To use the shell method, we need to integrate along the y-axis. The radius of each shell is y + 4, and the height of each shell is x. The limits of integration are y = 0 and y = 1.
The volume of the solid is given by:
V = 2π ∫[0,1] (y + 4) x dy
Using the equation y = x^2, we can express x in terms of y:
x = sqrt(y)
Substituting this into the integral, we get:
V = 2π ∫[0,1] (y + 4) sqrt(y) dy
We can simplify this integral by using u-substitution. Let u = y^(3/2), then du/dy = (3/2) y^(1/2) and dy = (2/3) u^(-2/3) du. Substituting these into the integral, we get:
V = 2π ∫[0,1] (y + 4) sqrt(y) dy
= 2π ∫[0,1] (u^(2/3) + 4) u^(-1/3) (2/3) du
= (4/3)π ∫[0,1] (u^(2/3) + 4) u^(-1/3) du
Integrating, we get:
V = (4/3)π [3u^(5/3)/5 + 12u^(2/3)/2] |_0^1
= (4/3)π [3/5 + 6]
= (22/5)π
Therefore, the volume of the solid generated by revolving R about the line y = -4 is (22/5)π cubic units.