Spheres, Cones, Pyramids and Composite Solids

Real objects are rarely a single neat shape: a grain silo is a cylinder with a dome on top, a medicine capsule is a cylinder with rounded ends, and a spinning top is a cone sitting on a hemisphere. This page adds spheres and hemispheres to the prisms, pyramids and cones you met in surface area and volume, and shows how to handle solids built from several pieces, the way IB questions do.

Key ideas

Spheres and hemispheres

For a sphere of radius $r$:

$$V = \frac{4}{3}\pi r^3 \qquad\qquad S = 4\pi r^2$$

A hemisphere is half a sphere. Its volume is half the sphere’s volume, but its surface area needs care, because cutting the sphere in half creates a new flat circular face:

Hemisphere of radius $r$	Formula
Volume	$\dfrac{2}{3}\pi r^3$
Curved surface area	$2\pi r^2$ (half of $4\pi r^2$)
Total surface area (solid hemisphere)	$2\pi r^2 + \pi r^2 = 3\pi r^2$

Read the question carefully: an open bowl only has the curved surface, while a solid paperweight also has the flat base.

Right pyramids and right cones

In a right pyramid or cone, the apex is directly above the centre of the base. Two different lengths matter:

• the height $h$, measured straight down from the apex to the centre of the base, is used for volume;
• the slant height $l$, measured down the sloping face, is used for surface area.

$$V_{\text{pyramid}} = \frac{1}{3} \times (\text{base area}) \times h \qquad\qquad V_{\text{cone}} = \frac{1}{3}\pi r^2 h$$

The curved surface of a cone has area $\pi r l$, so a solid cone has total surface area $\pi r^2 + \pi r l$. The height, the slant height and a horizontal distance across the base form a right triangle, so Pythagoras’ theorem connects them. For a cone, $l^2 = h^2 + r^2$.

A pyramid’s triangular faces don’t all need the same slant height. On a rectangular base, the faces on the long sides and the short sides have different slant heights, so you find each one separately (Example 3).

Composite solids

A composite solid is built from simpler solids.

• Volume: add the volumes of the pieces (or subtract a piece that has been cut out).
• Surface area: add only the surfaces on the outside. Where two pieces are joined, the faces that touch are hidden, so leave them out.

Two composite solids: a cone on a hemisphere (Example 4) and a cylinder with hemispherical ends (Practice 5).

Answers and units

Give exact answers in terms of $\pi$ when asked, and otherwise round to 3 significant figures (the IB default), using the full calculator value until the end. Volumes are in cubic units ($\text{cm}^3$, $\text{m}^3$) and areas in square units ($\text{cm}^2$, $\text{m}^2$).

On the SAT

The SAT reference sheet gives the volumes of a sphere, cone, and pyramid, but not the surface area of a sphere, so remember $SA = 4\pi r^2$. SAT answers are often exact, in terms of $\pi$, so don’t round to $3$ significant figures there the way this page does; compare with the choices instead. For a composite solid, add or subtract the pieces, and use Desmos (type pi) for the arithmetic if the choices are decimals. See using Desmos on the SAT.

Worked examples

Example 1: A sphere

A ball has a diameter of $22$ cm. Find its volume and its surface area.

Solution. The radius is half the diameter: $r = 11$ cm.

$$V = \frac{4}{3}\pi (11)^3 = \frac{5324\pi}{3} = 5575.27\ldots \approx 5580 \text{ cm}^3 \quad (\text{3 s.f.})$$ $$S = 4\pi (11)^2 = 484\pi = 1520.53\ldots \approx 1520 \text{ cm}^2 \quad (\text{3 s.f.})$$

Example 2: A solid hemisphere

A glass paperweight is a solid hemisphere of radius $6$ cm. Find its volume and its total surface area.

Solution.

$$V = \frac{2}{3}\pi (6)^3 = 144\pi \approx 452 \text{ cm}^3 \quad (\text{3 s.f.})$$

The paperweight has a curved surface and a flat circular base:

$$S = 2\pi (6)^2 + \pi (6)^2 = 72\pi + 36\pi = 108\pi \approx 339 \text{ cm}^2 \quad (\text{3 s.f.})$$

Check: using only $2\pi r^2 \approx 226 \text{ cm}^2$ would leave out the base you set on the desk.

Example 3: A pyramid on a rectangular base

A right pyramid has a rectangular base measuring $8$ cm by $6$ cm, and its apex is $12$ cm above the centre of the base. Find its volume and its total surface area.

Solution. Volume:

$$V = \frac{1}{3} \times (8 \times 6) \times 12 = 192 \text{ cm}^3$$

For the surface area you need the slant height of each pair of triangular faces. From the centre of the base, the midpoint of a long ($8$ cm) side is $3$ cm away, and the midpoint of a short ($6$ cm) side is $4$ cm away. Each slant height is the hypotenuse of a right triangle with the height $12$:

$$l_1 = \sqrt{12^2 + 3^2} = \sqrt{153} \qquad\qquad l_2 = \sqrt{12^2 + 4^2} = \sqrt{160}$$

There are two triangles with base $8$ and slant height $l_1$, and two with base $6$ and slant height $l_2$:

$$\begin{aligned} S &= 8 \times 6 + 2 \times \tfrac{1}{2}(8)\sqrt{153} + 2 \times \tfrac{1}{2}(6)\sqrt{160} \\ &= 48 + 8\sqrt{153} + 6\sqrt{160} \\ &= 222.84\ldots \approx 223 \text{ cm}^2 \quad (\text{3 s.f.}) \end{aligned}$$

Example 4: A cone on a hemisphere

A wooden spinning top is a cone of height $8$ cm sitting on a hemisphere of radius $3$ cm, with the same radius (see the figure). Find its volume and its surface area.

Solution. Volume: add the two pieces.

$$V = \frac{1}{3}\pi (3)^2(8) + \frac{2}{3}\pi (3)^3 = 24\pi + 18\pi = 42\pi \approx 132 \text{ cm}^3 \quad (\text{3 s.f.})$$

Surface area: the flat circles where the cone and hemisphere meet are hidden, so only the cone’s curved surface and the hemisphere’s curved surface count. First the slant height:

$$l = \sqrt{8^2 + 3^2} = \sqrt{73}$$ $$S = \pi (3)\sqrt{73} + 2\pi (3)^2 = 3\sqrt{73}\,\pi + 18\pi = 137.07\ldots \approx 137 \text{ cm}^2 \quad (\text{3 s.f.})$$

Common mistakes

Using $4\pi r^2$ or $2\pi r^2$ for a solid hemisphere. A solid hemisphere has a curved surface ($2\pi r^2$) and a flat base ($\pi r^2$), for a total of $3\pi r^2$. An open bowl has only the curved part. Decide which one the question describes before you calculate.

Mixing up height and slant height. Volume uses the vertical height $h$; the curved surface of a cone and the faces of a pyramid use the slant height $l$. If you’re given one, find the other with Pythagoras.

Counting hidden faces in a composite solid. When a cone sits on a hemisphere, the two circles where they join are inside the solid. Leave them out of the surface area, but do include both pieces in the volume.

Using the diameter as the radius. Many questions give a diameter (“a ball $22$ cm across”). Halve it first. Using $d$ instead of $r$ makes a volume $8$ times too big and an area $4$ times too big.

Rounding too early. In Example 4, rounding $\sqrt{73}$ to $8.5$ before multiplying gives $43.5\pi = 136.66\ldots$ instead of $137.07\ldots$. It happens to round to the same answer here, but often it won’t. Keep the exact value or the full calculator value until the last step.

Wrong units. Volume is in cubic units and area in square units. Writing $132 \text{ cm}^2$ for a volume loses marks even when the number is right.

Practice

1. (Warm-up) A spherical water tank has radius $4.5$ m. Find its volume and its surface area.

Solution

$$V = \frac{4}{3}\pi (4.5)^3 = 121.5\pi \approx 382 \text{ m}^3 \quad (\text{3 s.f.})$$$$S = 4\pi (4.5)^2 = 81\pi \approx 254 \text{ m}^2 \quad (\text{3 s.f.})$$

2. (Warm-up) A solid right cone has base radius $5$ cm and height $12$ cm. Find its slant height, volume and total surface area.

Solution

$$l = \sqrt{12^2 + 5^2} = \sqrt{169} = 13 \text{ cm}$$$$V = \frac{1}{3}\pi (5)^2 (12) = 100\pi \approx 314 \text{ cm}^3 \quad (\text{3 s.f.})$$$$S = \pi (5)^2 + \pi (5)(13) = 25\pi + 65\pi = 90\pi \approx 283 \text{ cm}^2 \quad (\text{3 s.f.})$$

3. (Warm-up) A hemisphere has radius $10$ cm. Find, in terms of $\pi$, (a) its curved surface area, and (b) its total surface area if it is solid.

Solution

(a) $2\pi (10)^2 = 200\pi \text{ cm}^2$ (about $628 \text{ cm}^2$).

(b) $3\pi (10)^2 = 300\pi \text{ cm}^2$ (about $942 \text{ cm}^2$).

4. (Core) A sphere has a volume of $1000 \text{ cm}^3$. Find its radius and its surface area.

Solution

$$\frac{4}{3}\pi r^3 = 1000 \quad\Rightarrow\quad r^3 = \frac{750}{\pi} \quad\Rightarrow\quad r = \sqrt[3]{\frac{750}{\pi}} = 6.2035\ldots \approx 6.20 \text{ cm}$$

Using the unrounded radius:

$$S = 4\pi r^2 = 483.59\ldots \approx 484 \text{ cm}^2 \quad (\text{3 s.f.})$$

(Using the rounded $r = 6.20$ gives $483$, which is why you should keep the full value.)

5. (Core) A medicine capsule is a cylinder with a hemisphere on each end. The capsule is $20$ mm long overall and $6$ mm in diameter (see the figure). Find its volume and its surface area.

Solution

The radius is $3$ mm. The two hemispheres take up $3 + 3 = 6$ mm of the length, so the cylinder is $20 - 6 = 14$ mm long. The two hemispheres together make one sphere.

$$V = \pi (3)^2 (14) + \frac{4}{3}\pi (3)^3 = 126\pi + 36\pi = 162\pi \approx 509 \text{ mm}^3 \quad (\text{3 s.f.})$$

The outside is the curved surface of the cylinder plus the surface of one whole sphere (the cylinder’s ends are hidden):

$$S = 2\pi (3)(14) + 4\pi (3)^2 = 84\pi + 36\pi = 120\pi \approx 377 \text{ mm}^2 \quad (\text{3 s.f.})$$

6. (Core) A grain silo is a cylinder of radius $4$ m and height $15$ m with a hemispherical roof.

• (a) Find the volume of the silo.
• (b) The outside of the silo, but not the floor, is to be painted. Find the area to be painted.

Solution

(a)

$$V = \pi (4)^2 (15) + \frac{2}{3}\pi (4)^3 = 240\pi + \frac{128\pi}{3} = \frac{848\pi}{3} \approx 888 \text{ m}^3 \quad (\text{3 s.f.})$$

(b) Paint the curved wall of the cylinder and the curved roof (the floor isn’t painted and the top of the cylinder is hidden under the roof):

$$S = 2\pi (4)(15) + 2\pi (4)^2 = 120\pi + 32\pi = 152\pi \approx 478 \text{ m}^2 \quad (\text{3 s.f.})$$

7. (Core) A right pyramid has a square base of side $10$ m. Each of its four sloping edges is $13$ m long.

• (a) Find the height of the pyramid.
• (b) Find its volume.
• (c) Find the total area of the four triangular faces.

Solution

(a) The centre of the base is half a diagonal from each corner. The diagonal is $10\sqrt{2}$, so half of it is $5\sqrt{2}$. The height, half-diagonal and edge form a right triangle:

$$h = \sqrt{13^2 - (5\sqrt{2})^2} = \sqrt{169 - 50} = \sqrt{119} \approx 10.9 \text{ m} \quad (\text{3 s.f.})$$

(b)

$$V = \frac{1}{3}(10^2)\sqrt{119} = 363.62\ldots \approx 364 \text{ m}^3 \quad (\text{3 s.f.})$$

(c) Each face is an isosceles triangle with sides $13$, $13$ and base $10$. Its slant height goes from the apex to the midpoint of the base:

$$l = \sqrt{13^2 - 5^2} = \sqrt{144} = 12 \text{ m}$$$$\text{Area} = 4 \times \frac{1}{2}(10)(12) = 240 \text{ m}^2$$

8. (Challenge) A solid metal sphere of radius $6$ cm is melted down and recast as a solid cone with base radius $6$ cm. No metal is lost.

• (a) Find the height of the cone.
• (b) Find the curved surface area of the cone. Give your answer exactly and to 3 s.f.

Solution

(a) The volumes are equal:

$$\frac{1}{3}\pi (6)^2 h = \frac{4}{3}\pi (6)^3 \quad\Rightarrow\quad 12\pi h = 288\pi \quad\Rightarrow\quad h = 24 \text{ cm}$$

(b)

$$l = \sqrt{24^2 + 6^2} = \sqrt{612} = 6\sqrt{17}$$$$\pi r l = \pi (6)(6\sqrt{17}) = 36\sqrt{17}\,\pi \approx 466 \text{ cm}^2 \quad (\text{3 s.f.})$$

9. (Challenge) A toy is made from a hemisphere of radius $r$ cm with a cone on top. The cone has the same radius and height $2r$ cm. The volume of the toy is $250 \text{ cm}^3$. Find $r$.

Solution

$$V = \frac{1}{3}\pi r^2 (2r) + \frac{2}{3}\pi r^3 = \frac{2}{3}\pi r^3 + \frac{2}{3}\pi r^3 = \frac{4}{3}\pi r^3$$

(Neat: the toy has exactly the volume of a whole sphere of radius $r$.) So

$$\frac{4}{3}\pi r^3 = 250 \quad\Rightarrow\quad r^3 = \frac{187.5}{\pi} \quad\Rightarrow\quad r = 3.9079\ldots \approx 3.91 \text{ cm} \quad (\text{3 s.f.})$$