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Spheres: Volume and Surface Area

Find the volume and surface area of spheres and hemispheres, and work backwards to the radius or diameter.

Practice this skill 22 min Spheres Leer en español
Key idea

What you'll learn

A sphere is a perfectly round 3-D object: a basketball, a globe, a gumball, a ball bearing, a round water tower. Every point on its surface is the same distance from the center.

In this lesson you will learn to:

  • find the volume of a sphere (how much it holds),
  • find the surface area of a sphere (how much material covers it),
  • handle hemispheres (half spheres, like a bowl or a dome),
  • work backwards from a volume or surface area to the radius or diameter,
  • predict how volume and surface area change when the radius changes.

Why it matters on the GED: Sphere questions are short but full of traps. The test likes to give the diameter when the formula needs the radius, to mix up volume and surface area, and to ask you to find the radius from a given volume. Both formulas are on the formula sheet; the skill is using them correctly.

Real life: How much air is in a ball? How much paint covers a dome? How many gallons fit in a round tank? All sphere problems.

Words to know

Key words

  • sphere — a round solid where every point on the surface is the same distance from the center (a ball).
  • center — the point in the exact middle of the sphere.
  • radius (r) — the distance from the center to the surface (a ball 10 in across has r = 5 in).
  • diameter (d) — the distance straight across the sphere through the center; d = 2r and r = \tfrac{d}{2}.
  • great circle — the largest circle you can draw on a sphere, around its "equator"; its radius equals the sphere's radius.
  • hemisphere — exactly half of a sphere, cut through the center (a dome or a bowl).
  • volume (V) — the space inside, in cubic units (in³, cm³).
  • surface area (SA) — the area of the curved outside, in square units (in², cm²).
  • cube (r^3) — a number times itself three times: 3^3 = 3 \times 3 \times 3 = 27.
  • cube root (\sqrt[3]{\ }) — the number that cubes to a given value: \sqrt[3]{27} = 3.

All Math words in the glossary

Sphere with its center marked. A radius r goes from the center to the surface. A diameter d passes through the center from one side to the other, so d = 2r. The dashed ellipse is the great circle (equator) around the middle. r d = 2r center

The radius r goes from the center to the surface. The diameter d goes all the way across through the center, so d = 2r.

Volume of a sphere

The volume of a sphere is

V = \tfrac{4}{3}\pi r^3

Three steps, in this order:

  1. Cube the radius: r^3 = r \times r \times r.
  2. Multiply by \tfrac{4}{3} (an easy way: divide by 3, then multiply by 4).
  3. Multiply by \pi — or leave \pi in the answer if the choices are written "in terms of \pi".

Example: r = 3. 3^3 = 27; 27 \div 3 = 9; 9 \times 4 = 36. So V = 36\pi \approx 113.1 cubic units.

Example: r = 6. 6^3 = 216; 216 \div 3 = 72; 72 \times 4 = 288. So V = 288\pi \approx 904.8.

Cube, don't square. 3^3 = 27, not 9. Not 3 \times 3, and not 3 \times 3 \times anything else.

Diameter given? Halve it first. A ball with diameter 6 has r = 3 and V = 36\pi. Using 6 as the radius gives 288\pi — 8 times too big.

Two spheres drawn to scale. Left: diameter 6, so radius 3 and volume 36 pi. Right: a sphere with radius 6 (diameter 12) has volume 288 pi, 8 times as much. Using the diameter as the radius gives an answer 8 times too big. r = 3 d = 6 d = 6, r = 3 V = 36π ✓ r = 6 r = 6 (twice as wide) V = 288π = 8 × 36π

Hemisphere. Half of a sphere has half the volume:

V_{\text{hemisphere}} = \tfrac{1}{2} \cdot \tfrac{4}{3}\pi r^3 = \tfrac{2}{3}\pi r^3

Example: a bowl with r = 4: \tfrac{2}{3}\pi(64) = \tfrac{128}{3}\pi \approx 134.0.

Hemisphere-shaped bowl with radius 4: half of a sphere, open on top. Volume = two-thirds pi times 4 cubed = 128 thirds pi, about 134.0 cubic units. r = 4 r = 4 hemisphere = ½ sphere V = ⅔π(4³) = ⅔π(64) ≈ 134.0

Surface area of a sphere

The surface area of a sphere is

SA = 4\pi r^2

Here's a nice fact: the surface of a sphere is exactly 4 times the area of its great circle (\pi r^2). Picture the peel of an orange covering four flat circles the same size as the orange's middle.

A sphere with radius 5 and four flat circles, each with radius 5 like the sphere's great circle (area 25 pi each). The surface of the sphere equals all four circles: 4 pi r squared = 100 pi. sphere r = 5 = 25π 25π 25π 25π 4 great circles, each π(5²) SA = 4πr² = 4 × 25π = 100π ≈ 314.2

Example: r = 5. 5^2 = 25; 4 \times 25 = 100. So SA = 100\pi \approx 314.2 square units.

Example: a globe with d = 16, so r = 8. SA = 4\pi(64) = 256\pi \approx 804.2 in².

Hemisphere surface area depends on the object:

ObjectSurfacesFormula
open bowl or dome (curved part only)half of the sphere\tfrac{1}{2}(4\pi r^2) = 2\pi r^2
solid half-ball (with flat circle)curved part + circle2\pi r^2 + \pi r^2 = 3\pi r^2

Example: a dome with r = 10 ft (no floor): 2\pi(100) = 200\pi \approx 628.3 ft² of roofing.

Two hemispheres with radius 10 feet. Left: an open dome with no floor; only the curved part is covered, 2 pi r squared = 200 pi, about 628.3 square feet. Right: a solid half-ball also has its flat circle, so its surface is 2 pi r squared + pi r squared = 3 pi r squared. r = 10 ft open dome (no floor): 2πr² = 200π ≈ 628.3 ft² r solid half-ball: 2πr² + πr² = 3πr²

Volume or surface area? Read the verbs. Fill, hold, contain, capacity, how much air/water → volume (cubic units). Paint, cover, wrap, coat, material → surface area (square units).

Working backwards and comparing sizes

Radius from surface area — divide by 4\pi, then take a square root:

r = \sqrt{\frac{SA}{4\pi}}

Example: SA = 196\pi. r^2 = 196 \div 4 = 49, so r = 7 and d = 14.

Radius from volume — multiply by 3, divide by 4\pi, then take a cube root:

r = \sqrt[3]{\frac{3V}{4\pi}}

Example: V = 972\pi. r^3 = 972 \times \tfrac{3}{4} = 729, so r = \sqrt[3]{729} = 9.

Example with a decimal: V \approx 523.6. r^3 = \dfrac{3(523.6)}{4\pi} \approx 125, so r = 5 and d = 10.

Helpful cubes to recognize: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000 (the cubes of 1 through 10).

How changes in radius affect a sphere. If the radius is multiplied by k:

RadiusSurface areaVolume
×2×4×8
×3×9×27
×\tfrac{1}{2}×\tfrac{1}{4}×\tfrac{1}{8}

So a ball twice as wide holds 8 times as much air, but needs only 4 times as much rubber.

Two spheres drawn to scale: one with radius r and one with radius 2r. Doubling the radius multiplies the surface area by 4 and the volume by 8. r 2r radius r radius × 2 surface × 4 volume × 8
Formula

Formulas (on the GED formula sheet)

V = \tfrac{4}{3}\pi r^3 \qquad SA = 4\pi r^2
  • r = radius (half the diameter), \pi \approx 3.14 (or the π key).
  • Volume uses r^3 → cubic units. Surface area uses r^2 → square units.

Both sphere formulas are on the GED formula sheet. The hemisphere versions (\tfrac{2}{3}\pi r^3, 2\pi r^2) and d = 2r are not — build them from the sphere formulas by halving.

Worked example

Example 1 (easy): volume of a small sphere

Find the volume of a sphere with radius 2 cm. Round to the nearest tenth.

  1. Cube the radius: 2^3 = 8.
  2. V = \tfrac{4}{3}\pi(8) = \tfrac{32}{3}\pi.
  3. \tfrac{32}{3} \approx 10.667, and 10.667 \times \pi \approx 33.51.
Example 1: sphere with radius 2 cm sitting inside a cube 4 cm on each side (the box it just fits in). Sphere volume four-thirds pi times 8, about 33.5 cubic cm; the cube holds 64 cubic cm. r = 2 cm 4 cm cube: 4³ = 64 cm³ sphere: ⁴⁄₃π(2³) ≈ 33.5 cm³ a little over half

Answer: V \approx 33.5\text{ cm}^3

Check: a cube 4 cm on each side (the box the sphere fits in) holds 64 cm³; 33.5 is a little over half ✓.

Worked example

Example 2: surface area of a soccer ball

A soccer ball has a diameter of 22 cm. About how much leather covers it?

  1. "Covers" → surface area. Radius: 22 \div 2 = 11 cm.
  2. SA = 4\pi r^2 = 4\pi(121) = 484\pi.
  3. 484\pi \approx 1{,}520.5.
Example 2: soccer ball with diameter 22 cm. The radius is half of that, 11 cm. Leather covering = surface area 4 pi times 11 squared = 484 pi, about 1,520.5 square cm. r = 11 d = 22 cm "covers" → surface area r = 22 ÷ 2 = 11 cm SA = 4π(11²) = 484π ≈ 1,520.5 cm²

Answer: about 1,520.5 cm²

Check: with \pi \approx 3: 4 \times 3 \times 121 = 1{,}452; our answer is a bit more ✓.

Worked example

Example 3 (real world): gallons in a spherical tank

A spherical water tank has a radius of 5 ft. One cubic foot holds about 7.48 gallons. About how many gallons does the tank hold?

  1. Volume: V = \tfrac{4}{3}\pi(5^3) = \tfrac{4}{3}\pi(125) = \tfrac{500}{3}\pi \approx 523.6\text{ ft}^3.
  2. Convert: 523.6 \times 7.48 \approx 3{,}916.5 gallons.
Example 3: spherical water tank with radius 5 feet on a stand. Volume four-thirds pi times 125, about 523.6 cubic feet; times 7.48 gallons per cubic foot is about 3,917 gallons. r = 5 ft V = ⁴⁄₃π(5³) ≈ 523.6 ft³ × 7.48 gal per ft³ ≈ 3,917 gallons

Answer: about 3,917 gallons

Check: units: \text{ft}^3 \times \dfrac{\text{gal}}{\text{ft}^3} = \text{gal} ✓.

Worked example

Example 4 (work backwards): diameter from volume

A sphere has a volume of about 4,188.8 in³. Find its diameter.

  1. r^3 = \dfrac{3V}{4\pi} = \dfrac{3(4{,}188.8)}{4\pi} = \dfrac{12{,}566.4}{12.566} \approx 1{,}000.
  2. Cube root: r = \sqrt[3]{1{,}000} = 10 in.
  3. Diameter: d = 2 \times 10 = 20 in.
Example 4: a sphere with volume about 4,188.8 cubic inches; the radius and diameter are unknown. Working backwards, r cubed is about 1,000, so r = 10 inches and d = 20 inches. r = ? d = ? V ≈ 4,188.8 in³ r³ = 3V ÷ 4π ≈ 1,000 r = ∛1,000 = 10 in d = 2 × 10 = 20 in

Answer: d = 20 in

Check: \tfrac{4}{3}\pi(1{,}000) \approx 4{,}188.8 ✓.

Worked example

Example 5 (GED-hard): surface area from volume

A ball bearing has a volume of 36\pi mm³. What is its surface area?

  1. Find the radius: \tfrac{4}{3}\pi r^3 = 36\pi → \tfrac{4}{3}r^3 = 36 → r^3 = 36 \times \tfrac{3}{4} = 27 → r = 3 mm.
  2. Surface area: SA = 4\pi(3^2) = 4\pi(9) = 36\pi.
Example 5: ball bearing with volume 36 pi cubic mm and unknown radius. Four-thirds r cubed = 36 gives r cubed = 27, so r = 3 mm, and the surface area is 4 pi times 9 = 36 pi square mm. r = ? V = 36π mm³ ⁴⁄₃r³ = 36 r³ = 27 → r = 3 mm SA = 4π(3²) = 36π ≈ 113.1 mm²

Answer: 36\pi \approx 113.1\text{ mm}^2

Check: volume with r = 3: \tfrac{4}{3}\pi(27) = 36\pi ✓. (Only r = 3 makes the two numbers equal — a fun coincidence, not a rule.)

Calculator tip

On the TI-30XS MultiView

  • Cubes: type the radius, then ^ 3. For \tfrac{4}{3}\pi(5^3): 4 n/d 3 ▶ × π × 5 ^ 3 enter → 523.598… (if the screen shows the answer in terms of \pi, press toggle for the decimal).
  • Or skip the fraction key: 4 × π × 5 ^ 3 ▶ ÷ 3 enter — same result.
  • Squares: use x²: 4 × π × 8 x² enter → 256\pi; toggle → 804.247…
  • Cube roots: type the index 3, then 2nd ^ (the \sqrt[n]{\ } key), then the number. 3 2nd ^ 729 enter → 9.
  • Square roots: 2nd x² then the number.
  • Exact answers: if the choices are "in terms of \pi", do the arithmetic without the π key. If the result shows as a fraction like \tfrac{128}{3}, press toggle to switch to a decimal.
  • Use parentheses for \dfrac{3V}{4\pi}: 3 × 523.6 ÷ ( 4 π ).
Common mistake

Common mistakes

  • Using the diameter as the radius. d = 10 → r = 5. Using 10 makes surface area 4 times too big and volume 8 times too big.
  • Squaring instead of cubing (or the reverse). Volume: r^3 (3^3 = 27). Surface area: r^2 (3^2 = 9).
  • Dropping part of \tfrac{4}{3}. \tfrac{4}{3}(27) = 36. Multiplying by 4 only gives 108; ignoring \tfrac{4}{3} gives 27.
  • Mixing up volume and surface area. "Paint" or "cover" is area; "fill" or "hold" is volume. Check the units in the answer choices.
  • Forgetting to halve for a hemisphere, or forgetting the flat circle when the half-ball is solid.
  • Stopping too early when working backwards. r^3 = 125 is not the answer; r = 5 is. Then check: radius or diameter?
Test strategy

GED test strategy

  • Underline "radius" or "diameter" in the question and in the figure. Then underline what is asked for at the end.
  • Match units to the formula: cubic units → \tfrac{4}{3}\pi r^3; square units → 4\pi r^2. Wrong-unit choices are usually traps.
  • Spot the factor-of-8 trap. If two answer choices differ by a factor of 8 (or 4 for area), one probably used the diameter.
  • Plug answers back in on work-backwards questions: cube each choice for the radius and see which gives the volume.
  • Estimate with \pi \approx 3. \tfrac{4}{3}\pi \approx 4, so a sphere's volume is about 4r^3. For r = 5: about 500 (exact: 523.6).
  • Round only at the end, and round to the place the question asks for. Round down when the question asks how many complete items fit or can be filled.

Check yourself

Spheres

What is the volume of the sphere shown, in terms of \pi?

Sphere with a radius of 6 cm drawn from the center to the surface. 6 cm
Answer choices
Hint

The sphere volume formula on the formula sheet is V = \tfrac{4}{3}\pi r^3. Cube the radius first.

Check yourself

SpheresFill in

A ball has the diameter shown.

Ball (sphere) with a diameter of 10 in drawn straight through the center. 10 in

What is the surface area of the ball, rounded to the nearest square inch?

in²

Type your answer in the box. Fractions like 3/4 are fine.

Hint

The figure shows the diameter. The formula SA = 4\pi r^2 needs the radius.

Summary

  • Sphere volume: V = \tfrac{4}{3}\pi r^3 (cubic units). Sphere surface area: SA = 4\pi r^2 (square units). Both are on the formula sheet.
  • Always use the radius: r = \tfrac{d}{2}.
  • Hemisphere: volume \tfrac{2}{3}\pi r^3; curved surface 2\pi r^2 (add \pi r^2 if the flat side is included).
  • Working backwards: surface area → square root; volume → cube root. Then check whether the question wants r or d.
  • Radius ×k → surface area ×k^2 → volume ×k^3.
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