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 .

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 Problem The volume of the solid can be calculated by finding the volume of the cone and subtracting the volume of the cylindrical hole. ### Step 3: Volume of the Cone The volume \( V_c \) of a cone is given by the formula: \[ V_c = \frac{1}{3} \pi R^2 H \] Substituting the values: \[ V_c = \frac{1}{3} \pi (8^2)(12) \] ### Step 4: Volume of the Cylindrical Hole The volume \( V_h \) of a cylinder is given by the formula: \[ V_h = \pi r^2 h \] Where \( h \) is the height of the cylinder. Since the hole goes through the entire height of the cone, \( h = 12 \): \[ V_h = \pi (6^2)(12) \] ### Step 5: Calculate the Volumes Now, we will calculate both volumes. 1. **Volume of the Cone**: \[ V_c = \frac{1}{3} \pi (8^2)(12) = \frac{1}{3} \pi (64)(12) = \frac{768}{3} \pi = 256 \pi \] 2. **Volume of the Cylindrical Hole**: \[ V_h = \pi (6^2)(12) = \pi (36)(12) = 432 \pi \] ### Step 6: Calculate the Volume of the Solid The volume of the solid \( V_s \) is given by: \[ V_s = V_c - V_h \] Substituting the calculated volumes: \[ V_s = 256 \pi - 432 \pi = -176 \pi \] ### Step 7: Interpret the Result Since the volume cannot be negative, this indicates that the cylindrical hole is larger than the cone's base, which is not physically possible in this context. Therefore, the problem as stated leads to an invalid scenario. ### Conclusion 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 is not feasible, as the hole's radius exceeds the cone's base radius. Thus, the solid cannot exist in this configuration.