Solids built by combining cuboids, cylinders, cones and spheres — their
surfaces, their capacities, and where every formula comes from.
1. Solid
A solid is a three-dimensional (3-D) object — one that has length,
breadth and height (all three dimensions), unlike a plane figure such as a square
or a circle, which is two-dimensional (2-D) and has only length and breadth. Because a solid
occupies space, every solid has two measurable properties: how much surface covers it,
and how much space it encloses.
Solids around us: a brick, a matchbox, a ball, an
ice-cream cone, a bucket, a pipe — every real object you can hold is a solid.
2. Surface Area
The surface area of a solid is the total area covered by its outer surface(s) —
flat or curved. It answers the question "how much material would it take to cover this solid
completely?" (paint, wrapping paper, cloth, metal sheet, and so on). Since it is an area, it is
always measured in square units ($\text{cm}^2$, $\text{m}^2$, etc.).
3. Volume
The volume of a solid is the amount of three-dimensional space it encloses — in
other words, how much it can hold, or how much material it is made of. Volume is measured in
cubic units ($\text{cm}^3$, $\text{m}^3$, etc.). For liquids, volume is often
measured in litres.
The Total Surface Area is the sum of the areas of every face of a solid
— every flat face and every curved surface, with nothing left out. It is the area you would need
to cover the solid's entire outer boundary, from top to bottom.
5. Lateral / Curved Surface Area (LSA / CSA)
The Lateral Surface Area (or Curved Surface Area) is the area
of only the side surface of a solid — excluding its top and bottom (the bases). The two
names refer to the same idea: "lateral" is the term generally used when the
side faces are flat (like the four walls of a cuboid), while "curved" is used
when the side surface bends around smoothly (like a cylinder or a cone). A solid such as a
sphere, which has no flat base at all, has only a curved surface — so its curved surface area
and total surface area are the same thing.
6. Cuboid
A cuboid is a solid bounded by six rectangular faces, placed at right angles to
each other. It is the 3-D shape of a box — described completely by its length
($l$), breadth ($b$) and height ($h$).
6 rectangular faces, in 3 pairs of identical, parallel opposite faces
12 edges and 8 vertices
Every face meets the next at a right angle ($90°$)
Has one space diagonal length joining any two opposite vertices (all four
space diagonals are equal)
Examples: a brick, a matchbox, a room, a book, a
shoebox.
Lateral Surface Area
$$\text{LSA} = 2h(l+b)$$
Total Surface Area
$$\text{TSA} = 2(lb+bh+hl)$$
Volume
$$V = l \times b \times h$$
Diagonal
$$d = \sqrt{l^2+b^2+h^2}$$
DERIVATION
The four lateral (side) faces come in two identical pairs: front & back, each of
area $l \times h$, and left & right, each of area $b \times h$. Adding all four:
$$\text{LSA} = 2(lh) + 2(bh) = 2h(l+b)$$
The total surface area also counts the top and bottom faces, each of area $l \times b$:
$$\text{TSA} = \text{LSA} + 2lb = 2h(l+b) + 2lb = 2(lb+bh+hl)$$
The volume is simply the area of the base ($l \times b$) swept through the height $h$:
$$V = (l \times b) \times h$$
For the diagonal, apply Pythagoras twice. First, within the base rectangle, the
diagonal joining two opposite corners has length $\sqrt{l^2+b^2}$. That base diagonal, the
vertical edge $h$, and the space diagonal $d$ themselves form a right triangle, so:
$$d = \sqrt{\left(\sqrt{l^2+b^2}\right)^2 + h^2} = \sqrt{l^2+b^2+h^2}$$
7. Cube
A cube is a special cuboid whose length, breadth and height are all equal. Every
edge has the same length, denoted $a$.
6 identical square faces
12 equal edges and 8 vertices
Every face meets the next at a right angle ($90°$)
Examples: a Rubik's cube, a die (dice), a sugar cube,
an ice cube.
Lateral Surface Area
$$\text{LSA} = 4a^2$$
Total Surface Area
$$\text{TSA} = 6a^2$$
Volume
$$V = a^3$$
Diagonal
$$d = a\sqrt{3}$$
DERIVATION
Every cube formula is just the cuboid formula with $l=b=h=a$ substituted in:
$$\text{LSA} = 2a(a+a) = 4a^2 \qquad \text{TSA} = 2(a^2+a^2+a^2) = 6a^2 \qquad V = a\times
a\times a = a^3$$
$$d = \sqrt{a^2+a^2+a^2} = \sqrt{3a^2} = a\sqrt{3}$$
8. Solid Cylinder
A (right circular) cylinder is a solid with two congruent, parallel circular
bases joined by a single curved lateral surface, with its axis perpendicular to both bases.
2 flat circular faces (radius $r$), and one curved lateral surface
Height $h$ is the perpendicular distance between the two circular faces
No vertices; 2 circular edges where the curved surface meets the bases
Examples: a tin can, a candle, a battery cell, a garden
roller, a wooden log.
Curved Surface Area
$$\text{CSA} = 2\pi r h$$
Total Surface Area
$$\text{TSA} = 2\pi r h + 2\pi r^2 = 2\pi r(h+r)$$
Volume
$$V = \pi r^2 h$$
DERIVATION
Imagine slicing the curved surface once (vertically) and unrolling it flat: it opens out into a
rectangle. The height of that rectangle is still $h$, and its width is exactly
the distance around the circular base — the circumference, $2\pi r$ — since the curved surface
wraps around the base exactly once. So:
$$\text{CSA} = \text{width} \times \text{height} = 2\pi r \times h$$
The total surface area adds the two circular bases, each of area $\pi r^2$:
$$\text{TSA} = \text{CSA} + 2\pi r^2 = 2\pi rh + 2\pi r^2 = 2\pi r(h+r)$$
The volume is, as with any prism-like solid, the base area swept through the height:
$$V = (\pi r^2) \times h$$
9. Hollow Cylinder
A hollow cylinder (a pipe or tube) is bounded by two coaxial (same-axis)
circular cylinders of different radii — an outer radius $R$ and an inner radius $r$ — sharing
the same height $h$. Its wall has a uniform thickness of $R-r$.
An outer curved surface and an inner curved surface
Two flat, ring-shaped (annular) end faces, each of area $\pi(R^2-r^2)$
Wall thickness $= R - r$
Examples: a water pipe, a metal tube, a drinking straw,
a cardboard roll.
Total Surface Area
$$\text{TSA} = 2\pi h(R+r) + 2\pi(R^2-r^2)$$
Volume (of the material)
$$V = \pi h(R^2-r^2)$$
DERIVATION
Just like a solid cylinder, the outer curved surface unrolls into a rectangle of width
$2\pi R$: area $=2\pi Rh$. The inner curved surface (the inside wall) similarly unrolls
to give area $2\pi rh$. Each of the two ends is a flat ring left over after removing a circle of
radius $r$ from a circle of radius $R$, so each end has area $\pi R^2 - \pi r^2 = \pi(R^2-r^2)$.
Adding the outer curve, the inner curve, and both ends:
$$\text{TSA} = 2\pi Rh + 2\pi rh + 2\pi(R^2-r^2) = 2\pi h(R+r) + 2\pi(R^2-r^2)$$
The volume of material is what's left after removing the inner (hollow) cylinder from the outer
one:
$$V = \underbrace{\pi R^2 h}_{\text{outer cylinder}} - \underbrace{\pi r^2 h}_{\text{inner
cylinder}} = \pi h(R^2-r^2)$$
10. Cone
A (right circular) cone is a solid with a flat circular base that tapers
smoothly to a single point, the apex, directly above the centre of the base.
1 flat circular base (radius $r$) and 1 curved lateral surface
Height $h$: the perpendicular distance from the apex to the centre of the base
Slant height $l$: the distance from the apex to any point on the base's rim
$h$, $r$ and $l$ always form a right triangle, so $l^2 = h^2 + r^2$
Examples: an ice-cream cone, a traffic cone, a birthday
party hat, a funnel.
Slant Height
$$l = \sqrt{h^2+r^2}$$
Curved Surface Area
$$\text{CSA} = \pi r l$$
Total Surface Area
$$\text{TSA} = \pi r l + \pi r^2 = \pi r(l+r)$$
Volume
$$V = \frac{1}{3}\pi r^2 h$$
DERIVATION
The apex, the centre of the base, and any point on the rim form a right triangle (right-angled
at the centre), with legs $h$ and $r$ and hypotenuse $l$ — giving $l^2 = h^2 + r^2$ directly
from Pythagoras.
To find the curved surface area, cut the cone along a slant line and flatten it out: it opens
into a sector of a circle of radius $l$. Its arc — the curved edge — is exactly
the base's circumference, $2\pi r$, since it used to wrap once around the base. A circular
sector with arc length $s$ and radius $l$ has area $\frac{1}{2}sl$ (the same "half base times
height" idea as a triangle, applied to a very thin pie-slice). Substituting $s = 2\pi r$:
$$\text{CSA} = \frac{1}{2}(2\pi r)(l) = \pi r l$$
The total surface area adds the circular base:
$$\text{TSA} = \pi rl + \pi r^2 = \pi r(l+r)$$
The volume can't be found by simply multiplying base area by height — a cone tapers away, unlike
a cylinder. It can be shown (by experiment, or by integration) that a cone's volume is always
exactly one-third of the cylinder sharing its base and height:
$$V = \frac{1}{3}\pi r^2 h$$
WHY ONE-THIRD? (a calculus view)
Slice the cone into thin discs perpendicular to its axis. At a distance $x$ from the apex,
similar triangles give the disc's radius as $\frac{r}{h}x$, so its area is
$\pi\left(\frac{r}{h}\right)^2 x^2$. Summing (integrating) these discs from $x=0$ to $x=h$:
$$V = \int_0^h \pi\left(\frac{r}{h}\right)^2 x^2\,dx = \pi\frac{r^2}{h^2}\cdot\frac{h^3}{3} =
\frac{1}{3}\pi r^2 h$$
11. Sphere
A sphere is a perfectly round solid — every single point on its surface lies at
the same fixed distance $r$ (the radius) from one central point.
One continuous curved surface; no flat faces, edges, or vertices
The largest circular cross-section through the centre is called a great
circle (area $\pi r^2$)
Since it has no flat face at all, its curved and total surface area are the same thing
Examples: a football, a globe, a marble, a ball
bearing.
Surface Area (Curved = Total)
$$\text{TSA} = 4\pi r^2$$
Volume
$$V = \frac{4}{3}\pi r^3$$
DERIVATION — Archimedes' cylinder
Picture a sphere of radius $r$ fitted exactly inside a cylinder of the same radius $r$
and height $2r$ — so the cylinder just touches the sphere on all sides, top and bottom.
Archimedes showed that the sphere's curved surface exactly equals the cylinder's curved surface:
$$\text{Sphere's TSA} = \text{Cylinder's CSA} = 2\pi r \times (2r) = 4\pi r^2$$
He also showed the sphere's volume is always exactly two-thirds of this circumscribing
cylinder's volume. Since the cylinder's volume is $\pi r^2(2r) = 2\pi r^3$:
$$V = \frac{2}{3}\times 2\pi r^3 = \frac{4}{3}\pi r^3$$
THE SAME RESULT, WITH CALCULUS
Slicing the sphere into thin discs along a diameter, the disc at position $x$ (measured from the
centre) has radius $\sqrt{r^2-x^2}$, giving area $\pi(r^2-x^2)$. Integrating from $x=-r$ to
$x=r$:
$$V = \int_{-r}^{r}\pi(r^2-x^2)\,dx = \pi\left[r^2x - \frac{x^3}{3}\right]_{-r}^{r} =
\frac{4}{3}\pi r^3$$
12. Hemisphere
A hemisphere is exactly half of a sphere, produced by slicing a sphere into two
equal halves through its centre.
1 curved surface (half the sphere's curved surface) and 1 flat circular face (the cut,
radius $r$)
The flat face and curved surface meet along a single circular rim
Examples: a mixing bowl, a dome, half an orange, a
hemispherical paperweight.
Curved Surface Area
$$\text{CSA} = 2\pi r^2$$
Total Surface Area
$$\text{TSA} = 2\pi r^2 + \pi r^2 = 3\pi r^2$$
Volume
$$V = \frac{2}{3}\pi r^3$$
DERIVATION
A hemisphere is exactly half a sphere, so its curved surface is exactly half the sphere's curved
surface, and its volume is exactly half the sphere's volume:
$$\text{CSA} = \frac{1}{2}(4\pi r^2) = 2\pi r^2 \qquad\qquad V = \frac{1}{2}\left(\frac{4}{3}\pi
r^3\right) = \frac{2}{3}\pi r^3$$
Slicing the sphere in half, however, exposes a brand-new flat circular face that wasn't
part of the original sphere's surface — of area $\pi r^2$. This must be added for the total
surface area (it plays no part in the volume, which is genuinely just halved):
$$\text{TSA} = \text{CSA} + \pi r^2 = 2\pi r^2 + \pi r^2 = 3\pi r^2$$
13. Hollow Sphere
A hollow sphere (a spherical shell) is bounded between two concentric spherical
surfaces — an outer radius $R$ and an inner radius $r$ — with a uniform wall thickness of $R-r$.
Unlike a bowl, it is normally a fully closed shell, so only its outer surface is ever exposed to
the outside.
Outer spherical surface (radius $R$) and inner spherical surface (radius $r$), sharing a
common centre
Uniform thickness $= R - r$
Overwhelmingly used for volume questions (melting & recasting the shell
material)
Examples: a hollow rubber ball, a ball bearing, a
hollow toy ball.
Outer Surface Area (exposed surface)
$$\text{TSA} = 4\pi R^2$$
Volume (of the shell material)
$$V = \frac{4}{3}\pi(R^3-r^3)$$
DERIVATION
The outside of the shell is just an ordinary sphere of radius $R$, so its surface area follows
the usual sphere formula, $4\pi R^2$. (The inner spherical surface exists mathematically, but
isn't exposed to the outside for a sealed shell — it only matters if the object is deliberately
open, like a bowl, in which case both curved surfaces would be added, the same way a hollow
cylinder's outer and inner curves are both counted.)
The material of the shell fills the gap between the outer sphere and the inner (empty) sphere,
so its volume is simply the difference of the two sphere volumes:
$$V = \underbrace{\frac{4}{3}\pi R^3}_{\text{outer sphere}} - \underbrace{\frac{4}{3}\pi
r^3}_{\text{inner sphere}} = \frac{4}{3}\pi(R^3-r^3)$$
14. Frustum of a Cone
A frustum is what remains of a cone after its top has been sliced off by a plane
parallel to the base — keeping the piece that still has the original base, and discarding the
small cone that contained the apex.
Two parallel circular faces of different radii: $R$ (the larger, original base) and
$r$ (the smaller, at the cut)
Height $h$: perpendicular distance between the two circular faces
Slant height $l$: distance along the lateral surface between the two rims, with
$l=\sqrt{h^2+(R-r)^2}$
Examples: a bucket, a lampshade, a drinking glass
(tumbler), a plant pot.
Slant Height
$$l = \sqrt{h^2+(R-r)^2}$$
Curved Surface Area
$$\text{CSA} = \pi(R+r)l$$
Total Surface Area
$$\text{TSA} = \pi(R+r)l + \pi R^2 + \pi r^2$$
DERIVATION
Dropping a perpendicular from the smaller rim down to the plane of the larger base gives a right
triangle with legs $h$ and $R-r$, and the frustum's own slant side as the hypotenuse:
$$l = \sqrt{h^2+(R-r)^2}$$
For the volume, imagine the frustum as part of the original, uncut cone. Let that full cone have
height $H$ (apex to the $R$-base) and let the small piece removed from its top have height $H-h$
and radius $r$. Because the removed cone is similar to the full cone (same shape, same
apex angle):
$$\frac{r}{R} = \frac{H-h}{H} \quad\Longrightarrow\quad H = \frac{Rh}{R-r}$$
The frustum's volume is the full cone's volume minus the removed cone's volume:
$$V = \frac{1}{3}\pi R^2 H - \frac{1}{3}\pi r^2(H-h) = \frac{1}{3}\pi\Big[H(R^2-r^2) +
r^2h\Big]$$
Substituting $H=\dfrac{Rh}{R-r}$ and using $R^2-r^2=(R-r)(R+r)$, the $(R-r)$ cancels neatly:
$$H(R^2-r^2) = \frac{Rh}{R-r}\times(R-r)(R+r) = Rh(R+r)$$
$$V = \frac{1}{3}\pi\Big[Rh(R+r)+r^2h\Big] = \frac{1}{3}\pi h\left(R^2+Rr+r^2\right)$$
The curved surface area follows the very same idea — CSA of the full cone minus CSA of the
removed cone — and simplifies (using the same similar-triangles ratio) to:
$$\text{CSA} = \pi(R+r)l$$
The total surface area simply adds both circular faces:
$$\text{TSA} = \pi(R+r)l + \pi R^2 + \pi r^2$$
15. Solved Examples
A spread of standard board-level problems — one plain example per solid, plus the two classic
problem types that make this chapter distinctive: combining solids, and converting one solid
into another.
Example 1 • Cuboid Volume
A matchbox measures $4\text{ cm} \times 2.5\text{ cm} \times
1.5\text{ cm}$. Find the volume of a packet containing $12$ such matchboxes.
Solution:
Volume of one matchbox:
$$V = l \times b \times h = 4 \times 2.5 \times 1.5 = 15\text{ cm}^3$$
Volume of $12$ matchboxes:
$$V_{12} = 15 \times 12 = 180\text{ cm}^3$$
Thus, the packet occupies $180\text{ cm}^3$.
Example 2 • Solid Cylinder
A solid right circular cylinder has radius $7\text{ cm}$ and
height $20\text{ cm}$. Find its curved surface area and its volume. (Use $\pi=\frac{22}{7}$)
A metal pipe has an outer radius of $10\text{ cm}$, an inner
radius of $8\text{ cm}$, and a length of $35\text{ cm}$. Find the volume of metal used to
make it. (Use $\pi=\frac{22}{7}$)
Find (a) the area of sheet metal needed for an open
hemispherical bowl, and (b) the total surface area of a solid hemispherical
paperweight — both of radius $3.5\text{ cm}$. (Use $\pi=\frac{22}{7}$)
Solution:
(a) An open bowl has only a curved surface (no flat face to cover):
$$\text{CSA} = 2\pi r^2 = 2\times\frac{22}{7}\times(3.5)^2 = 77\text{ cm}^2$$
(b) A solid paperweight resting on a table also exposes its flat circular base:
$$\text{TSA} = 3\pi r^2 = 3\times\frac{22}{7}\times(3.5)^2 = 115.5\text{ cm}^2$$
Example 7 • Hollow Sphere — Melt & Recast
A hollow sphere of external radius $6\text{ cm}$ and internal
radius $3\text{ cm}$ is melted and recast into a solid cylinder of radius $6\text{ cm}$.
Find the height of the cylinder. (Use $\pi=\frac{22}{7}$)
Solution:
Volume of the hollow sphere (the melted material):
$$V = \frac{4}{3}\pi(R^3-r^3) = \frac{4}{3}\times\frac{22}{7}\times(6^3-3^3) =
\frac{4}{3}\times\frac{22}{7}\times189 = 792\text{ cm}^3$$
This is recast into a cylinder of radius $6\text{ cm}$, so its volume stays the same:
$$\pi(6)^2h = 792 \implies \frac{22}{7}\times 36 \times h = 792 \implies h = 7\text{ cm}$$
Example 8 • Frustum — Bucket
A bucket, in the shape of a frustum, has a top radius of $20\text{
cm}$, a bottom radius of $12\text{ cm}$, and a height of $15\text{ cm}$. Find its capacity,
and the area of metal sheet needed to make it (it has no lid). (Use $\pi=\frac{22}{7}$)
Solution:
Slant height: $l=\sqrt{h^2+(R-r)^2}=\sqrt{15^2+8^2}=\sqrt{225+64}=\sqrt{289}=17\text{
cm}$.
Capacity (volume):
$$V = \frac{1}{3}\pi h(R^2+Rr+r^2) =
\frac{1}{3}\times\frac{22}{7}\times15\times(400+240+144) = 12320\text{ cm}^3 = 12.32\text{
L}$$
Metal sheet needed (curved surface, plus the solid bottom — the top is open):
$$\text{Sheet} = \pi(R+r)l + \pi r^2 = \frac{22}{7}\times32\times17 + \frac{22}{7}\times144
\approx 2162.3\text{ cm}^2$$
Example 9 • Combination of Solids — Ice-cream Cone
An ice-cream cone has a hemispherical top of the same radius as
the cone. If the radius is $3\text{ cm}$ and the height of the conical part is $9\text{
cm}$, find the total volume of ice cream. (Use $\pi=3.14$)
Solution:
Volume of the conical part:
$$V_{cone} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\times3.14\times3^2\times9 = 84.78\text{
cm}^3$$
Volume of the hemispherical top (same radius $r=3$):
$$V_{hemi} = \frac{2}{3}\pi r^3 = \frac{2}{3}\times3.14\times3^3 = 56.52\text{ cm}^3$$
Total volume of ice cream:
$$V = 84.78 + 56.52 = 141.3\text{ cm}^3$$
Example 10 • Conversion of Solids — Melt & Recast
A metallic sphere of radius $4.2\text{ cm}$ is melted and recast
into the shape of a solid cylinder of radius $6\text{ cm}$. Find the height of the cylinder.
(Use $\pi=\frac{22}{7}$)
Solution:
Since the same metal is melted and recast, the volume is conserved. Volume of sphere:
$$V = \frac{4}{3}\pi r^3 = \frac{4}{3}\times\frac{22}{7}\times(4.2)^3 = 310.464\text{
cm}^3$$
This becomes the cylinder's volume, so:
$$\pi(6)^2h = 310.464 \implies \frac{22}{7}\times36\times h = 310.464 \implies h \approx
2.74\text{ cm}$$