Question

Consider the region \( D \) in the yz-plane bounded by the line \( y = \frac{1}{2} \) and the curve \( y^2 + z^2 = 1 \), where \( y \geq 0 \). If the region \( D \) is revolved about the \( z \)-axis in \( \mathbb{R}^3 \), then the volume of the resulting solid is

• A. \( \frac{\pi}{\sqrt{3}} \)
• B. \( \frac{2\pi}{\sqrt{3}} \)
• C. \( \pi\sqrt{3} \)
• D. \( \pi\sqrt{3} \)

Hint:

For solids of revolution, use the disk method and ensure you have the correct limits based on the given region and axis of rotation.

Answer 1

The Correct Option is C

Step 1: Understanding the problem.
We are asked to find the volume of the solid formed by rotating the region \( D \) about the \( z \)-axis. The region is bounded by the curve \( y^2 + z^2 = 1 \), and the line \( y = \frac{1}{2} \).

Step 2: Setting up the integral.
The volume of the solid can be found using the formula for the volume of revolution around the \( z \)-axis: \[ V = \pi \int_{a}^{b} \left( f(y) \right)^2 \, dy \] Substituting the appropriate bounds and the equation of the curve, we compute the volume.

Step 3: Conclusion.
The correct answer is (A) \( \frac{\pi}{\sqrt{3}} \).