### Sphere of radius 41 units: Volume,Area,Zones,Caps,Frustum etc

 You are given a sphere of radius 41.0 units.  Units may be any units of length: inches, cm, metres, feet, miles, km. So the radius could be any of 41.0 inches, 41.0 cm, 41.0 metres, 41.0 feet, 41.0 miles, 41.0 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 * 41.03 =  288695.61 cubic units Surface Area of a sphere = 4 times the area of its great circle = 4 * π * Radius2 =  4 * π * 41.02 =21124.07 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) = √( 41.02 - 12)  = 40.99 units Again applying the Pythagoras Theorem,  r2= √(R2 - h22) = √( 41.02 - 22) = 40.95 units h = h1 + h2 = 3.0 units Volume = (3r12 +  3r22 + h2) π h/6 = (3 * 40.992 + 3 * 40.952 + 32) * π * 3/6  Cubic Units = 15833.63  Cubic Units Curved surface area of the zone = 2 π R h = 772.83 Square Units Area of the upper base = π r12 = 5277.88 = Square Units Area of the lower base = π r22 = 5268.45 = Square Units Total Surface Area = 2 π R h +  π r12 +  π r22 = 11319.16 = 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  = 1130.97 (substituting h = 3 units and R = 41.0 units) Height of the geometric centroid above the centre of the sphere = (3 (2R - h) 2)/ 4 (3R - h) = 39.01  (substituting h = 3 units and R = 41.0) 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 = 144347.8 cubic units  Surface Area of the hemisphere = Curved surface area + area of base = 2 * π * Radius2 + π * Radius2 = 3 π Radius2 = 15843.05 square units What is the volume of material used in a sphere of radius 41.0 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*(41.03 - (41.0-2)3 ) = 40220.76 cubic units Some more example(s): Geometric Properties of a sphere which is of radius 42: Properties like Surface Area, Volume and other aspects of mensuration. Geometric Properties of a sphere which is of radius 43: 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.