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The animation starts with a translucent sphere (pale orange). An dark red wedge is shown running from the centre of the sphere to its back surface. This wedge represents a small segment of the sphere volume. The volume of that wedge would be its base area (i.e. the part of the sphere surface that the wedge occupies) multiplied by the radius of the sphere (i.e. the length of the wedge) and divided by three. This is because the wedge is a thin pyramid whose volume = base x height/3. The volume of the sphere would be equal to the sum of the volumes of all such wedges that would fill the sphere (to make the result smooth an infinite number of such pyramids would be needed). Now, all such wedges would be pyramids whose height = r. Their combined base area would be the same as the surface area of the entire sphere. Consequently, the volume of the sphere would be the same as a pyramid whose base area = the sphere surface area and whose height = the radius of the sphere. This gives the formula: volume = 4/3 π r^{3}. The remainder of the animation is devoted to creating this equivalent pyramid.
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 (now in pale yellow green) has an area of 4 π r^{2}. This rectangle then morphs into a pyramid whose height is r. So we get: volume of pyramid = base (4 π r^{2}) x Height (r) / 3 = 4/3 π r^{3}
The two thirds relationships between volumes and areas of spheres and cylinders:The total volume of the cylinder (including its base and lid) is given by:
wall of cylinder = 4 π r^{2}so the area of the cylinder is 6 π r^{2} and that of its circumscribed sphere is 4 π r^{2}. 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 r^{3}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.