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. \( \dfrac{\pi}{\sqrt{3}} \)
• B. \( \dfrac{2\pi}{\sqrt{3}} \)
• C. \( \dfrac{\pi\sqrt{3}}{2} \)
• 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: Understand the region

The curve $y^2 + z^2 = 1$ is a circle of radius 1 centered at the origin in the $yz$ plane.

For $y \geq 0$, we have the upper semicircle.

The line $y = \frac{1}{2}$ intersects the circle where: $$\left(\frac{1}{2}\right)^2 + z^2 = 1$$ $$z^2 = 1 - \frac{1}{4} = \frac{3}{4}$$ $$z = \pm\frac{\sqrt{3}}{2}$$

So the region $D$ is bounded by:

• The line $y = \frac{1}{2}$ (horizontal line)
• The arc of the circle from $\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$ to $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$

Step 2: Set up the volume integral

When revolving around the $z$-axis, we use the washer method with $z$ as the variable of integration.

For a given $z \in \left[-\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}\right]$:

• Outer radius: $R(z) = y$ on the circle, so $y = \sqrt{1 - z^2}$
• Inner radius: $r(z) = \frac{1}{2}$

The volume is: $$V = \pi \int_{-\sqrt{3}/2}^{\sqrt{3}/2} \left[R(z)^2 - r(z)^2\right] dz$$

$$V = \pi \int_{-\sqrt{3}/2}^{\sqrt{3}/2} \left[(1-z^2) - \frac{1}{4}\right] dz$$

$$V = \pi \int_{-\sqrt{3}/2}^{\sqrt{3}/2} \left[\frac{3}{4} - z^2\right] dz$$

Step 3: Evaluate the integral

$$V = \pi \left[\frac{3z}{4} - \frac{z^3}{3}\right]_{-\sqrt{3}/2}^{\sqrt{3}/2}$$

By symmetry (the integrand is even): $$V = 2\pi \left[\frac{3z}{4} - \frac{z^3}{3}\right]_{0}^{\sqrt{3}/2}$$

$$V = 2\pi \left[\frac{3 \cdot \frac{\sqrt{3}}{2}}{4} - \frac{\left(\frac{\sqrt{3}}{2}\right)^3}{3}\right]$$

$$V = 2\pi \left[\frac{3\sqrt{3}}{8} - \frac{\frac{3\sqrt{3}}{8}}{3}\right]$$

$$V = 2\pi \left[\frac{3\sqrt{3}}{8} - \frac{3\sqrt{3}}{24}\right]$$

$$V = 2\pi \left[\frac{9\sqrt{3}}{24} - \frac{3\sqrt{3}}{24}\right]$$

$$V = 2\pi \left[\frac{6\sqrt{3}}{24}\right]$$

$$V = 2\pi \cdot \frac{\sqrt{3}}{4}$$

$$V = \frac{2\pi\sqrt{3}}{4} = \frac{\pi\sqrt{3}}{2}$$

Answer: (C)