Problem Analysis:
We are given a region D in the yz-plane bounded by the line y = 1/2 and the curve y^2 + z^2 = 1, where y ≥ 0. We are asked to find the volume of the solid obtained when this region is revolved about the z-axis.

Solution:
To find the volume of the solid, we can use the method of cylindrical shells. We will integrate the volume of each cylindrical shell over the region D.

Step 1: Determine the bounds of integration:
Since the region D is bounded by the line y = 1/2 and the curve y^2 + z^2 = 1, we need to find the limits of integration for y and z.

For y, the bounds are y = 0 and y = 1/2.

For z, we can rewrite the equation of the curve as z^2 = 1 - y^2. Since y ≥ 0, we have z = √(1 - y^2).

Step 2: Set up the integral:
The volume of each cylindrical shell is given by V = 2πrhΔy, where r is the distance from the z-axis to the shell, h is the height of the shell, and Δy is the thickness of the shell.

Since we are revolving the region about the z-axis, the distance from the z-axis to the shell is simply z.

The height of the shell is given by h = 1/2 - 0 = 1/2.

The thickness of the shell is Δy.

Therefore, the volume of each cylindrical shell is V = 2πz(1/2)Δy = πzΔy.

Step 3: Evaluate the integral:
To find the volume of the solid, we need to integrate the volume of each cylindrical shell over the region D.

The integral is given by:
V = ∫[y=0 to y=1/2] πzΔy.

Since z = √(1 - y^2), we have:
V = ∫[y=0 to y=1/2] π√(1 - y^2)Δy.

Step 4: Evaluate the integral:
To evaluate the integral, we can use a trigonometric substitution.

Let y = sin(θ), then Δy = cos(θ) dθ.

Substituting these values into the integral, we have:
V = ∫[θ=0 to θ=π/6] π√(1 - sin^2(θ)) cos(θ) dθ.

Simplifying the expression inside the square root, we get:
V = ∫[θ=0 to θ=π/6] π√(cos^2(θ)) cos(θ) dθ.

Since cos(θ) is positive in the given interval, we can remove the square root:
V = ∫[θ=0 to θ=π/6] πcos^2(θ) dθ.

Using the double angle formula for cosine, we have:
V = ∫[θ=0 to θ=π/6] π(1 + cos(2θ))/2 dθ.