The silo shown in the skecth is an air- and water-tight tower. It consists of a lower cylinder surmounted by a frustum of a cone whose lower base is the upper base of the cylinder. The Frustum in turn is surmounted by a cupola consisting a smaller cylinder whose lower base is the upper base of the frustum. This smaller cylinder is topped by a conical roof. The inside radii of the smaller and larger cylinders are 6ft. and 12ft., respectively. The altitudes of the frustum and larger cylinder are 6ft. and 21ft., respectively. If ensilage can be stored up to the cupola, find the storage capacity of the silo.
There's no diagram provided, but it seems the storage capacity of the silo is equal to the volume of the lower cylinder, plus the volume of the frustum. (Since the upper cylinder is part of the cupola, we won't count it as part of the silo's storage capacity.) Let C = the volume of the lower cylinder, and Let F = the volume of the frustum. Now, C = πr^2 h We're given that r = 12, and h = 21. So, C = π(12^2)(21) = 3,024π ft^3 The frustum is like a cone with its top part sliced off. The base-radius of the cone is 12 ft. The top part, sliced off, is a similar, smaller cone with base-radius 6 ft. If we let y = the height of the sliced-off cone, and consider the triangular cross-section of both cones, then, by similar triangles, (y + 6) / 12 = y / 6 => 2y = y + 6 => y = 6 So, the height of the larger cone is 6 + y = 12, and the height of the sliced-off cone is y = 6. (Recall again we're given that the base-radius of the larger cone is 12, and the base radius of the sliced-off cone is 6.) Then, the volume of the frustum is the volume of the larger cone, minus the volume of the sliced-off cone: F = (1/3)π(12^2)(12) - (1/3)π(6^2)(6) = 504π ft^3 Our total volume V is: V = C + F = 3,024π + 504π = 3,528π ft^3
