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 * Radius ^{3} = (4/3) * π * 59.0^{3} = 860289.54 cubic unitsSurface Area of a sphere = 4 times the area of its great circle = 4 * π * Radius ^{2 }= 4 * π * 59.0^{2 }=43743.54 square units
Great Circles and Small CirclesA 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 SphereA 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 r_{1}^{2} + π r_{2}^{2}Volume of a zone or frustum =
(3r_{1}^{2} + 3r_{2}^{2 }+ h^{2}) π h/6 Cubic UnitsConsider 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,
r_{1} = √( R√(^{2} - h_{1}^{2})^{ }= 59.0^{2} - 1^{2})^{ }= 58.99 unitsAgain applying the Pythagoras Theorem,
r_{2}= √(R^{2} - h_{2}^{2})^{ }= √( 59.0^{2} - 2^{2})^{ }= 58.97 unitsh = h_{1} + h_{2 }= 3.0 units_{
}Volume = (3r_{1}^{2} + 3r_{2}^{2 }+ h^{2}) π h/6 = (3 * 58.99^{2} + 3 * 58.97^{2} + 3^{2}) * π * 3/6 Cubic Units = 32798.23 Cubic UnitsCurved surface area of the zone = 2 π R h = 1112.12 Square UnitsArea of the upper base = π r_{1}^{2} = 10932.74 = Square UnitsArea of the lower base = π r_{2}^{2 }= 10923.32 = Square UnitsTotal Surface Area = 2 π R h + π r_{1}^{2} + π r_{2}^{2 }= 22968.18 = Square Units_{
}Mensuration for a Spherical CapWhat is the volume of a spherical cap of height h = 3 units?
As derived here, the required volume =
(3R - h) πh (substituting h = 3 units and R = 59.0 units)^{2}/3 = 1639.91Height of the geometric centroid above the centre of the sphere =
(3 (2R - h) (substituting h = 3 units and R = 59.0)^{2})/ 4 (3R - h) = 57.0 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 * Radius
^{3 }= 430144.77 cubic units Surface Area of the hemisphere = Curved surface area + area of base = 2 * π * Radius
^{2 }+ π * Radius^{2 }= 3 π Radius^{2 }= 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 unitsIn that case, the volume of material required will be (4/3)π*(59.0
^{3} - (59.0-2)^{3} ) = 84554.92 cubic unitsSome 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. |