Let's solve the problem step by step: **Problem restatement:** Find the volume \( V \) of the solid obtained by rotating the region bounded by the curves \( y = x \) and \( y = \sqrt{x} \) about the x-axis. #### Step 1: Find the Intersection Points Set \( y = x \) and \( y = \sqrt{x} \) equal: \[ x = \sqrt{x} \] \[ x^2 = x \] \[ x^2 - x = 0 \] \[ x(x-1) = 0 \] So, \( x = 0 \) and \( x = 1 \). --- #### Step 2: Sketch and Identify the Region For \( 0 \leq x \leq 1 \): - \( y = \sqrt{x} \) is **above** \( y = x \). --- #### Step 3: Volume by Washer Method The formula for the volume of revolution about the x-axis is: \[ V = \pi \int_{a}^{b} \left[ R_{out}^2 - R_{in}^2 \right] dx \] Where \( R_{out} \) is the outer radius and \( R_{in} \) is the inner radius. - Outer curve: \( y = \sqrt{x} \) (farther from x-axis) - Inner curve: \( y = x \) (closer to x-axis) So, \[ V = \pi \int_{0}^{1} \left[ (\sqrt{x})^2 - (x)^2 \right] dx \] \[ = \pi \int_{0}^{1} (x - x^2) dx \] --- #### Step 4: Integrate \[ \int_{0}^{1} (x - x^2) dx = \int_{0}^{1} x dx - \int_{0}^{1} x^2 dx \] \[ = \left[ \frac{x^2}{2} \right]_{0}^{1} - \left[ \frac{x^3}{3} \right]_{0}^{1} \] \[ = \left( \frac{1}{2} - 0 \right) - \left( \frac{1}{3} - 0 \right) \] \[ = \frac{1}{2} - \frac{1}{3} \] \[ = \frac{3}{6} - \frac{2}{6} \] \[ = \frac{1}{6} \] --- #### Step 5: Multiply by \(\pi\) \[ V = \pi \times \frac{1}{6} = \frac{\pi}{6} \] --- ### **Final Answer** \[ \boxed{\frac{\pi}{6}} \] This matches with one of the answer choices.

Let's analyze the problem carefully, understand each option, and see how the final answer is derived. --- ### **Problem Restatement:** Find the volume of the solid obtained by rotating the region bounded by the curves: - \( y = x \) - \( y = \sqrt{x} \) about the x-axis. --- ### **Step-by-step Solution with Explanation** --- #### **Step 1: Find the points of intersection** Set \( y = x \) and \( y = \sqrt{x} \): \[ x = \sqrt{x} \] Square both sides: \[ x^2 = x \] Rearranged: \[ x^2 - x = 0 \\ x(x - 1) = 0 \] So, \[ x = 0 \quad \text{or} \quad x = 1 \] At \( x=0 \), both curves meet at \((0,0)\). At \( x=1 \), both curves meet at \((1,1)\). --- #### **Step 2: Understand the region** - For \( x \in [0,1] \): - \( y = \sqrt{x} \) is **above** \( y = x \) - The region between these two curves forms a "band" from \( x=0 \) to \( x=1 \). --- #### **Step 3: Setup the volume integral** Since we're rotating around the x-axis, the **washer method** applies: \[ V = \pi \int_{a}^{b} \left[ R_{outer}^2 - R_{inner}^2 \right] dx \] - **Outer radius** \( R_{outer} \): distance from x-axis to the top curve \( y = \sqrt{x} \), so \( R_{outer} = \sqrt{x} \). - **Inner radius** \( R_{inner} \): distance from x-axis to the bottom curve \( y = x \), so \( R_{inner} = x \). Thus, \[ V = \pi \int_{0}^{1} \left( (\sqrt{x})^2 - (x)^2 \right) dx \] which simplifies to: \[ V = \pi \int_{0}^{1} \left( x - x^2 \right) dx \] --- #### **Step 4: Compute the integral** Break into parts: \[ \int_{0}^{1} x \, dx - \int_{0}^{1} x^2 \, dx \] Calculate each: \[ \int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2} \] \[ \int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1}{3} \] Subtract: \[ \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \frac{1}{6} \] Multiply by \(\pi\): \[ V = \pi \times \frac{1}{6} = \frac{\pi}{6} \] --- ### **Final Answer:** \[ \boxed{\frac{\pi}{6}} \] --- ### **Explanation of options:** Suppose the multiple-choice options are: a) \(\frac{\pi}{2}\) b) \(\frac{\pi}{3}\) c) \(\frac{\pi}{6}\) d) \(\frac{\pi}{4}\) - **Option a) \(\frac{\pi}{2}\):** This would be twice the correct volume, perhaps if the limits or the radii were misinterpreted. - **Option b) \(\frac{\pi}{3}\):** Slightly larger than the correct value; might result from not subtracting \( x^2 \). - **Option c) \(\frac{\pi}{6}\):** **Correct answer**, as derived. - **Option d) \(\frac{\pi}{4}\):** Not matching the integral result. --- ### **Summary:** - The intersection points are at \( x=0 \) and \( x=1 \). - The region is between \( y=x \) and \( y=\sqrt{x} \). - The volume is obtained by integrating the difference of squares of the radii. - The final volume is \(\boxed{\frac{\pi}{6}}\). This detailed explanation covers the derivation, the reasoning behind each step, and clarifies why the correct answer is \(\frac{\pi}{6}\).