Combined Volume and Surface Area

What fills it, and what covers it.

What fills it and what covers it

A solid can be measured in two different ways. Its volume is the space inside it, which is how much it holds. Volume is counted in cubes 1 cm along each edge, so it is measured in cubic centimeters, cm³. Its surface area is the total area of all its faces, which is how much material it takes to cover it. Surface area is counted in squares 1 cm along each side, so it is measured in square centimeters, cm².

A can of soup has both. The soup inside the can fills its volume. The metal of the can, its top, its bottom and its curved side together, makes up its surface area.

The volume: the circle times the height

Take a can with a radius of 3 cm and a height of 10 cm. A cylinder is shaped like a prism: every cut straight across it uncovers the same circle. So its volume is the area of that circle times the height, V = πr²h.

The circle has a radius of 3 cm, so its area is π × 3² = 9π cm². Each centimeter of height holds a layer of 9π cm³, and the cylinder is 10 cm tall, so its volume is 9π × 10 = 90π cm³.

39π10

The can is drawn lying on its side. Every cut across it uncovers the same circle, of radius 3 cm and area 9π cm², and the can is 10 cm from end to end: 9π × 10 = 90π cm³.

The surface area: lay it flat

To find what covers the cylinder, imagine cutting it open and laying every face flat. This flat shape is the net of the cylinder. It has two circles, the top and the bottom, and one rectangle, which is the curved side unrolled.

The rectangle is as tall as the cylinder, 10 cm. Its width is the distance around the circle, the circumference: 2πr = 2 × π × 3 = 6π cm. So the curved side has an area of 6π × 10 = 60π cm².

Each circle has an area of π × 3² = 9π cm². Add all three faces: 9π + 9π + 60π = 78π cm². For any closed cylinder, the surface area is 2πr² + 2πr × h.

The net of a cylinder: two circles of area πr², and a rectangle whose width is the circumference 2πr and whose height is the height of the cylinder, h.

In decimals, and in milliliters

The answers 90π cm³ and 78π cm² are exact, because they keep π as a symbol. For a decimal, multiply by π, which is about 3.14159. The volume is 90π ≈ 282.7, which is about 283 cm³ to the nearest whole number. The surface area is 78π ≈ 245.0, which is about 245 cm².

A milliliter is the same amount of space as a cubic centimeter: 1 ml = 1 cm³. So a can with a volume of about 283 cm³ holds about 283 ml of soup.

Two measures that cannot be compared

The can has a volume of 90π cm³ and a surface area of 78π cm². The two numbers count different things, cubes and squares, so it makes no sense to ask which one is bigger. Measure the same can in millimeters instead. A cubic centimeter is 10 × 10 × 10 = 1000 mm³, and a square centimeter is 10 × 10 = 100 mm², so the volume becomes 90,000π mm³ and the surface area becomes 7800π mm². In centimeters the volume’s number was a little larger, 90 against 78; in millimeters it is more than ten times larger. Keep each answer with its own unit: cm³ for a volume and cm² for an area.

Three slips

Giving 78π as the volume answers the wrong question: 78π cm² is what covers the can, not what fills it.

The curved side alone, 60π cm², is an area. Its unit is cm², so it cannot be the volume, and it is not the whole surface either.

A closed can has a top and a bottom. Counting one circle gives 9π + 60π = 69π cm², which is the surface of a can with no lid. Read the question to see which faces are there.

A solid made of two pieces

Many real objects are two solids joined together, such as a cylinder standing on a cone. Split the object where the two pieces meet, and work with each piece on its own.

Volumes add, because the two pieces fill separate spaces. Surface areas do not simply add. Where the pieces meet, a face of each one is pressed against the other, so those two faces are inside the object and are not part of its surface. Add only the outside surfaces.

For a cone, three facts are needed. Its volume is ⅓πr²h, a third of the cylinder with the same base and height. Its curved surface has area πrl, where l is the slant height, the length of the sloping side from the tip to the rim. The radius, the height and the slant height make a right-angled triangle inside the cone, so l² = r² + h² by Pythagoras.

Worked example: A Grain Silo on a Cone-Shaped Hopper: Its Volume and the Area to Paint

Question A grain silo is a cylinder of radius 2.5 m and height 12 m with a flat roof. It stands on a cone-shaped hopper of the same radius, whose point is 6 m below the bottom of the cylinder. (a) Find the total volume of the silo, as a multiple of π and correct to 3 significant figures. (b) The whole outside of the silo, meaning the roof, the curved wall and the hopper, is to be painted. Find the area to be painted, as a multiple of π and correct to 3 significant figures.

  1. 1.The cylinder has volume π r2 h = π × 2.52 × 12 = 75π m³. The hopper is a cone of radius 2.5 m and height 6 m, so its volume is 13 × π × 2.52 × 6 = 12.5π m³.

    2.5 m12 m6 mcylinder pi × 2.52× 12 = 75 pihopper 1/3 × pi × 2.52× 6 = 12.5 pi
    2.5 m12 m6 mcylinder pi × 2.52× 12 = 75 pihopper 1/3 × pi × 2.52× 6 = 12.5 pi
    The cylinder holds 75π m³ and the cone-shaped hopper 13π × 2.52 × 6 = 12.5π m³.

  2. 2.(a) The total volume is 75π + 12.5π = 87.5π m³. Since 87.5π ≈ 274.9, that is 275 m³ correct to 3 significant figures.

    2.5 m12 m6 mcylinder pi × 2.52× 12 = 75 pihopper 1/3 × pi × 2.52× 6 = 12.5 pi75 pi + 12.5 pi = 87.5 pi, about 275 m3
    2.5 m12 m6 mcylinder pi × 2.52× 12 = 75 pihopper 1/3 × pi × 2.52× 6 = 12.5 pi75 pi + 12.5 pi = 87.5 pi, about 275 m3
    (a) The silo holds 87.5π ≈ 275 m³.

  3. 3.The curved surface of the hopper needs its slant height l. The radius and the height of the cone are the two shorter sides of a right-angled triangle, so l2 = 2.52 + 62 = 6.25 + 36 = 42.25, and l = 6.5 m.

    2.5 m12 m6 m6.5 mcylinder pi × 2.52× 12 = 75 pihopper 1/3 × pi × 2.52× 6 = 12.5 pi75 pi + 12.5 pi = 87.5 pi, about 275 m3l2= 2.52+ 62= 42.25, l = 6.5 m
    2.5 m12 m6 m6.5 mcylinder pi × 2.52× 12 = 75 pihopper 1/3 × pi × 2.52× 6 = 12.5 pi75 pi + 12.5 pi = 87.5 pi, about 275 m3l2= 2.52+ 62= 42.25, l = 6.5 m
    The hopper's slant height is the hypotenuse: l2 = 2.52 + 62 = 42.25, so l = 6.5 m.

  4. 4.The roof is a circle of area π × 2.52 = 6.25π m². The curved wall has area 2π r h = 2 × π × 2.5 × 12 = 60π m². The hopper's curved surface has area π r l = π × 2.5 × 6.5 = 16.25π m². The circle where the cylinder meets the hopper is inside the silo, so it is not painted.

    2.5 m12 m6 m6.5 mcylinder pi × 2.52× 12 = 75 pihopper 1/3 × pi × 2.52× 6 = 12.5 pi75 pi + 12.5 pi = 87.5 pi, about 275 m3l2= 2.52+ 62= 42.25, l = 6.5 mroof 6.25 pi, wall 60 pi, hopper 16.25 pi
    2.5 m12 m6 m6.5 mcylinder pi × 2.52× 12 = 75 pihopper 1/3 × pi × 2.52× 6 = 12.5 pi75 pi + 12.5 pi = 87.5 pi, about 275 m3l2= 2.52+ 62= 42.25, l = 6.5 mroof 6.25 pi, wall 60 pi, hopper 16.25 pi
    The roof, the wall and the hopper are painted; the circle where the cylinder meets the hopper is inside.

  5. 5.(b) The area to be painted is 6.25π + 60π + 16.25π = 82.5π m². Since 82.5π ≈ 259.2, that is 259 m² correct to 3 significant figures.

    2.5 m12 m6 m6.5 mcylinder pi × 2.52× 12 = 75 pihopper 1/3 × pi × 2.52× 6 = 12.5 pi75 pi + 12.5 pi = 87.5 pi, about 275 m3l2= 2.52+ 62= 42.25, l = 6.5 mroof 6.25 pi, wall 60 pi, hopper 16.25 pi82.5 pi, about 259 m2
    2.5 m12 m6 m6.5 mcylinder pi × 2.52× 12 = 75 pihopper 1/3 × pi × 2.52× 6 = 12.5 pi75 pi + 12.5 pi = 87.5 pi, about 275 m3l2= 2.52+ 62= 42.25, l = 6.5 mroof 6.25 pi, wall 60 pi, hopper 16.25 pi82.5 pi, about 259 m2
    (b) The area to paint is 6.25π + 60π + 16.25π = 82.5π ≈ 259 m².

Answer: (a) 87.5π m³, which is 275 m³ to 3 significant figures; (b) 82.5π m², which is 259 m² to 3 significant figures

Common mistakes

  • Using the height of the hopper, 6 m, in π r l. The curved surface of a cone is measured along its slope, so it needs the slant height, 6.5 m.
  • Adding the two circles where the cylinder and the hopper meet. They are joined to each other inside the silo, so no paint goes on them.

More volume and surface area problems, worked step by step →

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