Volume & Surface Area of Cylinders, Cones, Spheres, Pyramids & Frustums
Calculating the volume and total surface area of 3D solids requires distinguishing between prisms and cylinders (where Volume = Cross-Sectional Area x Length) and pointed or curved 3D solids (pyramids, cones, spheres, hemispheres, and frustums). When finding the total surface area of a solid cone, hemisphere, or half-cylinder, you must always remember to add the flat circular base(s) to the curved surface area, and use Pythagoras' Theorem ($l^2 = r^2 + h^2$) to convert between a cone's vertical height $h$ and its slant height $l$. On the Higher Tier, examiners frequently test finding the volume of a frustum (a cone with its top sliced off) and equating the volumes of two melted solids to solve for an unknown radius or height in terms of $\pi$.
- Prisms & Cylinders (Volume & Surface Area)
- 1. Prism: Volume = Area of cross-section x Length. 2. Cylinder of radius r and height h: Volume = $\pi r^2 h$. The curved surface area of a cylinder unrolls into a rectangle of width $2\pi r$ (the circumference) and height $h$, so Curved Surface Area = $2\pi r h$, and Total Surface Area of a solid cylinder = $2\pi r h + 2\pi r^2$.
- Pyramids (Volume = ⅓ x Base Area x Height)
- For any pyramid (whether square-based, rectangular-based, or triangular-based): Volume = $\frac{1}{3} \times \text{Area of Base} \times \text{Perpendicular vertical height } (h)$. To find the surface area of a pyramid, add the area of the base to the areas of the triangular sloping faces (finding the slant height of each triangular face using Pythagoras).
- Cones (Volume & Surface Area)
- For a cone with base radius $r$, perpendicular vertical height $h$, and slant height $l$ (where $l^2 = r^2 + h^2$): 1. Volume = $\frac{1}{3}\pi r^2 h$. 2. Curved Surface Area = $\pi r l$. 3. Total Surface Area of a solid cone = $\pi r l + \pi r^2$ (curved area + circular base).
Volume & Surface Area of Cylinders, Cones, Spheres, Pyramids & Frustums: 3D Molecular, Lattice, DNA & Vector Explorer
Drag directly on the 3D model to rotate it through any angle, or switch tabs to inspect VSEPR bond angles, crystal lattices, DNA base pairing, and 3D space diagonals.
Position Vector r = (4i + 3j + 5k)
- 1. Base Diagonal √(x² + y²):
- √(4² + 3²) = 5.00
- 2. 3D Magnitude |r| = √(x² + y² + z²):
- √50 = 7.07
- 3. Angle with Base tan⁻¹(z / base):
- 45.0°
To find the angle between the space diagonal OP and the horizontal base plane, form the shaded vertical right-angled triangle with adjacent = √(x² + y²) and opposite = z.
Key Concepts & Exam Definitions (6)
- Prisms & Cylinders (Volume & Surface Area)
- 1. Prism: Volume = Area of cross-section x Length. 2. Cylinder of radius r and height h: Volume = $\pi r^2 h$. The curved surface area of a cylinder unrolls into a rectangle of width $2\pi r$ (the circumference) and height $h$, so Curved Surface Area = $2\pi r h$, and Total Surface Area of a solid cylinder = $2\pi r h + 2\pi r^2$.
- Pyramids (Volume = ⅓ x Base Area x Height)
- For any pyramid (whether square-based, rectangular-based, or triangular-based): Volume = $\frac{1}{3} \times \text{Area of Base} \times \text{Perpendicular vertical height } (h)$. To find the surface area of a pyramid, add the area of the base to the areas of the triangular sloping faces (finding the slant height of each triangular face using Pythagoras).
- Cones (Volume & Surface Area)
- For a cone with base radius $r$, perpendicular vertical height $h$, and slant height $l$ (where $l^2 = r^2 + h^2$): 1. Volume = $\frac{1}{3}\pi r^2 h$. 2. Curved Surface Area = $\pi r l$. 3. Total Surface Area of a solid cone = $\pi r l + \pi r^2$ (curved area + circular base).
- Spheres & Hemispheres (Volume & Surface Area)
- For a sphere of radius $r$: 1. Volume of a sphere = $\frac{4}{3}\pi r^3$ (so Volume of a hemisphere = $\frac{2}{3}\pi r^3$). 2. Surface Area of a sphere = $4\pi r^2$. 3. Curved Surface Area of a hemisphere = $2\pi r^2$, and Total Surface Area of a solid hemisphere = $2\pi r^2 + \pi r^2 = 3\pi r^2$.
- Frustums of a Cone (Higher Tier)
- A frustum is formed by removing a small top cone from a mathematically similar large cone: Volume of Frustum = $\frac{1}{3}\pi R^2 H - \frac{1}{3}\pi r^2 h$. Always sketch the 2D vertical cross-section first and use the ratio of the radii ($R / r$) to find the missing vertical height $h$ of the small cone ($H / h = R / r$).
- Composite Solids & Conservation of Volume
- 1. For composite 3D shapes (e.g., a silo made of a cylinder topped with a hemisphere): add the individual volumes together, and for external surface area, only add the exposed outer surfaces (do not include the hidden internal circle where the two shapes join). 2. When one solid is melted down to form $N$ smaller solids, equate Total Initial Volume = $N \times$ Volume of 1 small solid.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
State the formulas for the Volume and Total Surface Area of a solid cylinder of radius r and height h.
State the formulas for the Volume and Total Surface Area of a solid cone of radius r, vertical height h, and slant height l.
State the formulas for the Volume and Total Surface Area of a solid hemisphere of radius r.
A cone has a base radius of 5 cm and a perpendicular height of 12 cm. Calculate its exact total surface area in terms of π.
How do you find the volume of a frustum when given the bottom radius R, top radius r, and the height of the frustum itself?
A toy is made by attaching the flat circular base of a cone (radius r) to the flat circular base of a hemisphere (radius r). Which parts make up the external surface area of the toy?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
A solid metal cone has a base diameter of 16 cm and a perpendicular vertical height of 15 cm. Calculate: (a) the exact volume of the cone in terms of π, and (b) the total surface area of the solid cone in terms of π. [5 marks]
Click “Check Mark Scheme” to compare your answer, 1. Halve the diameter to find the radius: r = 16 ÷ 2 = 8 cm [1 mark]. 2. (a) Volume of cone = ⅓ x π x r² x h = ⅓ x π x 8² x 15 = 320π cm³ [1 mark]. 3. (b) Use Pythagoras' Theorem to find the slant height l of the cone: l = √(r² + h²) = √(8² + 15²) = √(64 + 225) = √289 = 17 cm [1 mark]. 4. Curved surface area = π x r x l = π x 8 x 17 = 136π cm² [1 mark]. 5. Add the area of the flat circular base (πr² = π x 8² = 64π): Total surface area = 136π + 64π = 200π cm² [1 mark].
- Q2
A frustum is made by removing a small cone of base radius 4 cm from the top of a large cone of base radius 10 cm. The vertical height of the frustum is 9 cm. Calculate the volume of the frustum, giving your answer in terms of π. (Higher Tier) [5 marks]
Click “Check Mark Scheme” to compare your answer, 1. Let h be the vertical height of the small removed cone, so the total height of the large original cone is H = h + 9 [1 mark]. 2. Because the small cone and large cone are mathematically similar, the ratio of their heights equals the ratio of their radii: (h + 9) / h = 10 / 4 [1 mark]. 3. Cross-multiply and solve for h: 4(h + 9) = 10h -> 4h + 36 = 10h -> 6h = 36 -> h = 6 cm, so the large cone has height H = 6 + 9 = 15 cm [1 mark]. 4. Volume of large cone = ⅓ x π x 10² x 15 = 500π cm³; Volume of small cone = ⅓ x π x 4² x 6 = 32π cm³ [1 mark]. 5. Volume of frustum = 500π - 32π = 468π cm³ (or 1,470 cm³ to 3 s.f) [1 mark].
- Q3
A solidwax sphere of radius 6 cm is melted down and all the wax is used to make a solid cylinder of radius 4 cm and height h cm. Calculate the exact value of h. [3 marks]
Click “Check Mark Scheme” to compare your answer, 1. Calculate the volume of the sphere in terms of π: V_sphere = (4/3) x π x r³ = (4/3) x π x 6³ = 288π cm³ [1 mark]. 2. Write the volume of the cylinder in terms of π and equate the two volumes: π x 4² x h = 16πh = 288π [1 mark]. 3. Divide both sides by 16π: h = 288 ÷ 16 = 18 cm [1 mark].
- Q4
A solid glass paperweight is in the shape of a hemisphere of radius 5 cm. Calculate the total surface area of the paperweight to 3 significant figures. [3 marks]
Click “Check Mark Scheme” to compare your answer, 1. Curved surface area of the hemisphere = ½ x (4πr²) = 2 x π x 5² = 50π cm² [1 mark]. 2. Area of the flat circular base = π x r² = π x 5² = 25π cm² [1 mark]. 3. Total surface area = 50π + 25π = 75π = 235.619. = 236 cm² (to 3 s.f) [1 mark].
Common Exam Questions (4)
What is the difference between 'h' and 'l' in the cone formulas, and how are they linked?
In a cone of base radius r: 'h' is the perpendicular vertical height (used in the volume formula $V = \frac{1}{3}\pi r^2 h$), whereas 'l' is the slant height up the sloping side (used in the curved surface area formula $CSA = \pi r l$). Because the radius r, vertical height h, and slant height l form a right-angled triangle inside the cone, they are always linked by Pythagoras' Theorem: $l^2 = r^2 + h^2$!
Why is the total surface area of a solid hemisphere equal to 3πr² and not 2πr²?
This is the #1 surface area exam trap. The surface area of a full sphere is $4\pi r^2$, so halving it gives the curved surface area of the hemisphere as $2\pi r^2$. However, a solid hemisphere also has a flat circular base on the bottom of area $\pi r^2$! Adding the curved top and the flat circular base together gives: Total Surface Area of Solid Hemisphere = $2\pi r^2 + \pi r^2 = 3\pi r^2$.
How do you calculate the volume of a frustum of a cone? (Higher Tier)
A frustum is the bottom piece left when a smaller similar cone is sliced off the top of a large original cone parallel to its base. 1. Use similar triangles (scale factor of the two radii) to find the missing vertical height of the small removed cone (and the full height of the large original cone). 2. Calculate Volume of Large Cone = $\frac{1}{3}\pi R^2 H$. 3. Calculate Volume of Small Removed Cone = $\frac{1}{3}\pi r^2 h$. 4. Subtract them: Volume of Frustum = Volume of Large Cone: Volume of Small Cone.
Why should you leave your intermediate working in terms of π when a metal sphere is melted down and recast into a cone?
When a 3D solid is melted down into a new shape, total volume is conserved (Volume of Sphere = Volume of Cone), and both formulas contain $\pi$ (e.g., $\frac{4}{3}\pi r^3 = \frac{1}{3}\pi R^2 h$). Leaving both sides in terms of $\pi$ allows you to divide both sides by $\pi$ immediately, eliminating rounding errors and making the algebra works cleanly even on a Non-Calculator paper.
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