Use the shell method to find the volume of the solid formed when a hole of radius 6 is drilled symmetrically along the axis of a right circular cone of radius 8 and height 12 . Model the situation on a set of axes by placing the center of the base of the cone at the origin and the cone's axis along the positive \( y \)-axis. Set up the integral that gives the volume of the solid using the shell method. Use nonnegative and increasing limits of integration. Select the correct choice below and fill in the answer boxes to complete your choice.

To find the volume of the solid formed when a hole of radius 6 is drilled symmetrically along the axis of a right circular cone of radius 8 and height 12 using the shell method, we can follow these steps: ### Step 1: Understand the Geometry 1. **Cone Dimensions**: - Radius \( R = 8 \) - Height \( H = 12 \) 2. **Drilled Hole**: - Radius of the hole \( r = 6 \) ### Step 2: Set Up the Cone Equation The equation of the cone can be derived from its geometry. The cone's vertex is at the origin (0,0) and the base is at \( y = 12 \). The radius of the cone at any height \( y \) can be expressed as: \[ x = \frac{R}{H} y = \frac{8}{12} y = \frac{2}{3} y \] ### Step 3: Shell Method Setup Using the shell method, we will integrate with respect to \( y \). The volume of a cylindrical shell is given by: \[ V = 2\pi \int_{a}^{b} (radius)(height) \, dy \] - **Radius**: The distance from the axis of rotation (the \( y \)-axis) to the shell, which is \( x \). - **Height**: The height of the shell, which is the difference between the outer radius of the cone and the inner radius of the hole. ### Step 4: Determine Limits of Integration The limits of integration will be from \( y = 0 \) to \( y = 12 \) (the height of the cone). ### Step 5: Set Up the Integral 1. **Outer Radius**: The outer radius of the cone at height \( y \) is \( x = \frac{2}{3} y \). 2. **Inner Radius**: The inner radius (the hole) is constant at \( r = 6 \). Thus, the height of the shell is: \[ \text{Height} = \frac{2}{3} y - 6 \] ### Step 6: Write the Integral The volume integral using the shell method is: \[ V = 2\pi \int_{0}^{12} \left( y \left( \frac{2}{3} y - 6 \right) \right) \, dy \] ### Final Integral Expression The integral that gives the volume of the solid is: \[ V = 2\pi \int_{0}^{12} \left( y \left( \frac{2}{3} y - 6 \right) \right) \, dy \] This integral can be evaluated to find the volume of the solid. Would you like me to calculate the volume?