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 find 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 using triangular cross-sections perpendicular to the x-axis, we can use the method of disks. The formula for the volume of a solid formed by revolving a region around the x-axis using the method of disks is given by: \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \] In this case, the function \( f(x) = 2x \) represents the line \( y = 2x \), and the limits of integration are from \( x = 0 \) to \( x = 1 \). Substitute \( f(x) = 2x \) into the formula for the volume of the solid: \[ V = \pi \int_{0}^{1} (2x)^2 \, dx \] Now, we can calculate the integral to find the volume of the solid. Evaluate the integral by following steps: - step0: Evaluate using partial integration formula: \(\pi \times \int_{0}^{1} \left(2x\right)^{2} dx\) - step1: Evaluate the integral: \(\pi \times \frac{4}{3}\) - step2: Multiply: \(\frac{\pi \times 4}{3}\) - step3: Multiply: \(\frac{4\pi }{3}\) 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 using triangular cross-sections perpendicular to the x-axis is \( \frac{4\pi}{3} \) cubic units.