Mensuration (3D)

medium importance~2 Q in Tier 121 formulas⚡ 15 shortcuts5 subtopics

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Sphere and hemisphere

⏱ 3 min read🧩 5 question types🎯 12 practice Q

The idea in one minute

A sphere of radius r has volume four-thirds pi r cubed and surface four pi r squared. A hemisphere is half a sphere for volume, but its total surface is three pi r squared because it adds the flat circle. Radius scaling cubes for volume and squares for surface.

The sphere pair

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

Radius $7$ : surface $4\times154=616$ sq cm; volume $\dfrac43\times\dfrac{22}{7}\times343=\dfrac{4312}{3}\approx1437.3$ cu cm.

Radius $14$ : surface $4\times616=2464$ sq cm; volume $\dfrac43\times\dfrac{22}{7}\times2744=\dfrac{34496}{3}$ cu cm, exactly $8$ times the radius- $7$ volume. Radius $3.5$ : surface $154$ sq cm, volume $\dfrac{539}{3}\approx179.7$ cu cm.

The surface formula is exactly four circles of the same radius. Half of it, $2\pi r^2$ , is the hemisphere's curved part. A sphere also fills exactly two-thirds of the cylinder that tightly contains it.

Rule: Surface grows with $r^2$ , volume with $r^3$ . Doubling the radius gives $4$ times the surface and $8$ times the volume, as the $7\to14$ line above shows.

Hemisphere: three halves and three pi

Volume: $\dfrac{2}{3}\pi r^3$ (half the sphere).
Curved surface: $2\pi r^2$ (half the shell).
Total surface: $3\pi r^2$ because the flat circular face $\pi r^2$ joins the shell.

Radius $7$ : curved $308$ , total $3\times154=462$ , volume $\dfrac{2156}{3}$ cu cm. Radius $21$ : curved $2772$ , total $4158$ , all from the $1386$ base circle.

Watch: 'Total surface of a hemisphere' is $3\pi r^2$ , never $2\pi r^2$ . The flat face counts once it is solid.

Melting and dividing

A sphere of radius $6$ melted into $8$ equal small spheres: each carries $\dfrac{216}{8}=27$ of the cubed length, so each radius is $3$ .

Count multiplies the volume, so the radius shrinks by the cube root of the count. Eight pieces mean half the radius; $27$ pieces mean one-third; counts $1,8,27$ pair with radius factors $1,\dfrac12,\dfrac13$ .

Tip: When a big solid becomes $n$ equal small ones, divide the cubed length by $n$ , then take the cube root.

Displacing water

A solid dropped in a full tub pushes out its own volume. Sphere of radius $3$ in a full tank: overflow $=\dfrac43\pi\times27=36\pi$ cu cm.

In a partly filled cylinder the level rises instead: sphere of radius $7$ into a cylinder of radius $14$ raises the water by $\dfrac{4312/3}{616}=\dfrac73$ cm. Read the story: 'water level rose' means the rise's cylinder volume equals the sphere's volume.

Radius ratios

Radii $3:4$ give volumes $27:64$ and surfaces $9:16$ . Cube the ratio for volumes, square it for surfaces, and say which one the question wants. Equal surfaces force equal radii; equal volumes likewise.

Remember: Ratio questions never need pi. Cancel it first, then compare powers of the ratio.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Surface area and volume of a sphere

How to spot it:

One radius given, surface or volume asked.

Method

Surface: $4\pi r^2$ .
Volume: $\dfrac{4}{3}\pi r^3$ .
Keep pi as $\dfrac{22}{7}$ for radius multiples of $7$ .

Why it works:

Two direct formulas; radius $7$ alone answers many papers since $4 imes154=616$ .

Try this

Find the surface area of a sphere of radius 7 cm (take pi = 22/7).

Show solution

$S=4\times\dfrac{22}{7}\times49$ .
$S=616$ sq cm.

Answer

616 sq cm

Type 2very common2 practice Q

Hemisphere: curved vs total vs volume

How to spot it:

A solid or hollow hemisphere; a surface or volume asked.

Method

Curved surface: $2\pi r^2$ .
Solid total: $3\pi r^2$ .
Volume: $\dfrac23\pi r^3$ .

Why it works:

The only trap is which surface: curved for hollow, total (plus flat face) for solid.

Try this

Find the total surface area of a solid hemisphere of radius 7 cm (take pi = 22/7).

Show solution

Total $=3\pi r^2$ .
$=3\times154=462$ sq cm.

Answer

462 sq cm

Type 3common2 practice Q

Recasting and water displacement

How to spot it:

A sphere melted into smaller ones, or dropped into water.

Method

Write the big sphere's volume.
Divide by the count for equal pieces.
Overflow equals the submerged volume.

Why it works:

Both stories reduce to comparing volumes; the pi cancels on the way.

Try this

A solid metal sphere of radius 6 cm is melted into 8 equal smaller spheres. The radius of each small sphere is:

Show solution

$r^3=\dfrac{216}{8}=27$ .
$r=3$ cm.

Answer

3 cm

Type 4common2 practice Q

Radius-scaling multipliers

How to spot it:

The radius scaled; new volume or surface asked.

Method

Read the scale factor $k$ .
Volume: multiply by $k^3$ .
Surface: multiply by $k^2$ .

Why it works:

The formulas' exponents convert any scale factor directly, no recomputation.

Try this

If the radius of a sphere is doubled, its volume becomes:

Show solution

$k=2$ .
Volume $\times 2^3=8$ times.

Answer

8 times

Type 5common

Two spheres: ratio questions

How to spot it:

Radii in a ratio; volume or surface ratio asked.

Method

Cancel $\pi$ and constants.
Volumes: cube the radius ratio.
Surfaces: square it.

Why it works:

With pi cancelled, a ratio question is a pure power of the given ratio.

Try this

The radii of two spheres are in the ratio 3 : 4. The ratio of their surface areas is:

Show solution

Square the ratio: $3^2:4^2$ .
$9:16$ .

Answer

9 : 16

Formula sheet

Sphere

V=\frac{4}{3}\pi r^3,\quad S=4\pi r^2

One radius drives both.

Hemisphere

V=\frac{2}{3}\pi r^3,\quad \text{curved}=2\pi r^2,\quad \text{total}=3\pi r^2

Total adds the flat circle.

Melting into n parts

r_{\text{small}}^3=\frac{R^3}{n}

Divide the cubed length, then cube-root.

Radius scaling

V\to k^3V,\quad S\to k^2S

k = radius scale factor.

Shortcuts that save time

⚡ Total hemisphere = 3 circles

Curved shell 2 pi r squared plus flat face pi r squared equals 3 pi r squared. Never 2.

Example

Find the total surface area of a solid hemisphere of radius 7 cm (take pi = 22/7).

Show solution

Total $=3\pi r^2$ .
$=3\times\dfrac{22}{7}\times49$ .
$=462$ sq cm.

Answer

462 sq cm

⚡ Cube-root the split

One sphere into 8 equal spheres: each cubed radius is one-eighth, so each radius is half.

Example

A solid metal sphere of radius 6 cm is melted into 8 equal smaller spheres. The radius of each small sphere is:

Show solution

$r^3=\dfrac{6^3}{8}=\dfrac{216}{8}=27$ .
$r=\sqrt[3]{27}$ .
$r=3$ cm.

Answer

3 cm

⚡ Ratio: cube it or square it

Radii 3:4 mean volumes 27:64 and surfaces 9:16. Cancel pi, apply the right power.

Example

The radii of two spheres are in the ratio 3 : 4. The ratio of their volumes is:

Show solution

Cube the ratio: $3^3:4^3$ .
$27:64$ .

Answer

27 : 64

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Total hemisphere surface taken as $2\pi r^2$ .

That is the curved part only. Solid hemisphere total is $3\pi r^2=462$ for $r=7$ .

Mistake 02

Squaring instead of cubing for volume ratios.

Volume goes with $r^3$ : radii $3:4$ give $27:64$ .

Mistake 03

Forgetting $\dfrac43$ when melting.

Sphere volume is $\dfrac43\pi r^3$ ; drop it and the sphere comes out too big.

Mistake 04

Dividing the radius by $n$ for $n$ small spheres.

Divide the cubed radius: $n=8$ halves the radius, not eighth-ing it.

Mistake 05

Doubling radius thought to double volume.

Volume scales by $2^3=8$ ; surface by $2^2=4$ .

Quick revision

Read this the night before the exam.

Sphere: $V=\dfrac43\pi r^3$ , $S=4\pi r^2$ .
Hemisphere: volume $\dfrac23\pi r^3$ , curved $2\pi r^2$ , total $3\pi r^2$ .
$n$ equal pieces: divide $r^3$ by $n$ , then cube-root.
Overflow $=$ submerged volume; no new formula.
Ratios: volumes cube, surfaces square; cancel $\pi$ first.

Practice: 12 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 12 questions

Suggested time 7 min · wrong answers go to your mistake notebook automatically.

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