Let's solve the problem step by step: **Given:** - The curve is \( y = x^{1/2} \) (i.e., \( y = \sqrt{x} \)). - The shaded area is bounded by \( x = 0 \), \( x = 1 \), \( y = 0 \), and \( y = \sqrt{x} \). - The region is from \( x = 0 \) to \( x = 1 \). - We need the **moment of inertia about the x-axis (\( I_x \))**. ### 1. **Moment of Inertia Formula** The moment of inertia of a region about the x-axis is: \[ I_x = \int_A y^2 dA \] In strip form (vertical strips): \[ I_x = \int_{x=0}^{x=1} y^2 \, dA = \int_{x=0}^{x=1} y^2 [\text{height}] \, dx \] But for this region, the "height" is \( y \) and "width" is \( dx \). So \( dA = y \, dx \). So, \[ I_x = \int_{x=0}^{x=1} y^2 \, dA = \int_{x=0}^{x=1} (y^2) y \, dx = \int_{x=0}^{x=1} y^3 \, dx \] But let's clarify: - For a horizontal strip at height \( y \), the width is \( x \), but for a vertical strip at \( x \), the height is \( y \). So for a vertical strip of width \( dx \) at position \( x \), the area element is \( dA = y \, dx \). The distance from the x-axis is \( y \), so the moment of inertia contribution is \( y^2 \, dA \): \[ dI_x = y^2 \, dA = y^2 (y \, dx) = y^3 dx \] So, \[ I_x = \int_{x=0}^{x=1} y^3 dx \] But \( y = \sqrt{x} \), so \( y^3 = (\sqrt{x})^3 = x^{3/2} \). ### 2. **Set Up the Integral** \[ I_x = \int_{x=0}^{x=1} x^{3/2} dx \] ### 3. **Evaluate the Integral** \[ \int x^{3/2} dx = \frac{2}{5} x^{5/2} \] So, \[ I_x = \left[ \frac{2}{5} x^{5/2} \right]_{0}^{1} \] \[ I_x = \frac{2}{5} (1)^{5/2} - \frac{2}{5} (0)^{5/2} \] \[ I_x = \frac{2}{5} \] ### **Final Answer** \[ \boxed{I_x = \frac{2}{5} \ \text{m}^4} \] This is the moment of inertia of the shaded area about the x-axis.

**Step 1: Conceptual Introduction** To find the moment of inertia of a region about the x-axis, we consider how the area is distributed relative to that axis. Here, the region is bounded by the curve \( y = \sqrt{x} \), the x-axis, and the lines \( x=0 \) and \( x=1 \). The moment of inertia about the x-axis measures how the area resists bending or rotation about that axis. It involves integrating the square of the distance from the axis (which is \( y \)) over the area. --- **Step 2: Relevant Formulae** For a region bounded by \( y = f(x) \), the moment of inertia about the x-axis is given by: \[ I_x = \int_A y^2 \, dA \] Using vertical strips at position \( x \), the differential area is: \[ dA = y \, dx \] because the width of the strip is \( dx \) and its height is \( y \). Thus, \[ dI_x = y^2 \, dA = y^2 \times y \, dx = y^3 \, dx \] Since \( y = \sqrt{x} \), we have: \[ I_x = \int_{x=0}^{1} y^3 \, dx \] which becomes: \[ I_x = \int_0^1 (\sqrt{x})^3 \, dx = \int_0^1 x^{3/2} \, dx \] --- **Step 3: Step-by-Step Calculation** 1. **Express \( y \) in terms of \( x \):** \( y = \sqrt{x} = x^{1/2} \) 2. **Write \( y^3 \) in terms of \( x \):** \( y^3 = (x^{1/2})^3 = x^{3/2} \) 3. **Set up the integral for \( I_x \):** \[ I_x = \int_0^1 x^{3/2} dx \] 4. **Integrate \( x^{3/2} \):** \[ \int x^{n} dx = \frac{x^{n+1}}{n+1} \] Here, \( n = \frac{3}{2} \), so: \[ \int x^{3/2} dx = \frac{x^{(3/2)+1}}{(3/2)+1} = \frac{x^{5/2}}{\frac{5}{2}} = \frac{2}{5} x^{5/2} \] 5. **Evaluate the definite integral from 0 to 1:** \[ I_x = \left[\frac{2}{5} x^{5/2}\right]_0^1 = \frac{2}{5} \times (1)^{5/2} - \frac{2}{5} \times (0)^{5/2} \] 6. **Calculate the values:** \[ (1)^{5/2} = 1, \quad (0)^{5/2} = 0 \] So, \[ I_x = \frac{2}{5} \times 1 - 0 = \frac{2}{5} \] **Final Answer:** \[ \boxed{I_x = \frac{2}{5}} \] This is the moment of inertia of the shaded area about the x-axis.

**Step 1: Conceptual Introduction** To find the moment of inertia of a region about the x-axis, we analyze how the area is distributed relative to that axis. In this problem, the region is bounded by the curve \( y = \sqrt{x} \), the x-axis, and the lines \( x=0 \) and \( x=1 \). The moment of inertia about the x-axis quantifies the region's resistance to bending or rotation about that axis, which involves integrating the square of the distance from the axis (\( y \)) over the area. --- **Step 2: Relevant Formulae** *Explanation:* For a region bounded by \( y = f(x) \), the moment of inertia about the x-axis is calculated by integrating \( y^2 \) over the area \( A \): \[ I_x = \int_A y^2 \, dA \] When we consider vertical strips (thin slices of the region at a specific \( x \)), each strip has a width \( dx \) and extends from the x-axis (\( y=0 \)) up to \( y = \sqrt{x} \). The area of each strip is: \[ dA = y \, dx \] because the height of the strip is \( y \), and its width is \( dx \). The contribution of each strip to the moment of inertia is: \[ dI_x = y^2 \times dA = y^2 \times y \, dx = y^3 \, dx \] Since \( y = \sqrt{x} \), we substitute: \[ y^3 = (\sqrt{x})^3 = x^{3/2} \] Therefore, the total moment of inertia about the x-axis is: \[ I_x = \int_{x=0}^{1} y^3 \, dx = \int_0^1 x^{3/2} \, dx \] --- **Step 3: Step-by-Step Calculation** *Explanation:* 1. **Express \( y \) in terms of \( x \):** Given \( y = \sqrt{x} = x^{1/2} \). 2. **Compute \( y^3 \):** \[ y^3 = (x^{1/2})^3 = x^{3/2} \] 3. **Set up the integral for \( I_x \):** \[ I_x = \int_0^1 x^{3/2} dx \] 4. **Integrate \( x^{3/2} \):** Recall the power rule for integration: \[ \int x^n dx = \frac{x^{n+1}}{n+1} \] Here, \( n = \frac{3}{2} \), so: \[ \int x^{3/2} dx = \frac{x^{(3/2)+1}}{(3/2)+1} = \frac{x^{5/2}}{\frac{5}{2}} = \frac{2}{5} x^{5/2} \] 5. **Evaluate the definite integral from 0 to 1:** \[ I_x = \left[\frac{2}{5} x^{5/2}\right]_0^1 = \frac{2}{5} \times (1)^{5/2} - \frac{2}{5} \times (0)^{5/2} \] 6. **Calculate the values:** Since \( (1)^{5/2} = 1 \) and \( (0)^{5/2} = 0 \), \[ I_x = \frac{2}{5} \times 1 - 0 = \frac{2}{5} \] **Final Answer:** \[ \boxed{I_x = \frac{2}{5}} \] This is the moment of inertia of the shaded region about the x-axis.