Use the shell method to find the volume of the solid formed when a hole of radius 1 is driled symmetrically along the axis of a right circular cone of radius 6 and height 9 . 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. (Type exact answers.)

To find the volume of the solid formed by drilling a cylindrical hole of radius 1 along the axis of a right circular cone with radius 6 and height 9, we'll use the **shell method**. Here's a step-by-step setup for the integral: 1. **Coordinate System Setup:** - Place the center of the base of the cone at the origin \((0, 0)\). - Align the cone's axis along the positive \( y \)-axis, extending from \( y = 0 \) at the base to \( y = 9 \) at the apex. 2. **Cone Equation:** - The radius of the cone at any height \( y \) can be determined using similar triangles. - At \( y = 0 \), the radius \( r = 6 \). - At \( y = 9 \), the radius \( r = 0 \). - Therefore, the radius as a function of \( y \) is: \[ r(y) = 6 - \left(\frac{6}{9}\right)y = 6 - \frac{2}{3}y \] 3. **Shell Method Setup:** - Consider a vertical cylindrical shell at a radius \( x \) from the \( y \)-axis. - **Radius of the Shell:** \( x \) - **Height of the Shell:** Given by the cone's equation \( y(x) = 9 - \frac{3}{2}x \) - **Thickness of the Shell:** \( dx \) 4. **Limits of Integration:** - The hole has a radius of 1, so \( x \) ranges from 1 to 6. 5. **Integral Expression:** - The volume \( V \) is given by integrating the volume of each cylindrical shell from \( x = 1 \) to \( x = 6 \): \[ V = 2\pi \int_{1}^{6} x \left(9 - \frac{3}{2}x\right) \, dx \] **Final Integral Setup:** \[ V = 2\pi \int_{1}^{6} x \left(9 - \frac{3}{2}x\right) \, dx \]