Unizor - Geometry3D - Spherical Sectors

Unizor - Creative Minds through Art of Mathematics - Math4Teens

Notes to a video lecture on http://www.unizor.com

Spherical Sector

Consider a spherical cap that is a part of a sphere of radius R with a center O.
Connect each point of a base circle of a cap with center O by a radius of a sphere. All these radiuses form a conical surface with a base circle as a directrix.
An original spherical cap and a cone formed as described above constitute a spherical sector.


A spherical sector is defined by two parameters - radius of a main sphere R, from which it is a part, and the height of a cap on its top H. For "true" spherical sectors, the ones we will be considering here, height H is smaller or equal to radius R of a main sphere.

Since perpendicular from a center of a sphere onto a base circle of a cap falls into a center of this circle, our cone is a right circular one - the only type we considered in this course.

Volume of a Sector

The radius of the base circle (in terms of the radius of a main sphere R and the height of a spherical cap H) equals, by Pythagorean Theorem, to
L = √R²−(R−H)² = √2RH−H²
This is a cone's radius.
The cone's height is, obviously, R−H.
So, the cone's volume is
Vcone = πL²(R−H)/3 =
= π(2RH−H²)(R−H)/3 =
= π(2R²H−3RH²+H³)/3

As we know from the previous lecture about spherical caps, the volume of a cap is
Vcap = = πH²(3R−H)/3

From the two formulas for volumes of a cap and a cone we can determine the volume of a spherical sector:
Vsector = = πH²(3R−H)/3 +
+ π(2R²H−3RH²+H³)/3 =
= 2πR²H/3
A very short formula indeed!

Just for checking, if H=R, our spherical sector occupies exactly half a sphere. The volume of a sphere, as we have determined before, is 4πR³/3.
So, half a sphere has a volume 2πR³/3.
This is exactly a formula we get from the volume of a spherical sector if H=R.

Area of a Dome

Recall the technique we used to calculate the area of a sphere Ssphere, knowing its volume Vsphere.
We considered a sphere's surface as approximated by an infinitely large number of inscribed polyhedrons, whose faces are infinitely small. Then the volume of a sphere would be approximated by a sum of volumes of all the pyramids obtained by connecting each vertex of each face of each polyhedron with a center of a sphere.
Then the volume of a sphere Vsphere would be equal to an area of its surface Ssphere multiplied by one third of its radius R:
Vsphere = Ssphere·R/3.
This consideration allowed us to derive from a formula for the volume of a sphere -
Vsphere=4πR³/3
- its surface area:
Ssphere=4πR².

Exactly the same considerations connect the area of a dome with a volume of a spherical sector: Vsector = Sdome·R/3.
Therefore,
2πR²H/3 = Sdome·R/3
Now we can calculate the area of a dome:
Sdome = 2πRH