Sphere of radius 59 units: Volume,Area,Zones,Caps,Frustum etc

You are given a sphere of radius 59.0 units. 
Units may be any units of length: inches, cm, metres, feet, miles, km. So the radius could be any of 59.0 inches, 59.0 cm, 59.0 metres, 59.0 feet, 59.0 miles, 59.0 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 * 59.03 =  860289.54 cubic units
Surface Area of a sphere = 4 times the area of its great circle = 4 * π * Radius2 =  4 * π * 59.02 =43743.54 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) = √( 59.02 - 12)  = 58.99 units

Again applying the Pythagoras Theorem,  r2= √(R2 - h22) = √( 59.02 - 22) = 58.97 units

h = h1 + h2 = 3.0 units

Volume = (3r12 +  3r22 + h2) π h/6 = (3 * 58.992 + 3 * 58.972 + 32) * π * 3/6  Cubic Units = 32798.23  Cubic Units

Curved surface area of the zone = 2 π R h = 1112.12 Square Units
Area of the upper base = π r12 = 10932.74 = Square Units
Area of the lower base = π r22 = 10923.32 = Square Units
Total Surface Area = 2 π R h +  π r12 +  π r22 = 22968.18 = 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  = 1639.91 (substituting h = 3 units and R = 59.0 units)

Height of the geometric centroid above the centre of the sphere = (3 (2R - h) 2)/ 4 (3R - h) = 57.0  (substituting h = 3 units and R = 59.0)



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 = 430144.77 cubic units 
Surface Area of the hemisphere = Curved surface area + area of base = 2 * π * Radius2 + π * Radius2 = 3 π Radius2 = 32807.65 square units



What is the volume of material used in a sphere of radius 59.0 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*(59.03 - (59.0-2)3 ) = 84554.92 cubic units


Some more example(s):


Geometric Properties of a sphere which is of radius 60: Properties like Surface Area, Volume and other aspects of mensuration.



Geometric Properties of a sphere which is of radius 61: 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.
 


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