The volume of a frustum of a cone is given by .. V = (1/3)*π*h*(R^2 +r^2 +R*r) We will assume the diameter at the bottom is 1 1/2 inch. Then the slanted part has volume .. V = (1/3)*π*4*(2.5^2 +.75^2 +2.5*.75) ≈ 36.39 . . . . in^3
The volume of the 1-inch cylinder at the bottom is .. V = π*(1/2)^2*1 = π/4 ≈ 0.79 . . . . in^3
Then the total volume of the cone is approximately 37.18 in^3.
Well, if it is increasing at a fixed rate of 2 mm/s, then the volume would still be increasing at a rate of 2 mm/s when the diameter is 80 mm, if I understand your question correctly.
