Mensuration: volumes and surface areas

In short

This topic deals with the space a three-dimensional solid occupies (volume) and the area of the surfaces that enclose it (curved, lateral and total surface area). The solids examined are the cuboid and cube, right prisms, right circular cylinders, cones and pyramids, frustums, spheres, hemispheres and shells, plus solids built by combining, melting or cutting these. It carries guaranteed marks in university engineering mathematics and in every aptitude test - CAT, XAT, SSC CGL, IBPS/SBI PO, RRB, GATE Civil earthwork questions and the quantitative round of TCS NQT, Infosys, Wipro, Capgemini and Accenture. On the job the same formulas price concrete and earthwork, size water tanks and pressure vessels, and estimate paint, sheet metal and packaging.

Explained

Units, basic ideas and the three "areas"

• Volume is measured in cubic units - m³, cm³, mm³, litres. Key conversions: 1 m³ = 1,000 litres = 10⁶ cm³; 1 litre = 1,000 cm³ = 1,000 ml; 1 cm³ = 1 ml. In Indian construction a brass = 100 cubic feet ≈ 2.832 m³ for sand and aggregate.
• Curved surface area (CSA) - only the curved part (cylinder, cone, sphere). Lateral surface area (LSA) - all faces except the top and bottom (cuboid, prism, pyramid). Total surface area (TSA) - everything that can be touched from outside, including bases.
• Read the question carefully: *painting a room* = four walls + ceiling; *an open tank* = walls + base, no top; *a closed box* = full TSA; *a pipe* = both curved surfaces plus two annular ends only if the ends are exposed.
• Mass = volume × density. Iron ≈ 7.8 g/cm³, concrete ≈ 2,400 kg/m³, water = 1 g/cm³ (1 kg/litre).

The prism family - volume = base area × height

• A right prism has two parallel congruent polygonal bases joined by rectangular faces. V = (area of base) × h; LSA = (perimeter of base) × h; TSA = LSA + 2 × base area.
• A cuboid is a rectangular prism; a cube is a cuboid with l = b = h. Their space diagonal is the longest straight rod that fits inside - the classic "longest pole in a room" question.
• A right circular cylinder is a prism with a circular base: V = πr²h, CSA = 2πrh, TSA = 2πr(r + h). A hollow cylinder (pipe) of outer radius R, inner radius r: V = πh(R² − r²); thickness t = R − r.
• Rotating a rectangle of sides a and b about side b generates a cylinder of radius a and height b - so the same rectangle gives two different cylinders depending on the axis.

The pyramid and cone family - volume = (1/3) × base area × height

• A pyramid has a polygonal base and triangular faces meeting at an apex. V = (1/3) × base area × h; LSA = (1/2) × base perimeter × slant height of a face.
• For a square pyramid of base side a and vertical height h: slant height of a face = √(h² + (a/2)²); lateral (sloping) edge = √(h² + a²/2). Keep the two distinct.
• A right circular cone is a pyramid with a circular base: V = (1/3)πr²h, slant height l = √(r² + h²), CSA = πrl, TSA = πr(r + l). A right triangle rotated about a leg generates a cone.
• Why the factor 1/3: a cone (or pyramid) of the same base and height as a cylinder (or prism) holds exactly one-third of it - demonstrable by filling, and derivable by integrating πr(x)²dx with r(x) = rx/h.
• A frustum is what remains after the top of a cone or pyramid is cut off by a plane parallel to the base - a bucket, a lampshade, a dam block. For a conical frustum of radii R and r and vertical height h: V = (πh/3)(R² + Rr + r²), slant l = √(h² + (R − r)²), CSA = π(R + r)l.
• Similar-cone property: cutting a cone at height h/2 gives a small cone with half the radius, so 1/8 of the volume; the frustum keeps 7/8.

Spheres, hemispheres and shells

• Sphere: V = (4/3)πr³, surface area = 4πr² (there is no "flat" face, so CSA = TSA).
• Hemisphere: V = (2/3)πr³, CSA = 2πr², TSA = 3πr² (curved part plus the flat circular face). A hemispherical bowl's capacity uses the inner radius; the metal in it uses R³ − r³.
• Spherical shell of outer radius R and inner radius r: V = (4/3)π(R³ − r³).
• Archimedes' results: a sphere inscribed in a cylinder of the same radius and height 2r has exactly 2/3 of its volume, and exactly the same curved surface area (4πr²). With a cone of the same r and h = 2r, cone : sphere : cylinder = 1 : 2 : 3.
• Spherical cap of height h cut from a sphere of radius r: V = (πh²/3)(3r − h), curved area = 2πrh (asked in higher-level papers only).

Combination, conversion and cutting problems - the exam favourites

1. Combination solids: a toy (cone + hemisphere), a capsule (cylinder + 2 hemispheres), a tent (cylinder + cone), a water tank (cylinder + hemispherical bottom). Add volumes; add only the exposed surfaces - the circle where two parts join belongs to neither total surface.
2. Melting / recasting: metal is conserved, so volume before = volume after. Surface area is *not* conserved - breaking one solid into many small ones increases total surface sharply (why atomisation needs more coating, and why small ice cubes melt faster).
3. Number of small solids from a big one = (volume of big) ÷ (volume of one small), only when the material is melted. If pieces are merely *packed* inside, the answer is a packing/fitting count using dimensions, not volumes.
4. Digging and spreading: earth from a well or pit is spread as an embankment (annular ring) or over a rectangular field; equate volumes to find the rise in level.
5. Water flow: volume delivered = (cross-section area) × (speed) × (time). Keep speed and dimensions in the same unit system before multiplying.
6. Immersion: a solid dropped into a liquid raises the level by its own volume (if fully submerged); water displaced = volume of the solid.
7. Cutting: a cube of side a cut into unit cubes gives a³ pieces and the surface area multiplies by a; a cylinder cut along its axis exposes two new rectangles of area 2rh each.

Scaling, ratios and optimisation

• If every linear dimension is multiplied by k, then lengths ×k, areas ×k², volumes ×k³. A 20% rise in radius raises surface area by 44% and volume by 72.8%.
• Hence equal-volume solids are compared by k = (ratio of volumes)^1/3, and equal-surface solids by k = (ratio of areas)^1/2.
• For a fixed volume, the sphere has the least surface area - which is why droplets, pressure vessels and LPG storage bullets are curved; among cuboids of fixed volume the cube is the most economical.
• For a closed cylinder of given volume, material is least when h = 2r (height equals diameter); for an open-top cylinder, when h = r.
• Largest solid inside another: sphere in a cube of side a → r = a/2; cube in a sphere of radius R → side = 2R/√3; cone in a cylinder of same r, h → 1/3 of its volume.

Polyhedra, standards and practice in India

• Euler's formula for any simple polyhedron: F + V = E + 2 (cube: 6 + 8 = 12 + 2). There are exactly five regular (Platonic) solids - tetrahedron, cube, octahedron, dodecahedron, icosahedron.
• Prismoidal (Simpson's) formula: V = (h/6)(A₁ + 4A_m + A₂), where A_m is the mid-section area. It is exact when the cross-sectional area varies quadratically, and it reduces to the cone, frustum, sphere and prism formulas as special cases.
• In surveying and estimation, the average-end-area (trapezoidal) method V = h(A₁ + A₂)/2 is simpler but less accurate; the prismoidal formula is the standard correction for earthwork quantities.
• IS 1200 (Methods of Measurement of Building and Civil Engineering Works) fixes how such volumes are measured for billing - Part I earthwork, Part II concrete; CPWD/State PWD rate schedules (DSR) then price them per m³. Rates and code editions are revised periodically, so always quote the current schedule.
• Consumer products are sold by these volumes: water tanks in 500/1,000/2,000 litres (BIS-marked), domestic LPG cylinders of 14.2 kg, tankers in kilolitres, cement bags of 50 kg ≈ 0.0347 m³ loose.

Key points

• Volume - space occupied, cubic units; surface area - enclosing area, square units; never mix the two in one answer.
• 1 m³ = 1,000 litres and 1 cm³ = 1 ml - the single most used conversion in tank and capacity questions.
• Prism rule - volume = base area × height for cuboid, cube, cylinder and any right prism.
• Cone/pyramid rule - volume = one-third of the prism on the same base and height.
• Slant height - cone: l = √(r² + h²); frustum: l = √(h² + (R − r)²); it is never the vertical height.
• Sphere - V = (4/3)πr³, SA = 4πr²; hemisphere TSA = 3πr², not 2πr².
• Frustum volume = (πh/3)(R² + Rr + r²) - the middle term Rr is the one most often dropped.
• Cone : sphere : cylinder = 1 : 2 : 3 for the same radius with cylinder/cone height = 2r.
• Recasting - volume is conserved, surface area is not; splitting a solid increases total surface.
• Scaling - ×k in length means ×k² in area and ×k³ in volume.
• Least surface for a given volume - the sphere; for a closed cylinder, h = 2r.
• Cuboid diagonal = √(l² + b² + h²); cube diagonal = a√3; face diagonal = a√2.
• Hollow solids - pipe V = πh(R² − r²); shell V = (4/3)π(R³ − r³); capacity uses inner radius, material uses the difference.
• Open vs closed - subtract πr² (cylinder/cone) or lb (box) when a lid or base is absent.
• Flow problems - volume = area × velocity × time, after unit matching.
• Euler - F + V = E + 2; exactly five Platonic solids.
• Prismoidal formula V = (h/6)(A₁ + 4A_m + A₂) is the general volume rule used for earthwork.
• π choice - use 22/7 when the radius is a multiple of 7 or 3.5, otherwise 3.1416.

Formulas and examples

Example 1 - open tank: capacity, painting and filling time. A rectangular tank is 2.5 m × 1.6 m × 1.2 m deep, open at the top.

• Capacity = 2.5 × 1.6 × 1.2 = 4.8 m³ = 4,800 litres.
• Surface to be painted inside (4 walls + base) = 2 × 1.2 × (2.5 + 1.6) + 2.5 × 1.6 = 9.84 + 4.00 = 13.84 m². At ₹35/m², cost = ₹484.40.
• A pipe of internal diameter 4 cm (r = 0.02 m) delivers water at 2 m/s. Rate = π × 0.02² × 2 = 0.0008π ≈ 0.0025133 m³/s. Time = 4.8 ÷ 0.0025133 ≈ 1,910 s ≈ 31.8 minutes.

Example 2 - recasting, and what happens to surface area. A solid metal cylinder of radius 6 cm and height 32 cm is melted into solid spheres of radius 3 cm.

• Cylinder volume = π × 6² × 32 = 1,152π cm³. One sphere = (4/3)π × 3³ = 36π cm³.
• Number of spheres = 1,152π ÷ 36π = 32.
• Cylinder TSA = 2π × 6 × (6 + 32) = 456π ≈ 1,432.6 cm². The 32 spheres together = 32 × 4π × 9 = 1,152π ≈ 3,619.1 cm² - about 2.53 times the original surface, although the metal is unchanged. Hence plating or painting cost jumps even though material cost does not.

Example 3 - frustum bucket. A bucket is a frustum with top radius 20 cm, bottom radius 10 cm and vertical depth 24 cm.

• V = (π × 24/3)(20² + 20 × 10 + 10²) = 8π × 700 = 5,600π ≈ 17,593 cm³ ≈ 17.59 litres.
• Slant height l = √(24² + (20 − 10)²) = √(576 + 100) = √676 = 26 cm.
• Sheet needed (curved surface + base) = π(20 + 10) × 26 + π × 10² = 780π + 100π = 880π ≈ 2,764.6 cm² = 0.2765 m². At ₹200/m², metal cost ≈ ₹55.30.

Example 4 - combination solid (cone on a hemisphere). A wooden toy is a cone mounted on a hemisphere of radius 3.5 cm; total height is 15.5 cm. Take π = 22/7.

• Cone height = 15.5 − 3.5 = 12 cm; slant l = √(12² + 3.5²) = √156.25 = 12.5 cm.
• Volume = (1/3)πr²(h + 2r) = (1/3) × (22/7) × 12.25 × 19 = 243.83 cm³.
• Total surface = πrl + 2πr² = πr(l + 2r) = (22/7) × 3.5 × (12.5 + 7) = 11 × 19.5 = 214.5 cm² (the joining circle is not counted).
• Packed in a cylindrical box of radius 3.5 cm and height 15.5 cm: box volume = (22/7) × 12.25 × 15.5 = 596.75 cm³, so empty space = 596.75 − 243.83 = 352.92 cm³.

Example 5 - well and embankment (earthwork). A well of diameter 3 m is dug 14 m deep and the earth is spread evenly to form a 4 m wide embankment around it. Take π = 22/7.

• Earth excavated = π × 1.5² × 14 = (22/7) × 2.25 × 14 = 99 m³.
• Embankment is an annulus with inner radius 1.5 m and outer radius 5.5 m: area = π(5.5² − 1.5²) = (22/7) × 28 = 88 m².
• Height of embankment = 99 ÷ 88 = 1.125 m (112.5 cm).

Where it is used

• Civil engineering and estimation: concrete, brickwork, plaster, excavation and embankment quantities in m³ as per IS 1200, priced from CPWD/PWD rate schedules; cut-and-fill volumes on highway and canal projects via the prismoidal formula.
• Water and sanitation: sizing overhead and underground tanks in litres, septic tanks, sumps, water towers, reservoir capacity and hydraulic flow in pipes (Jal Jeevan Mission type works).
• Manufacturing and process plants: pressure vessels, cylindrical and spherical storage (LPG bullets and Horton spheres at refineries), silos, hoppers (conical frustums), tanker and drum capacity, metal weight of castings and pipes.
• Packaging, FMCG and sheet metal: carton and bottle design, filling volumes printed on labels, sheet-metal blank area for cans and buckets, paint and coating coverage per m².
• Job roles: civil estimator and quantity surveyor, site/billing engineer, mechanical design and sheet-metal engineer, packaging designer, production planner, and any bank/SSC/PSU written test with quantitative aptitude.
• Everyday commerce: sand and aggregate bought in brass or m³, LPG cylinder volumes, tanker water supply billed in kilolitres.

Interview questions

• Q: Why is the volume of a cone one-third that of a cylinder with the same base and height? A: Because the cross-sectional radius shrinks linearly to zero, so integrating π(rx/h)² dx from 0 to h gives πr²h/3 - the same 1/3 factor applies to every pyramid.
• Q: A sphere and a cube have the same volume - which has more surface area? A: The cube; for a given volume the sphere has the minimum possible surface area, so any other shape has more.
• Q: If a cylinder's radius is halved and its height doubled, what happens to the volume? A: V = πr²h, so the volume becomes (1/4) × 2 = half the original.
• Q: How does the surface area change when one big sphere is melted into many small ones? A: Volume stays the same but surface area increases in proportion to 1/r - n spheres of radius r/n^1/3 give n^1/3 times the original area.
• Q: What is the difference between the slant height and the vertical height of a cone? A: The vertical height is the perpendicular from apex to base centre; the slant height is the distance from apex to the base circumference, l = √(r² + h²), and it is the one used for curved surface area.
• Q: Why is the curved surface area of a sphere equal to that of its circumscribing cylinder? A: Both equal 4πr² - Archimedes showed the radial narrowing of the sphere exactly compensates for its vertical stretch, which is also the basis of equal-area map projections.
• Q: For a closed cylindrical tank of fixed volume, what proportions use the least metal? A: Height equal to diameter (h = 2r); for an open-top cylinder the optimum is h = r.
• Q: A cone is cut halfway up by a plane parallel to the base - how do the two parts compare? A: The small cone is similar with ratio 1/2, so it has 1/8 of the volume; the frustum has the remaining 7/8.
• Q: How many cubes of side 2 cm fit inside a box 9 cm × 7 cm × 5 cm? A: By fitting, 4 × 3 × 2 = 24 cubes - not 315/8, because unmelted solids must be counted dimension-wise, leaving waste space.
• Q: Why do engineers prefer the prismoidal formula over the average-end-area method for earthwork? A: The prismoidal formula uses the mid-section area as well and is exact when area varies quadratically along the length, so it corrects the systematic error of the trapezoidal method.
• Q: A hemispherical bowl has 1 cm thick walls - which radius gives its capacity? A: The inner radius; the volume of material is (2/3)π(R³ − r³), which is a separate calculation.
• Q: Convert 2.5 m³ into litres and explain its practical use. A: 2.5 m³ = 2,500 litres - roughly the capacity of a small domestic sump or a mini water tanker.

Common mistakes

• Using the vertical height instead of the slant height in πrl, or forgetting that for a frustum l = √(h² + (R − r)²), not √(h² + R²).
• Dropping the middle term Rr in the frustum volume, or using the average of the two radii as if it were a cylinder.
• Adding the full surface areas of two joined solids - the hidden interface circle must be excluded from the combination's total surface.
• Assuming surface area is conserved in melting and recasting problems; only volume is.
• Mixing units - cm with m, or litres with m³ - and forgetting that a ×100 length factor is ×10⁶ in volume.
• Giving hemisphere TSA as 2πr² (that is only the curved part) or including a lid/base that the question says is open.
• Dividing volumes to count solid blocks that are merely packed, not melted, which ignores the unusable gaps.

Related topics

Mensuration: areas and perimeters · Geometry and coordinate basics · Squares, cubes, roots and surds · Ratio, proportion and partnership · Percentages · Time, speed and distance (flow and filling rates).