Mensuration • Topic 1 of 3

Surface Area and Volume of Combination of Solids

What is a combination of solids? A combination of solids is a three-dimensional object formed by joining two or more basic solids. Very few real objects are pure shapes. An ice-cream cone is a cone topped by a hemisphere; a medicine capsule is a cylinder with a hemisphere on each end; a circus tent is often a cylinder capped by a cone; a childs spinning top is a cone fixed to a hemisphere. To handle such objects we break them into the standard solids we already know and recombine the relevant pieces.

The two golden rules. Surface area and volume behave very differently when solids are joined, and confusing them is the single most common board-exam mistake.

  • Volume adds, always. Space inside two joined solids is simply the sum of the two volumes. Nothing is lost at the joint, because the joint is just a shared internal boundary: $V_{\text{total}}=V_1+V_2$.
  • Surface area counts only what is exposed. When two solids are stuck together the faces that meet are sealed inside and vanish from view. So you add only the outer visible surfaces, never the full TSAs of both parts.

Standard formulas you must know cold.

SolidCSA / LSATSAVolume
Cube (side $a$)$4a^2$$6a^2$$a^3$
Cuboid ($l,b,h$)$2h(l+b)$$2(lb+bh+hl)$$lbh$
Cylinder ($r,h$)$2\pi rh$$2\pi r(r+h)$$\pi r^2 h$
Cone ($r,h,l$)$\pi r l$$\pi r(r+l)$$\tfrac13\pi r^2 h$
Sphere ($r$)$4\pi r^2$$4\pi r^2$$\tfrac43\pi r^3$
Hemisphere ($r$)$2\pi r^2$$3\pi r^2$$\tfrac23\pi r^3$

The cone always needs its slant height $l=\sqrt{r^2+h^2}$ for any surface-area work, while volume needs the vertical height $h$. Keep the two straight: never feed a slant height into a volume formula.

How joints change the surface area — worked reasoning.

  • Cone on a hemisphere (a toy). The flat base of the cone sits exactly on the flat face of the hemisphere; both circular faces are hidden. Exposed surface $=$ CSA of cone $+$ CSA of hemisphere $=\pi r l + 2\pi r^2$.
  • Cylinder with a hemisphere on each end (a capsule). Two flat ends of the cylinder are covered. Exposed surface $=$ CSA of cylinder $+ 2\times$ CSA of hemisphere $=2\pi r h + 2(2\pi r^2)=2\pi r h + 4\pi r^2$.
  • Cylinder topped by a cone (a tent / a rocket). The top face of the cylinder is sealed by the cone base and the cone base is itself hidden. Exposed surface $=$ CSA of cylinder $+$ CSA of cone $+$ the bottom circle if the tent has a floor (usually it does not). For canvas of a tent: $2\pi r h + \pi r l$.
  • Hemisphere scooped out of a cube / cuboid. A hemispherical cavity removes the circle $\pi r^2$ from one face but adds the inner curved surface $2\pi r^2$. Net surface $=6a^2-\pi r^2+2\pi r^2=6a^2+\pi r^2$.
  • Two cubes joined. Joining two cubes of side $a$ into a cuboid hides one face of each cube ($2a^2$ in total): new TSA $=2(6a^2)-2a^2=10a^2$, not $12a^2$.

Capacity vs. volume. For containers (tanks, vessels, bottles) the volume in cubic centimetres can be converted to capacity: $1000\text{ cm}^3 = 1$ litre, and $1\text{ m}^3 = 1000$ litres. Always state the final answer in the unit the question asks for.

Common mistakes to avoid. (i) Adding TSAs instead of only the exposed surfaces. (ii) Forgetting that a hemisphere on a flat solid adds $2\pi r^2$ while only the circle $\pi r^2$ is lost. (iii) Mixing slant height $l$ into a volume. (iv) Using diameter where the formula wants radius — halve it first. (v) Forgetting the second hemisphere in a capsule.

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Solid toy: a cone surmounted on a hemisphere with radius r, cone height h and slant height l Toy = cone on a hemisphere h r l flat faces meet & are hidden → exposed = πrl + 2πr^2 l = √(r^2 + h^2) Capsule: a cylinder of height h with a hemisphere of radius r on each end Capsule = cylinder + 2 hemispheres h (cylinder) r exposed = 2πrh + 4πr^2
1
Worked Example
A solid toy is a hemisphere surmounted by a right circular cone. The radius of both the hemisphere and the cone base is 3 cm, and the height of the cone is 4 cm. Find the total volume of the toy. (Take $\pi=22/7$.)
Solution
  1. Step 1: List the parts. Hemisphere radius $r=3$ cm; cone radius $r=3$ cm, height $h=4$ cm.
  2. Step 2: Volume of toy $=$ volume of hemisphere $+$ volume of cone.
  3. Step 3: Hemisphere $=\tfrac23\pi r^3=\tfrac23\cdot\tfrac{22}{7}\cdot 27=\tfrac{22}{7}\cdot 18=\tfrac{396}{7}$ cubic cm.
  4. Step 4: Cone $=\tfrac13\pi r^2 h=\tfrac13\cdot\tfrac{22}{7}\cdot 9\cdot 4=\tfrac{22}{7}\cdot 12=\tfrac{264}{7}$ cubic cm.
  5. Step 5: Add: $\tfrac{396}{7}+\tfrac{264}{7}=\tfrac{660}{7}\approx 94.29$ cubic cm.

Answer: $\tfrac{660}{7}\approx 94.29$ cubic cm.

2
Worked Example
A test tube is a cylinder with a hemispherical bottom. The common radius is 7 cm and the total length of the tube is 27 cm. Find the outer surface area available for painting. (Take $\pi=22/7$.)
Solution
  1. Step 1: Total length $=$ cylinder height $h$ $+$ hemisphere radius $r$, so $27=h+7\Rightarrow h=20$ cm.
  2. Step 2: The open top is not painted; exposed surface $=$ CSA of cylinder $+$ CSA of hemisphere.
  3. Step 3: CSA cylinder $=2\pi r h=2\cdot\tfrac{22}{7}\cdot 7\cdot 20=880$ sq cm.
  4. Step 4: CSA hemisphere $=2\pi r^2=2\cdot\tfrac{22}{7}\cdot 49=308$ sq cm.
  5. Step 5: Total $=880+308=1188$ sq cm.

Answer: $1188$ sq cm.

3
Worked Example
A wooden block is a cube of side 5 cm with a hemisphere of diameter 4.2 cm carved out on top, with the flat circular face flush with the top of the cube. Find the surface area of the remaining solid. (Take $\pi=22/7$.)
Solution
  1. Step 1: Hemisphere radius $r=4.2/2=2.1$ cm; cube side $a=5$ cm.
  2. Step 2: Carving removes the circle $\pi r^2$ from the top face but exposes the inner curved surface $2\pi r^2$.
  3. Step 3: Surface $=6a^2-\pi r^2+2\pi r^2=6a^2+\pi r^2$.
  4. Step 4: $6a^2=6\cdot 25=150$ sq cm.
  5. Step 5: $\pi r^2=\tfrac{22}{7}\cdot 2.1\cdot 2.1=22\cdot 0.3\cdot 2.1=13.86$ sq cm.
  6. Step 6: Surface $=150+13.86=163.86$ sq cm.

Answer: $163.86$ sq cm.

4
Worked Example
A medicine capsule is a cylinder with two hemispheres, one stuck to each end. The diameter of the capsule is 5 mm and its total length is 14 mm. Find its surface area. (Take $\pi=22/7$.)
Solution
  1. Step 1: Radius $r=5/2=2.5$ mm. The two hemispheres add a length of $2r=5$ mm.
  2. Step 2: Cylinder height $h=14-5=9$ mm.
  3. Step 3: Surface $=$ CSA cylinder $+2\times$ CSA hemisphere $=2\pi r h+2(2\pi r^2)=2\pi r(h+2r)$.
  4. Step 4: $=2\cdot\tfrac{22}{7}\cdot 2.5\cdot(9+5)=2\cdot\tfrac{22}{7}\cdot 2.5\cdot 14$.
  5. Step 5: $=2\cdot 22\cdot 2.5\cdot 2=220$ sq mm.

Answer: $220$ sq mm.

5
Worked Example
A tent is a cylinder of height 3 m and radius 14 m, topped by a cone of the same radius and slant height 25 m. Find the area of canvas needed for the tent (no floor). (Take $\pi=22/7$.)
Solution
  1. Step 1: Canvas $=$ CSA of cylinder $+$ CSA of cone (the floor is open).
  2. Step 2: CSA cylinder $=2\pi r h=2\cdot\tfrac{22}{7}\cdot 14\cdot 3=264$ sq m.
  3. Step 3: CSA cone $=\pi r l=\tfrac{22}{7}\cdot 14\cdot 25=1100$ sq m.
  4. Step 4: Total canvas $=264+1100=1364$ sq m.

Answer: $1364$ sq m.

6
Worked Example
A solid is a cone standing on a hemisphere; both have radius 7 cm and the cone height is 24 cm. Find the volume of the solid. (Take $\pi=22/7$.)
Solution
  1. Step 1: Volume $=$ volume of cone $+$ volume of hemisphere.
  2. Step 2: Cone $=\tfrac13\pi r^2 h=\tfrac13\cdot\tfrac{22}{7}\cdot 49\cdot 24=\tfrac{22}{7}\cdot 49\cdot 8=22\cdot 7\cdot 8=1232$ cubic cm.
  3. Step 3: Hemisphere $=\tfrac23\pi r^3=\tfrac23\cdot\tfrac{22}{7}\cdot 343=\tfrac{22}{7}\cdot\tfrac{686}{3}=\tfrac{2156}{3}\approx 718.67$ cubic cm.
  4. Step 4: Total $=1232+718.67=1950.67$ cubic cm (i.e. $\tfrac{5852}{3}$).

Answer: $\tfrac{5852}{3}\approx 1950.67$ cubic cm.

7
Worked Example
A juice glass is a cylinder of radius 3.5 cm and height 10 cm with a hemispherical cavity scooped out of the bottom (radius 3.5 cm). Find the actual capacity of the glass. (Take $\pi=22/7$.)
Solution
  1. Step 1: Apparent capacity $=$ full cylinder $=\pi r^2 h=\tfrac{22}{7}\cdot 3.5\cdot 3.5\cdot 10=385$ cubic cm.
  2. Step 2: The hemispherical cavity at the bottom occupies space, so subtract its volume.
  3. Step 3: Hemisphere $=\tfrac23\pi r^3=\tfrac23\cdot\tfrac{22}{7}\cdot 3.5\cdot 3.5\cdot 3.5=\tfrac23\cdot 134.75\approx 89.83$ cubic cm.
  4. Step 4: Actual capacity $=385-89.83=295.17$ cubic cm.

Answer: $\approx 295.17$ cubic cm.

8
Worked Example
Two cubes each of volume 64 cubic cm are joined end to end. Find the surface area of the resulting cuboid.
Solution
  1. Step 1: Side of each cube $a=\sqrt[3]{64}=4$ cm.
  2. Step 2: Joining end to end gives a cuboid of length $4+4=8$ cm, breadth 4 cm, height 4 cm.
  3. Step 3: TSA $=2(lb+bh+hl)=2(8\cdot 4+4\cdot 4+4\cdot 8)=2(32+16+32)$.
  4. Step 4: $=2\cdot 80=160$ sq cm. (Check: $2\times 6a^2=192$ minus two hidden faces $2a^2=32$ gives $160$.)

Answer: $160$ sq cm.

9
Worked Example
A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, surmounted by another cylinder of height 60 cm and radius 8 cm. Find the volume of iron. (Take $\pi=3.14$.)
Solution
  1. Step 1: Lower cylinder: radius $r_1=12$ cm, height $h_1=220$ cm.
  2. Step 2: Volume$_1=\pi r_1^2 h_1=3.14\cdot 144\cdot 220=3.14\cdot 31680=99475.2$ cubic cm.
  3. Step 3: Upper cylinder: radius $r_2=8$ cm, height $h_2=60$ cm.
  4. Step 4: Volume$_2=\pi r_2^2 h_2=3.14\cdot 64\cdot 60=3.14\cdot 3840=12057.6$ cubic cm.
  5. Step 5: Total iron $=99475.2+12057.6=111532.8$ cubic cm.

Answer: $111532.8$ cubic cm.

10
Worked Example
A gulab jamun is modelled as a cylinder of length 5 cm with a hemisphere on each end; the full length is 5 cm and the diameter is 2.8 cm. Find the volume of one piece. (Take $\pi=22/7$.)
Solution
  1. Step 1: Radius $r=2.8/2=1.4$ cm. The two hemispheres take up $2r=2.8$ cm of the length.
  2. Step 2: Cylinder length $h=5-2.8=2.2$ cm.
  3. Step 3: Volume $=$ cylinder $+2\times$ hemisphere $=\pi r^2 h+\tfrac43\pi r^3$ (two hemispheres make one sphere).
  4. Step 4: Cylinder $=\tfrac{22}{7}\cdot 1.96\cdot 2.2=\tfrac{22}{7}\cdot 4.312\approx 13.55$ cubic cm.
  5. Step 5: Sphere $=\tfrac43\cdot\tfrac{22}{7}\cdot 1.4^3=\tfrac43\cdot\tfrac{22}{7}\cdot 2.744\approx 11.50$ cubic cm.
  6. Step 6: Total $\approx 13.55+11.50=25.05$ cubic cm.

Answer: $\approx 25.05$ cubic cm.

11
Worked Example
A wooden pen stand is a cuboid 15 cm by 10 cm by 3.5 cm with four conical holes drilled into it. Each hole has radius 0.5 cm and depth 1.4 cm. Find the volume of wood left. (Take $\pi=22/7$.)
Solution
  1. Step 1: Volume of cuboid $=15\cdot 10\cdot 3.5=525$ cubic cm.
  2. Step 2: Volume of one conical hole $=\tfrac13\pi r^2 h=\tfrac13\cdot\tfrac{22}{7}\cdot 0.25\cdot 1.4=\tfrac13\cdot\tfrac{22}{7}\cdot 0.35$.
  3. Step 3: $=\tfrac13\cdot 1.1=0.3667$ cubic cm per hole.
  4. Step 4: Four holes remove $4\cdot 0.3667=1.4667$ cubic cm.
  5. Step 5: Wood left $=525-1.4667\approx 523.53$ cubic cm.

Answer: $\approx 523.53$ cubic cm.

12
Worked Example
A vessel is a hollow cylinder of radius 7 cm with a hemispherical bowl of the same radius forming its closed bottom. The straight cylindrical part is 6 cm tall. Find the inner surface area available to hold liquid (the top is open). (Take $\pi=22/7$.)
Solution
  1. Step 1: The liquid touches the cylindrical wall and the hemispherical bottom; the open top is not counted.
  2. Step 2: Inner surface $=$ CSA of cylinder $+$ CSA of hemisphere $=2\pi r h+2\pi r^2$, with $r=7$, $h=6$.
  3. Step 3: $2\pi r h=2\cdot\tfrac{22}{7}\cdot 7\cdot 6=264$ sq cm.
  4. Step 4: $2\pi r^2=2\cdot\tfrac{22}{7}\cdot 49=308$ sq cm.
  5. Step 5: Inner surface $=264+308=572$ sq cm.

Answer: $572$ sq cm.

Key Points

  • A combined solid is made by joining basic shapes like cylinders, cones, and spheres together.
  • The total volume of a combined solid is the sum of the volumes of its parts.
  • The total surface area includes only the outer, visible surfaces. Any face caught inside the joint must be subtracted.
  • When finding a cone's surface area, always calculate its slant height first using the formula $l = \sqrt{r^2 + h^2}$.
  • Carefully read problem dimensions to distinguish between the radius and the full diameter of an object.
  • For tents/canvas remember there is usually no floor — exclude the base circle.
  • Capacity conversion: $1000\text{ cm}^3=1$ litre and $1\text{ m}^3=1000$ litres.
  • For tents/canvas remember there is usually no floor — exclude the base circle.
  • Capacity conversion: $1000\text{ cm}^3=1$ litre and $1\text{ m}^3=1000$ litres.
Tap an option to check your answer0 / 4
Q1.The volume of a cylinder is:
Explanation: $\pi r^2h$.
Q2.The volume of a cone is:
Explanation: One-third of the cylinder.
Q3.The volume of a sphere is:
Explanation: $\tfrac43\pi r^3$.
Q4.The curved surface area of a cylinder is:
Explanation: $2\pi rh$.