Surface Areas and Volumes

A cuboid is a box with length l, breadth b and height h. Its total surface area (TSA) is the sum of all six rectangular faces, so TSA = 2(lb + bh + hl), while its volume is l × b × h. A cube is a special cuboid with every edge equal to a, giving TSA = 6a² and volume = a³.

A cylinder is generated by rotating a rectangle; it has two circular ends of radius r and a curved side of height h. The curved surface area (CSA) is the label wrapped around it, 2πrh, and the total surface area adds the two circular ends: 2πr(r + h). Its volume is the base area times height, πr²h.

Worked example. A solid cylinder has r = 7 cm and h = 10 cm (take π = 22/7).

• CSA = 2πrh = 2 × (22/7) × 7 × 10 = 440 cm².
• TSA = 2πr(r + h) = 2 × (22/7) × 7 × 17 = 748 cm².
• Volume = πr²h = (22/7) × 49 × 10 = 1540 cm³.

Always keep area in square units and volume in cubic units, and convert all measurements to a single unit before computing.

A right circular cone has base radius r, vertical height h and slant height l. The three are linked by a right triangle, so l = √(r² + h²). Its curved surface area is πrl, its total surface area adds the circular base to give πr(l + r), and its volume is one-third of the cylinder on the same base: (1/3)πr²h.

A sphere is a perfectly round solid of radius r. It has a single surface, so its surface area is 4πr² and its volume is (4/3)πr³. A hemisphere is exactly half a sphere; its curved part is 2πr², and because it also has a flat circular face its total surface area is 2πr² + πr² = 3πr². Its volume is half the sphere, (2/3)πr³.

Worked example (cone). For r = 6 cm and h = 8 cm, slant height l = √(6² + 8²) = √100 = 10 cm. Taking π = 3.14, CSA = πrl = 3.14 × 6 × 10 = 188.4 cm² and volume = (1/3)πr²h = (1/3) × 3.14 × 36 × 8 = 301.44 cm³.

Worked example (sphere). For r = 7 cm with π = 22/7, surface area = 4πr² = 4 × (22/7) × 49 = 616 cm² and volume = (4/3)πr³ = (4/3) × (22/7) × 343 = 1437.33 cm³.

Many everyday objects are built by joining two basic solids — a cone on a hemisphere (a toy top), a cylinder with hemispherical ends (a capsule), or a cylinder with a conical roof (a tent). To find the surface area of such a combination, we add up only the surfaces that are visible from outside.

The key idea: when two solids are stuck together, their meeting faces are pressed against each other and disappear from view. So we never add the joined (hidden) faces. For a cone sitting on a hemisphere of the same radius, the flat base of the cone and the flat face of the hemisphere both vanish, and the total surface area is simply CSA of cone + CSA of hemisphere = πrl + 2πr².

Worked example (tent). A tent has a cylindrical base of radius r = 7 m and height 3 m, with a conical top of the same radius and slant height 25 m. The canvas needed is the cylinder's curved surface plus the cone's curved surface (the flat top of the cylinder is covered by the cone and the ground base is not canvas):

• CSA of cylinder = 2πrh = 2 × (22/7) × 7 × 3 = 132 m².
• CSA of cone = πrl = (22/7) × 7 × 25 = 550 m².
• Total canvas = 132 + 550 = 682 m².

So identify each exposed piece, use the correct curved-surface formula, and add.

Volume behaves more simply than surface area. Since volume is just the amount of space filled, the total volume of a combined solid is the sum of the volumes of the parts. Nothing is hidden or lost, so there is no adjustment for joined faces.

If material is scooped out (for example a conical or hemispherical cavity drilled into a solid), we subtract the removed volume from the whole. This is the only sign change you need to watch for.

Worked example. A solid is a cylinder of radius r = 7 cm and height 10 cm, surmounted by a cone of the same radius and height 6 cm (take π = 22/7).

• Volume of cylinder = πr²h = (22/7) × 49 × 10 = 1540 cm³.
• Volume of cone = (1/3)πr²h = (1/3) × (22/7) × 49 × 6 = 308 cm³.
• Total volume = 1540 + 308 = 1848 cm³.

For a capsule (a cylinder with a hemisphere at each end), the two hemispheres together make one full sphere, so its volume is πr²h + (4/3)πr³. Recognising such pairings keeps the working short.

Every combination question yields to the same routine.

1. Decompose: split the object into the basic solids you know (cone, cylinder, hemisphere, etc.) and note the shared radius or height.
2. Decide the quantity: for surface area, list only the exposed curved/flat surfaces and drop the joined faces; for volume, add every part (subtract any cavity).
3. Find missing lengths: most cone problems first need the slant height l = √(r² + h²).
4. Choose π smartly: use π = 22/7 when a radius is a multiple of 7, and π = 3.14 otherwise; always state which you used.
5. Units: convert everything to one unit first; give area in square units and volume in cubic units.

Common shapes to recognise:

• Cone on a hemisphere → ice-cream / spinning top.
• Cylinder with a hemisphere at each end → medicine capsule.
• Cylinder with a conical top → tent or rocket.
• Cylinder with a hemispherical or conical cavity scooped out → wooden article, glass with raised bottom.

Worked example. A solid is a cone (r = 1 cm, h = 1 cm) standing on a hemisphere (r = 1 cm). Volume = (1/3)πr²h + (2/3)πr³ = (1/3)π(1)(1) + (2/3)π(1) = (1/3)π + (2/3)π = π cm³. Expressing an answer in terms of π is perfectly acceptable when the question asks for it.