Consider a region bounded by the x-axis and the line \( y = 2x \) from \( x = 0 \) to \( x = 1 \). If this shape is revolved around the x-axis, what is the volume of the resulting solid using triangular cross-sections perpendicular to the x-axis?

To determine the volume of the solid formed by revolving the region bounded by the \( x \)-axis and the line \( y = 2x \) from \( x = 0 \) to \( x = 1 \) around the \( x \)-axis with triangular cross-sections perpendicular to the \( x \)-axis, we'll follow these steps: ### 1. Understanding the Region and Its Rotation - **Region:** The region is a right triangle with vertices at \( (0,0) \), \( (1,0) \), and \( (1,2) \). - **Rotation:** When this region is revolved around the \( x \)-axis, each cross-section perpendicular to the \( x \)-axis is a triangle. ### 2. Defining the Triangular Cross-Sections - **Base of Triangle:** For a given \( x \), the height of the original triangle is \( y = 2x \). When rotated around the \( x \)-axis, this creates a circular base with diameter \( 2y = 4x \). - **Height of Triangle:** Assume the height of the triangular cross-section is also \( 4x \) (making it an isosceles right triangle for simplicity). ### 3. Calculating the Area of a Triangular Cross-Section The area \( A(x) \) of each triangular cross-section at position \( x \) can be calculated as: \[ A(x) = \frac{\text{base} \times \text{height}}{2} = \frac{4x \times 4x}{2} = 8x^2 \] ### 4. Integrating to Find the Volume To find the total volume \( V \), integrate the area of the cross-sections from \( x = 0 \) to \( x = 1 \): \[ V = \int_{0}^{1} 8x^2 \, dx = 8 \int_{0}^{1} x^2 \, dx = 8 \left[ \frac{x^3}{3} \right]_0^1 = 8 \left( \frac{1}{3} \right) = \frac{8}{3} \] ### 5. Conclusion The volume of the resulting solid is: \[ V = \frac{8}{3} \text{ cubic units} \] **Answer:** Its volume is 8⁄3 cubic units