### Sphere of radius 49 units: Volume,Area,Zones,Caps,Frustum etc

 You are given a sphere of radius 49.0 units.  Units may be any units of length: inches, cm, metres, feet, miles, km. So the radius could be any of 49.0 inches, 49.0 cm, 49.0 metres, 49.0 feet, 49.0 miles, 49.0 km, etc.  A Sphere is a 3D locus of points which are all equidistant from the centre of the sphere.  Volume of a sphere = (4/3) * π * Radius3 =  (4/3) * π * 49.03 =  492806.98 cubic units Surface Area of a sphere = 4 times the area of its great circle = 4 * π * Radius2 =  4 * π * 49.02 =30171.86 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) = √( 49.02 - 12)  = 48.99 units Again applying the Pythagoras Theorem,  r2= √(R2 - h22) = √( 49.02 - 22) = 48.96 units h = h1 + h2 = 3.0 units Volume = (3r12 +  3r22 + h2) π h/6 = (3 * 48.992 + 3 * 48.962 + 32) * π * 3/6  Cubic Units = 22619.47  Cubic Units Curved surface area of the zone = 2 π R h = 923.63 Square Units Area of the upper base = π r12 = 7539.82 = Square Units Area of the lower base = π r22 = 7530.4 = Square Units Total Surface Area = 2 π R h +  π r12 +  π r22 = 15993.85 = 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  = 1357.17 (substituting h = 3 units and R = 49.0 units) Height of the geometric centroid above the centre of the sphere = (3 (2R - h) 2)/ 4 (3R - h) = 47.01  (substituting h = 3 units and R = 49.0) 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 = 246403.49 cubic units  Surface Area of the hemisphere = Curved surface area + area of base = 2 * π * Radius2 + π * Radius2 = 3 π Radius2 = 22628.89 square units What is the volume of material used in a sphere of radius 49.0 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*(49.03 - (49.0-2)3 ) = 57914.21 cubic units Some more example(s): Geometric Properties of a sphere which is of radius 50: Properties like Surface Area, Volume and other aspects of mensuration. Geometric Properties of a sphere which is of radius 51: 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.