BUY THIS ANIMATION of surface area of a sphere!

The surface area of a sphere is opened up to illustrate how the formula surface area = 4 π r2 can be understood.

The animation starts with a sphere (orange) with its equator shown in yellow. Latitudes north and south are shown as horizontal red rings. A meridian is shown in green. These latitude rings all open up creating a curved surface shown in pink. This then uncurls to map on to an imaginary vertical plane that touches the back of the sphere. The mapped surface (i.e. the sphere surface area) looks vaguely leaf shaped. It is actually formed from cosine curves. The challenge is then to convert this complex shape into a rectangle to determine its area. To do this, those parts of the mapped surface that are north and south of the sphere are replaced by translucent red boxes. The remaining leaf shape thus has a maximum height that is the same as the height of the sphere (i.e. 2r) and a maximum length that is the same as the circumference of the sphere (i.e. 2 π r). These boxes then migrate across to fill up the gaps and show that the area of the map above the height r is the same as the gap below. The resulting rectangle has an area of 4 π r2. Since it is the same height as the sphere and is as long as the sphere is around, this rectangle can curl around the sphere and precisely enclose it forming an open ended cylinder.

The two thirds relationships between volumes and areas of spheres and cylinders:

The total surface area of the cylinder (including its base and lid) is given by:

wall of cylinder = 4 π r2
lid of cylinder = π r2 (see area of circle animation)
base of cylinder = π r2
TOTAL AREA = 6 π r2

so the area of the cylinder is 6 π r2 and that of its circumscribed sphere is 4 π r2. In other words, the sphere has 4/6 or two thirds the area of its enclosing cylinder. Now, this is interesting because it is the same ratio as the volume a sphere to the volume of its circumscribing cylinder:

volume of sphere = 4/3 π r3
volume of cylinder = lid (π r2) x height (2 π r) = 2 π r3

Archimedes discovered these relationships between a cylinder and its enclosed (circumscribed) sphere.

Try our circle and area calculator to derive various values from different starting points.

Russell Kightley Media
PO Box 9150, Deakin, ACT 2600, Australia. Mobile phone Australia 0405 17 64 71
email RKM