Use the disk method or the shell method to find the volume of the solid generated by revolving the region bounded by the graphs of the equations about each given line. \[ \begin{array}{r}y=x^{3} \\ y=0 \\ x=2\end{array} \] (a) \( x \)-axis (b) \( y \)-axis (c) \( x=4 \) (b)

To find the volume of the solid generated by revolving the region bounded by the graphs of the equations about each given line, we can use the disk method or the shell method. Given equations: \[ y = x^3 \] \[ y = 0 \] \[ x = 2 \] (a) Revolving about the \( x \)-axis: The volume of the solid generated by revolving the region about the \( x \)-axis can be found using the disk method. The formula for the volume of a solid of revolution about the \( x \)-axis is: \[ V = \pi \int_{a}^{b} [f(x)]^2 dx \] In this case, the function is \( f(x) = x^3 \), and the limits of integration are from \( x = 0 \) to \( x = 2 \). Let's calculate the volume using the disk method. Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\pi \times \int_{0}^{2} \left(x^{3}\right)^{2} dx\) - step1: Multiply the exponents: \(\pi \times \int_{0}^{2} x^{3\times 2} dx\) - step2: Multiply the numbers: \(\pi \times \int_{0}^{2} x^{6} dx\) - step3: Evaluate the integral: \(\pi \times \frac{128}{7}\) - step4: Multiply: \(\frac{\pi \times 128}{7}\) - step5: Multiply: \(\frac{128\pi }{7}\) The volume of the solid generated by revolving the region about the \( x \)-axis is \( \frac{128\pi}{7} \) cubic units. (b) Revolving about the \( y \)-axis: To find the volume of the solid generated by revolving the region about the \( y \)-axis, we can use the shell method. The formula for the volume of a solid of revolution about the \( y \)-axis is: \[ V = 2\pi \int_{a}^{b} x \cdot f(x) dx \] In this case, the function is \( f(x) = x^3 \), and the limits of integration are from \( x = 0 \) to \( x = 2 \). Let's calculate the volume using the shell method. Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(2\pi \times \int_{0}^{2} x\times x^{3} dx\) - step1: Multiply the terms: \(2\pi \times \int_{0}^{2} x^{4} dx\) - step2: Evaluate the integral: \(2\pi \times \frac{32}{5}\) - step3: Multiply the terms: \(\frac{64}{5}\pi \) - step4: Multiply: \(\frac{64\pi }{5}\) The volume of the solid generated by revolving the region about the \( y \)-axis is \( \frac{64\pi}{5} \) cubic units. (c) Revolving about the line \( x = 4 \): To find the volume of the solid generated by revolving the region about the line \( x = 4 \), we can use the shell method. The formula for the volume of a solid of revolution about the line \( x = a \) is: \[ V = 2\pi \int_{a}^{b} (a - x) \cdot f(x) dx \] In this case, the function is \( f(x) = x^3 \), the line is \( x = 4 \), and the limits of integration are from \( x = 0 \) to \( x = 2 \). Let's calculate the volume using the shell method. Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(2\pi \times \int_{0}^{2} \left(4-x\right)x^{3} dx\) - step1: Multiply the terms: \(2\pi \times \int_{0}^{2} \left(4x^{3}-x^{4}\right) dx\) - step2: Evaluate the integral: \(2\pi \times \frac{48}{5}\) - step3: Multiply the terms: \(\frac{96}{5}\pi \) - step4: Multiply: \(\frac{96\pi }{5}\) The volume of the solid generated by revolving the region about the line \( x = 4 \) is \( \frac{96\pi}{5} \) cubic units.