Calculus question. How to use the shell method to find the volume of a solid like this

Use the shell method to find the volume of the solid generated by revolving the region bounded by the given curves about the given lines.
y =16-x^2, y=16, x=4, revolve about the line y =16
Be detailed.

The axis of rotation is horizontal, so slice horizontally, implying dy. Note that the parabola intersects y = 16 at y = 16 and intersects x = 4 at y = 0.

V = 2π ∫ rh(y) dy from 0 to 16

First find r. When y = 16, the radius is 0 and when y = 0 the radius is 16, so r = 16 - y. Now determine the height. the line x = 4 is to the right of the parabola, so subtract the parabola from the line. First:

y = 16 - x^2
x = ± sqrt(16 - y)

But we know that the plus sign must be correct, because the line intersects the positive half of the parabola. So h(y) = 4 - sqrt(16 - y). Now sub all of that in:

V = 2π ∫ [16 - y][4 - sqrt(16 - y)] dy from 0 to 16
V = 8π ∫ (16 - y) dy - 2π ∫ (16 - y)(sqrt(16 - y))dy
V = 8π[16y - (1/2)y^2] - 2π ∫ (16 - y)(sqrt(16 - y))dy

In the other integral:

let u = 16 - y
du = -dy

+ 2π ∫ (u)(sqrt(u))dy
2π ∫ u^(3/2)dy
2π[(2/5)u^(5/2)]
(4π/5)[16 - y]^(5/2)

V = (8π)[16y - (1/2)y^2] + (4π/5)[16 - y]^(5/2) from 0 to 16
V = {8π[16(16) - (1/2)(16)^2] + (4π/5)[16 - (16)]^(5/2)} - {8π[16(0) - (1/2)(0)^2] + (4π/5)[16 - (0)]^(5/2)}
V = (1024/5)π

Done!