You are given a sphere of radius 102.0 units.
Units may be any units of length: inches, cm, metres, feet, miles, km. So the radius could be any of 102.0 inches, 102.0 cm, 102.0 metres, 102.0 feet, 102.0 miles, 102.0 km, etc.
Volume of a sphere = (4/3) * π * Radius3 = (4/3) * π * 102.03 = 4445177.68 cubic units Surface Area of a sphere = 4 times the area of its great circle = 4 * π * Radius2 = 4 * π * 102.02 =130740.52 square units ![]() Great Circles and Small Circles
A great circle is a circular ring on the sphere, the centre for which, coincides with the centre of the sphere.
So the radius of the Great Circle is same as that for the sphere.
A small circle of a sphere, is a circle drawn on the sphere, with lesser radius than the sphere itself.
An example of a Zone of a Sphere or Frustum of a Sphere
![]() A zone or frustum of a sphere is The portion of a sphere intercepted between two parallel planes. Total surface area of a zone or frustum = 2 π R h + π r12 + π r22
Volume of a zone or frustum = (3r12 + 3r22 + h2) π h/6 Cubic Units
Consider two parallel planes cutting through the sphere. The first one cuts through 1 unit above the centre. The other one cuts though 2 units below the centre.
What is the volume and surface area of the frustum so formed?
(a) Let's compute the volume.
Applying the Pythagorean Theorem, r1 = √( R2 - h12) = √( 102.02 - 12) = 102.0 units
Again applying the Pythagoras Theorem, r2= √(R2 - h22) = √( 102.02 - 22) = 101.98 units
h = h1 + h2 = 3.0 units
Volume = (3r12 + 3r22 + h2) π h/6 = (3 * 102.02 + 3 * 101.982 + 32) * π * 3/6 Cubic Units = 98045.97 Cubic Units
Curved surface area of the zone = 2 π R h = 1922.65 Square Units
Area of the upper base = π r12 = 32681.99 = Square Units
Area of the lower base = π r22 = 32672.56 = Square Units
Total Surface Area = 2 π R h + π r12 + π r22 = 67277.21 = Square Units
Mensuration for a Spherical Cap
![]() What is the volume of a spherical cap of height h = 3 units?
As derived here, the required volume = (3R - h) πh2/3 = 2855.71 (substituting h = 3 units and R = 102.0 units)
Height of the geometric centroid above the centre of the sphere = (3 (2R - h) 2)/ 4 (3R - h) = 100.0 (substituting h = 3 units and R = 102.0)
Mensuration for a Hemisphere
![]() Let's cut the sphere into two hemispheres. What is the volume and total surface area, for either of the hemispheres?
Volume of the hemisphere = (2/3) * π * Radius3 = 2222588.84 cubic units
Surface Area of the hemisphere = Curved surface area + area of base = 2 * π * Radius2 + π * Radius2 = 3 π Radius2 = 98055.39 square units
What is the volume of material used in a sphere of radius 102.0 units a hollow sphere of thickness 2 units?
Inner radius = Outer Radius - Thickness. So the volume of the spherical gap inside = (4/3)π*(Inner Radius)3 cubic units
In that case, the volume of material required will be (4/3)π*(102.03 - (102.0-2)3 ) = 256387.47 cubic units
A demonstration of Scale Factors and SimilarityAll spheres are similar. Compare this current sphere (A) with a sphere B of radius 51 units. Ratio of radii of A and B = Scale Factor = 2:1, ratio of their surface areas = Square of scale factor = 4:1 and ration of their volumes = Cube of Scale Factors = 8:1. You may read more about scale factors here.Some more example(s): Geometric Properties of a sphere which is of radius 103: Properties like Surface Area, Volume and other aspects of mensuration. Geometric Properties of a sphere which is of radius 104: Properties like Surface Area, Volume and other aspects of mensuration. To understand more about the geometric features and properties of spheres, formulas related to mensuration and the principles of symmetry - you might find it useful to read the properties of a Sphere tutorial over here. Many of these concepts are a part of the Grade 9 and 10 Mathematics syllabus of the UK GCSE curriculum, Common Core Standards in the US, ICSE/CBSE/SSC and NTSE syllabus in India. You may check out our free and printable worksheets for Common Core and GCSE. |