What the Join Hides Surface Areas and Volumes

Class 9 covered four basic solids: the cuboid, the cone, the cylinder, and the sphere, each with its own surface area and volume formula. But almost nothing built or manufactured is purely one of these. A grain silo is a cylinder with a cone for a roof. A test tube is a cylinder capped with a hemisphere. A capsule of medicine is a short cylinder with a hemisphere stuck on each end. None of these fit any single basic-solid formula, yet each one is clearly built from two or more basic solids joined together. This chapter's whole content is one idea applied twice: once to surface area, once to volume, for exactly these combined, everyday shapes.

Take a wooden bead shaped like a cylinder with a hemisphere glued onto each flat end, so it looks like a rounded capsule. To paint its outside, do you need the total surface area of the cylinder plus the total surface area of both hemispheres, all added up? No. Where the hemisphere is glued to the cylinder, that flat circular face of the cylinder is now hidden inside the join; it is not part of the outside at all anymore, and neither is the flat face of the hemisphere doing the gluing. Only the curved surfaces are still visible from outside: the curved surface of the cylinder in the middle, and the curved (dome) surface of each hemisphere at the two ends. So the total surface area of the combined solid is CSA of cylinder + CSA of hemisphere + CSA of hemisphere, using curved surface area for every piece, never the total surface area of the individual solids on their own. This is the one idea that makes combined-solid surface area different from simply adding up each piece's own full surface area: joining two solids face to face always removes exactly the two faces that got glued together from the final visible total.

A decorative glass stopper is shaped like a cone sitting on top of a cylinder, the cone's flat circular base exactly covering the cylinder's flat top. The cylinder has radius 3 cm and height 8 cm; the cone on top has the same radius, 3 cm, and a slant height of 5 cm. Find the total surface area to be polished (the flat bottom of the cylinder rests on a stand and is not polished). The visible surfaces are: the curved surface of the cylinder, the flat circular base is excluded (resting on the stand), and the curved surface of the cone (its own flat base is hidden, glued to the cylinder's top). CSA of cylinder = 2 pi r h = 2 x (22/7) x 3 x 8 = 1056/7, about 150.9 square cm. CSA of cone = pi r l = (22/7) x 3 x 5 = 330/7, about 47.1 square cm. Total surface area to polish = 150.9 + 47.1 = 198 square cm (approximately). Notice neither the cylinder's top (hidden under the cone) nor the cone's base (the same hidden circle, seen from the other side) was ever added in; only the two curved surfaces, the parts a hand could actually touch from outside, were counted.

Surface area needed care because gluing two solids together hides some faces. Volume needs no such care at all. The amount of material (or air, or liquid) a combined solid can hold is always exactly the volume of one piece plus the volume of the other, full stop, because volume measures the space filled, and joining two solids does not remove any of that space, it only changes the shape of the outer boundary. A grain silo's total capacity is exactly the cylinder's volume plus the cone-roof's volume added together, with nothing subtracted for the join. This asymmetry, care needed for surface area, none needed for volume, is the single most important thing to remember from this whole chapter, and it is very often the exact point a rushed answer gets wrong, by treating volume and surface area as if the same subtlety applied to both.

Suppose the same glass stopper shape (cylinder radius 3 cm height 8 cm, cone on top same radius, this time given a height of 4 cm rather than a slant height) were hollow and needed to be filled with scent instead of polished from outside. Volume of cylinder = pi r squared h = (22/7) x 3 x 3 x 8 = 1584/7, about 226.3 cubic cm. Volume of cone = (1/3) pi r squared h = (1/3) x (22/7) x 3 x 3 x 4 = 264/7, about 37.7 cubic cm. Total volume = 226.3 + 37.7 = 264 cubic cm (approximately), simply added, no term removed for the shared circular join between them. Compare how differently the two calculations behaved: the surface area calculation deliberately excluded the join's flat circle from both pieces; the volume calculation used the complete volume of both pieces without excluding anything at all.

Some real objects do not join two solids outward, but hollow one solid out from inside another; a bowl-shaped glass with a raised bump at the bottom is one such case. If a cylindrical glass of radius 4 cm and height 12 cm has a solid hemisphere of the same radius, 4 cm, bulging up from its base, the glass's true holding capacity is not the full cylinder's volume; it is the cylinder's volume with the hemisphere's volume removed, since that hemisphere-shaped bump takes up space liquid could otherwise fill. Apparent capacity (ignoring the bump) = pi r squared h = (22/7) x 4 x 4 x 12 = 8448/7, about 1207 cubic cm. Volume of the hemisphere bump = (2/3) pi r cubed = (2/3) x (22/7) x 4 x 4 x 4 = 2816/21, about 134.1 cubic cm. True capacity = apparent capacity minus the bump = 1207 - 134.1, about 1073 cubic cm. This is the one case in the whole chapter where a combined-solid volume genuinely does need a subtraction, not because of any hidden surface, but because one shape is carved INTO another rather than glued onto its outside, physically displacing space the liquid would otherwise occupy.