Learn first: Prisms, parallelepipeds and pyramids
Solids of revolution are produced by rotating a plane figure about an axis: a rectangle rotated about a side gives a cylinder, a right-angled triangle rotated about a leg gives a cone, and a semicircle rotated about its diameter gives a ball. They are everywhere: tins, pipes, funnels, balls. The most convenient way to study them is the axial section: the plane figure obtained by cutting the solid with a plane through its axis shows all its dimensions and turns a 3D problem into a plane one. This lesson concentrates on surface areas; volumes are treated in detail in the topic “Volumes”.
A cylinder is the solid obtained by rotating a rectangle about one of its sides. Its two bases are equal discs of radius R; the height h is the distance between them. The axis passes through the centres of the bases; the rotating side is called a generator, and in a right cylinder it equals the height.
The axial section is a rectangle with sides 2R and h. If it is a square, then h = 2R and the cylinder is called equilateral.
Cut the lateral surface along a generator and unroll it: you get a rectangle with one side equal to the circumference of the base 2πR and the other equal to h. Hence:
Lateral area S_lat = 2πRh
Total area S = 2πRh + 2πR² = 2πR(R + h)
Volume V = πR²h
The total area includes both bases — the most commonly forgotten detail.
The axial section of a cylinder is a square with side 6. Find the lateral and total areas.
A cone is obtained by rotating a right-angled triangle about a leg. The axis leg is the height h, the other leg is the base radius R, the hypotenuse is the slant height l. By Pythagoras:
l² = R² + h²
The axial section is an isosceles triangle with base 2R and equal sides l. If it is equilateral, then l = 2R.
Cutting the lateral surface along a slant line and unrolling it gives a circular sector of radius l whose arc equals the base circumference 2πR. Using the sector area lR/2 (arc length × radius / 2):
Lateral area S_lat = πRl; total area S = πR(R + l)
The net is a sector of radius l with angle φ = 360° · R/l
Volume V = πR²h/3
The angle formula comes from 2πR = πlφ/180°. Since R < l, the angle is always less than 360°.
A cone has R = 3, h = 4. Find the slant height, lateral and total areas, and the angle of the net.
Cut a cone with a plane parallel to the base and remove the top: what remains is a frustum. Its bases are discs of radii R and r, height h, slant height l. The axial section is an isosceles trapezium with bases 2R and 2r. From the trapezium:
l² = h² + (R − r)²
Lateral area S_lat = π(R + r)l; total S = π(R + r)l + πR² + πr²
The formula is easy to check: with r = 0 it becomes the cone’s πRl, with r = R the cylinder’s 2πRl.
A frustum has base radii 5 and 2 and height 4. Find its total area.
A ball is the set of points at distance at most R from the centre; a sphere is its surface, the points at distance exactly R. A ball is produced by rotating a semicircle about its diameter.
Sphere area S = 4πR²; ball volume V = 4πR³/3
A plane section is a disc of radius r = √(R² − d²) (d is the distance from the centre to the plane)
The section formula is Pythagoras: the ball’s radius is the hypotenuse, d and r are the legs. The section through the centre is a great circle (r = R), the largest section. A tangent plane to a sphere is perpendicular to the radius at the point of contact, just as for a circle.
A nice fact (Archimedes): the area of a sphere equals the area of four great circles.
A ball of radius 13 is cut by a plane at distance 5 from the centre. Find the area of the section.
In combination problems the axial section helps again:
Cylinder circumscribed about a ball: S_cyl : S_sphere = 6πR² : 4πR² = 3 : 2
A beam of maximum volume with a rectangular cross-section is cut from a cylindrical log of length 6 m and radius 5 dm. Find the beam’s volume and the share of waste (π ≈ 3).
A cone has R = 3, h = 4. Find the radius of the inscribed ball.