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 is drilled through the cone, we can consider the entire cone's volume and subtract the volume of the cylindrical hole. The volume \( V_{cone} \) of the cone is given by: \[ V_{cone} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (8^2)(12) = \frac{1}{3}\pi (64)(12) = 256\pi. \] The volume \( V_{cylinder} \) of the hole, which is a cylinder with a radius of 6 and the same height as the cone (12), is calculated as follows: \[ V_{cylinder} = \pi r^2 h = \pi (6^2)(12) = \pi (36)(12) = 432\pi. \] Finally, the volume of the solid after drilling the hole is: \[ V_{solid} = V_{cone} - V_{cylinder} = 256\pi - 432\pi = -176\pi. \] It appears that drilling a hole of radius 6 through a cone with radius 8 is not possible since the volume calculated would become negative, indicating the hole exceeds the dimensions of the cone! So, you would end up with no physical solid, which is rather a “conical conundrum.”