of each shell is x, the volume of the solid is:

V = 2π ∫ x(4x - 4x^2) dx (from x=0 to 1)

= 2π ∫ (4x^2 - 4x^3) dx (from x=0 to 1)

= 2π[(4/3)x^3 - x^4] (evaluated from x=0 to 1)

= 2π[(4/3 - 1) - (0 - 0)]

= 2π/3.

I hope this helps!