Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the \( y \)-axis. \[ \begin{aligned} y=4(3-x) \\ y=0 \\ x=0\end{aligned} \]

To find the volume of the solid generated by revolving the region bounded by the given equations about the \( y \)-axis, we can use the method of cylindrical shells. ### Step 1: Identify the region The equations given are: 1. \( y = 4(3 - x) \) 2. \( y = 0 \) (the x-axis) 3. \( x = 0 \) (the y-axis) First, we need to find the points of intersection of the line \( y = 4(3 - x) \) with the x-axis. Setting \( y = 0 \): \[ 0 = 4(3 - x) \implies 3 - x = 0 \implies x = 3 \] Thus, the region is bounded by \( x = 0 \), \( x = 3 \), and the line \( y = 4(3 - x) \). ### Step 2: Set up the volume integral Using the method of cylindrical shells, the volume \( V \) is given by the formula: \[ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \] where \( f(x) \) is the height of the shell, and \( a \) and \( b \) are the bounds of integration. In this case: - \( f(x) = 4(3 - x) \) - The bounds are \( a = 0 \) and \( b = 3 \). ### Step 3: Write the integral The volume integral becomes: \[ V = 2\pi \int_{0}^{3} x \cdot 4(3 - x) \, dx \] ### Step 4: Simplify the integral Now, we simplify the integrand: \[ V = 2\pi \int_{0}^{3} (12x - 4x^2) \, dx \] ### Step 5: Calculate the integral Now we compute the integral: \[ V = 2\pi \left[ \int_{0}^{3} 12x \, dx - \int_{0}^{3} 4x^2 \, dx \right] \] Calculating each integral separately: 1. \(\int 12x \, dx = 6x^2\) 2. \(\int 4x^2 \, dx = \frac{4}{3}x^3\) Now evaluate from 0 to 3: \[ \int_{0}^{3} 12x \, dx = 6(3^2) - 6(0^2) = 6 \cdot 9 = 54 \] \[ \int_{0}^{3} 4x^2 \, dx = \frac{4}{3}(3^3) - \frac{4}{3}(0^3) = \frac{4}{3} \cdot 27 = 36 \] ### Step 6: Combine results Now substituting back into the volume formula: \[ V = 2\pi \left( 54 - 36 \right) = 2\pi \cdot 18 = 36\pi \] ### Final Answer The volume of the solid generated by revolving the region about the \( y \)-axis is: \[ \boxed{36\pi} \]