Surface area of combination of solids

When solid A is joined to solid B along a shared face, that shared face is sealed inside the new solid — it can no longer be touched, painted, or measured from outside, so it drops out of the surface-area count entirely. This single idea replaces a long list of special cases.

The formulae you already know, reused here. With radius $r$, height $h$, and slant height $l$ for a cone:

where: $r$ is the base radius and $h$ is the vertical height of the cone; $l$ is the distance from the apex to a point on the base circle, needed in every cone surface-area formula.

• Cylinder: curved surface area (CSA) $= 2\pi rh$, total surface area (TSA) $= 2\pi r(r+h)$.
• Cone: CSA $= \pi rl$, TSA $= \pi r(l+r)$.
• Sphere: surface area $= 4\pi r^2$.
• Hemisphere: CSA $= 2\pi r^2$, TSA $= 3\pi r^2$.
• Cuboid: TSA $= 2(lb + bh + hl)$.

The four-step method. First, split the composite figure into its basic solids and label each one's own radius and height — a toy that is "a cone on a hemisphere" gives you a cone and a hemisphere sharing one radius. Second, mark the face where they touch: this is always hidden and must be dropped from both solids' formulae. Third, add up only the curved (or otherwise still-exposed) surfaces of every part. Fourth, check whether any flat face of the composite figure — not the internal joint — remains open to the air, such as the bottom of a cylinder standing on a table; if so, add that plane area, $\pi r^2$, once.

where: the sum runs over every basic solid in the composite, and "visible" excludes any face that coincides with another solid's face at a joint.

The goal is to derive, from the visible-faces rule stated above, a single reusable formula for the total surface area of the extremely common composite of a cone mounted on a hemisphere of the same radius, so that any problem of this shape can be solved by substitution instead of re-reasoning from scratch each time.

Step 1. Identify the two solids and their shared dimension. The composite is a cone of base radius $r$ and height $h$ sitting exactly on a hemisphere of the same radius $r$. Because the cone's circular base and the hemisphere's flat circular face have equal radius $r$, they coincide perfectly when joined — there is no rim or gap.

Step 2. Mark the hidden face. The cone's flat base and the hemisphere's flat top are the same circle, now sealed inside the solid. Neither the flat area $\pi r^2$ of the cone's base nor the flat area $\pi r^2$ of the hemisphere's face contributes to the outer surface — both are dropped entirely, not averaged or halved.

Step 3. List what remains visible. Only two curved surfaces are left exposed to the outside: the cone's slanted curved surface, $\pi r l$, and the hemisphere's rounded curved surface, $2\pi r^2$. Nothing else of either solid touches the outside air.

Step 4. Add the visible parts to get the composite formula.

where: $r$ is the shared radius, $l = \sqrt{r^2+h^2}$ is the cone's slant height computed from its own height $h$, and the formula holds only when the cone and hemisphere share exactly the same radius, which is why that condition is always stated in such problems.

A toy is in the shape of a cone mounted on a hemisphere of the same radius. The diameter of the base of the cone is 6 cm and its height is 4 cm. Find the total surface area of the toy. (Use $\pi = 3.14$.)

Step 1. Identify dimensions. Radius $r = 3$ cm (half of the 6 cm diameter), cone height $h = 4$ cm. Slant height $l = \sqrt{r^2+h^2} = \sqrt{3^2+4^2} = \sqrt{9+16} = \sqrt{25} = 5$ cm.

Step 2. Identify visible surfaces. The hemisphere's flat face is glued to the cone's base, so both flat circles are hidden. Visible: the cone's curved surface $\pi r l$ and the hemisphere's curved surface $2\pi r^2$.

Step 3. Compute each piece.

Step 4. Add the two visible surfaces. $47.1 + 56.52 = 103.62$. The total surface area of the toy is $103.62\text{ cm}^2$.

Competitive stretch

A cylindrical wooden block of radius 7 cm and height 10 cm has hemispherical depressions of the same radius scooped out from both flat circular ends. Find the total surface area of the remaining solid. (Use $\pi = \tfrac{22}{7}$.)

Step 1. The full curved surface of the cylinder stays exposed: $2\pi rh = 2 \times \frac{22}{7} \times 7 \times 10 = 440\text{ cm}^2$.

Step 2. At each end, the flat disc $\pi r^2$ is gone, replaced by the curved inner surface of a hemispherical dent, which has the same area formula as an added hemisphere, $2\pi r^2 = 2 \times \frac{22}{7} \times 49 = 308\text{ cm}^2$, even though the material has been removed rather than added.

Step 3. Two depressions contribute $2 \times 308 = 616\text{ cm}^2$; total $= 440 + 616 = 1056\text{ cm}^2$. The total surface area of the remaining solid is $1056\text{ cm}^2$.

• Double-counting the joint. Adding the full TSA of each solid separately counts the hidden circular face on both sides — always subtract or simply never include it.
• Wrong radius at the joint. The two solids sharing a face must have identical radius; do not mix a cone's own radius with a mismatched cylinder or sphere radius from elsewhere in the problem.
• Skipping the slant height. Plugging the cone's height $h$ directly into $\pi r l$ instead of first computing $l = \sqrt{r^2+h^2}$ is the single most common numeric slip in this chapter.
• Treating a depression like an added bump. A hemispherical dent uses the same area formula, $2\pi r^2$, as a hemisphere sitting on top — but a dent leaves no flat disc behind at that end, while a solid with nothing scooped out still has one.

Ask the tutor: Not sure which faces vanish in a three-solid composite like a cylinder capped with a cone and dented with a hemisphere — ask the AI tutor to sketch it and mark every joint with you.