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Mensuration (3D)

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medium importance~2 Q in Tier 120 formulas⚡ 10 shortcuts5 subtopics

Sphere and hemisphere

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  • Sphere (radius rr): volume =43πr3=\frac43\pi r^3, surface =4πr2=4\pi r^2.
  • Hemisphere: volume =23πr3=\frac23\pi r^3; curved surface =2πr2=2\pi r^2; total surface =3πr2=3\pi r^2 (adds the flat circular face).
  • Melting and recasting conserves volume (surface area almost never matches — that option is a trap).

Detailed notes

The sphere family

For a sphere of radius rr: V=43πr3,SA=4πr2V = \frac{4}{3}\pi r^3,\qquad \text{SA} = 4\pi r^2 For a hemisphere of the same radius: curved surface =2πr2= 2\pi r^2 (half the sphere's), flat face =πr2= \pi r^2, and total =3πr2= 3\pi r^2 — the famous "3" that examiners love. Volume of a hemisphere =23πr3= \frac{2}{3}\pi r^3. The radius is the only input: every question is "get rr, then apply one formula". Given a diameter? Halve it first. Given the surface area? r=SA/4πr = \sqrt{\text{SA}/4\pi}. Given the volume? r=3V4π3r = \sqrt[3]{\frac{3V}{4\pi}}.

Melting, scaling and submersion

  • Melting preserves volume: a big sphere recast into nn small equal spheres gives rsmall=rn3r_{\text{small}} = \frac{r}{\sqrt[3]{n}} — for n=8n = 8 the radius halves, for n=27n = 27 it divides by 3.
  • Scaling: radius ×k\times k → surface ×k2\times k^2, volume ×k3\times k^3. "Radius doubled" means SA ×4\times 4 and V×8V \times 8; "radius halved" gives 14\frac14 and 18\frac18.
  • Submersion: a solid dropped into a (partly filled) cylinder of base radius RR raises the water level by h=VsolidπR2h = \frac{V_{\text{solid}}}{\pi R^2} — the displaced volume spreads over the base.

Same-base relations

Sphere vs cylinder of radius rr, height 2r2r (the sphere fits exactly inside): both have volume 43πr3\frac{4}{3}\pi r^3. Sphere vs hemisphere: hemisphere is half the volume, 34\frac34 of the total surface (3πr23\pi r^2 vs 4πr24\pi r^2).

The 22/7 number family

  • r=7r = 7: SA =616= 616, V=43×227×343=4312/3V = \frac{4}{3} \times \frac{22}{7} \times 343 = 4312/3 — exams avoid it; they use r=21r = 21: r3=9261r^3 = 9261, V=4×22×441=38808V = 4 \times 22 \times 441 = 38808, SA =4×227×441=5544= 4 \times \frac{22}{7} \times 441 = 5544.
  • Hemisphere at r=7r = 7: CSA 308308, TSA 462462; at r=21r = 21: CSA 27722772, TSA 41584158.
  • The displacement pair worth memorising: a sphere of radius 3 displaces 36π36\pi; a cylinder of radius 4 therefore rises 3616=2.25\frac{36}{16} = 2.25 cm.

Scaling in percent form

Radius +10%+10\%: SA +21%+21\%, VV +33.1%+33.1\% (from 1.121.1^2 and 1.131.1^3). Radius −50%-50\%: SA 14\frac14, VV 18\frac18 — volume swings harder than area in both directions. Percent-change questions in this subtopic are always these two exponent rules; if you find yourself expanding 43π\frac{4}{3}\pi, stop and write k2k^2 or k3k^3 instead.

Quick revision

  • V=43πr3V = \frac43\pi r^3; SA =4πr2= 4\pi r^2; hemisphere: CSA 2πr22\pi r^2, TSA 3πr23\pi r^2, V=23πr3V = \frac23\pi r^3.
  • Radius ×k\times k: SA ×k2\times k^2, V×k3V \times k^3.
  • Melt into nn equal spheres: radius ÷n3\div \sqrt[3]{n}.
  • Water rise: h=VsolidπR2h = \frac{V_{\text{solid}}}{\pi R^2} of the container.

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Surface area / volume of a spherevery common2 practice Q
How to spot it:

Radius (or diameter) given; SA or volume asked — or one of them given and rr asked.

SA=4πr2,V=43πr3SA = 4\pi r^2,\qquad V = \frac{4}{3}\pi r^3
  1. Halve diameters first; then rr drives both formulas.
  2. With π=22/7\pi = 22/7, the 7-cube of r=21r = 21 gives the classic V=38808V = 38808.
  3. Reverse: r=SA/4πr = \sqrt{\text{SA}/4\pi} or r=3V/4π3r = \sqrt[3]{3V/4\pi}.

Why: the sphere has a single parameter, so forward and reverse questions differ only by which side you solve for.

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

SA =4×227×49=616= 4 \times \frac{22}{7} \times 49 = 616 sq cm.

Type 2: Hemisphere: curved vs total vs volumevery common2 practice Q
How to spot it:

A hemisphere (dome, bowl) — the curved surface, total surface, or volume asked; the TSA = 3pir^2 fact is the trap-setter.

CSA=2πr2,TSA=3πr2,V=23πr3\text{CSA} = 2\pi r^2,\quad \text{TSA} = 3\pi r^2,\quad V = \frac{2}{3}\pi r^3
  1. Dome/tent/scoop of a sphere → hemisphere.
  2. Painted outside only → CSA 2πr22\pi r^2; a solid hemisphere's whole surface → TSA 3πr23\pi r^2; capacity → 23πr3\frac23\pi r^3.
  3. Keep the three straight; the difference is exactly one πr2\pi r^2.

Why: half a sphere's skin plus one circular lid — the arithmetic is deliberately easy once the pieces are named.

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

TSA =3×227×49=462= 3 \times \frac{22}{7} \times 49 = 462 sq cm.

Type 3: Recasting and water displacementcommon2 practice Q
How to spot it:

A sphere melted into smaller spheres — or dropped into a cylindrical vessel, where the water rise is asked.

Vtotal=n Vsmall;hrise=VsphereπR2V_{\text{total}} = n\,V_{\text{small}};\quad h_{\text{rise}} = \frac{V_{\text{sphere}}}{\pi R^2}
  1. Melting: equate volumes; for nn equal spheres divide the radius by n3\sqrt[3]{n}.
  2. Displacement: the rise times the container's base area equals the sphere's volume.
  3. Watch which radius is which — the sphere's rr vs the vessel's RR.

Why: displaced water is just the volume wearing a different shape; recasting is the same statement in metal.

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

43π63=8⋅43πr3⇒r3=27⇒r=3\frac{4}{3}\pi 6^3 = 8 \cdot \frac43 \pi r^3 \Rightarrow r^3 = 27 \Rightarrow r = 3 cm.

Type 4: Radius-scaling multiplierscommon2 practice Q
How to spot it:

'Radius doubled/halved/increased 50%' — how do surface area and volume change?

SA×k2,V×k3\text{SA} \times k^2,\qquad V \times k^3
  1. Write the multiplier: r→krr \to kr.
  2. SA scales k2k^2; volume scales k3k^3 — never the same number.
  3. Percent form: radius +50%+50\% → k=1.5k = 1.5 → SA ×2.25\times 2.25, V×3.375V \times 3.375.

Why: area is a product of two lengths, volume of three — the exponents do all the work.

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

k3=8k^3 = 8 times the original volume.

Formulas

Sphere
V=43πr3,S=4πr2V=\frac43\pi r^3,\quad S=4\pi r^2
Hemisphere
V=23πr3,Scurved=2πr2,Stotal=3πr2V=\frac23\pi r^3,\quad S_{\text{curved}}=2\pi r^2,\quad S_{\text{total}}=3\pi r^2
Melting rule
Vbefore=VafterV_{\text{before}}=V_{\text{after}}
Count of small spheres
count=(Rr)3\text{count}=\left(\frac{R}{r}\right)^3

Shortcut tricks

⚡ Count small balls by the cube of the radius ratio

Number of small spheres =(Rr)3=\left(\frac{R}{r}\right)^3. Radii in ratio 8:1 → 512 pieces, no π needed.

Example: A solid metallic sphere of radius 8 cm is melted and drawn into small spheres of radius 1 cm. How many are formed?

(81)3=512\left(\frac81\right)^3=512.

⚡ Hemisphere TSA is three circles

3πr23\pi r^2: two for the curved half (recall the sphere is 4πr24\pi r^2) plus one for the base.

Example: Find the total surface area of a solid hemisphere of radius 7 cm. [Use π=227\pi=\frac{22}{7}]

3×227×49=4623\times\frac{22}{7}\times49=462 sq cm.

Where students lose marks

  • Hemisphere TSA taken as 2πr22\pi r^2 (that is the curved part only).

  • Melting questions answered with equality of surface areas.

  • 43πr3\frac43\pi r^3 misremembered as 4πr33 or 2\frac{4\pi r^3}{3\text{ or }2} variants — write the fraction before substituting.

Practice sets — 12 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 12 questions

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