What is the volume of the solid obtained by rotating the region bounded by the curves; y=x^2 & y=1, about y=2

I'd go with the washer method. The axis of rotation is horizontal, so slice vertically, implying dx. Note that the curves intersect when x = -1 and x = 1.

V = ∫ A(x) dx from -1 to 1

A(x) = πR^2 - πr^2
A(x) = π(R^2 - r^2)

First determine the outer radius R. The outer radius is defined by the parabola. When x = -1, the radius is 1, when x = 0, the radius is 2 and when x = 1 the radius is 1 again, so R = 2 - x^2 .

Now determine the inner radius r. The inner radius is defined by the line, so r = 1. Sub those in:

A(x) = π((2 - x^2)^2 - (1)^2)
A(x) = π(x^4 - 4x^2 + 3)

V = π ∫ (x^4 - 4x^2 + 3) dx from -1 to 1
V = π [(1/5)x^5 - (4/3)x^3 + 3x] from -1 to 1
V = π [(1/5)(1)^5 - (4/3)(1)^3 + 3(1)] - π [(1/5)(-1)^5 - (4/3)(-1)^3 + 3(-1)]
V = (56/15)π

Done!