Derivation of Geometric Formulas
Date: 5/29/96 at 18:35:16 From: Anonymous Subject: Derivation of Geometric Formulas Could you please answer these questions? Your help would be very much appreciated. How were the formulas for surface area, total surface area, and volume of a sphere derived? How were the formulas for the volume of a pyramid and cone derived? Please give an explanation of the answers to the above questions and include equations if there are any. Thank you. SwimDuck
Date: 5/31/96 at 11:45:38 From: Doctor Anthony Subject: Re:Derivation of Geometric Formulas In the case of the surface area of a sphere, consider an elementary element of arc at position (a,theta) (in polar coordinates) rotated through 360 degrees about the x axis, thereby forming a hoop of radius a.sin(theta) and surface area a.d(theta).2.pi.a.sin(theta) = 2.pi.a^ 2.sin(theta).d(theta) To get the total surface area we integrate this between theta = 0 and pi: Surface area = 2.pi.a^2.INT[sin(theta).d(theta)] between 0 and pi. = 2.pi.a^2.[-cos(theta)] between 0 and pi = -2.pi.a^2[-1 - 1] = 4.pi.a^2 For volume consider the circle x^2 + y^2 = a^2, and take a slice of thickness dx, and radius y (= sqrt(a^2-x^2)). Rotate this about the x axis to get a disk of volume pi.y^2.dx. Integrate between x = -a and +a: Volume = pi.INT[(a^2 - x^2).dx] = pi.[a^2.x - (1/3)x^3] between -a and +a = pi.[a^3 - (1/3)a^3 + a^3 - (1/3)a^3] = pi.[2a^3 - (2/3)a^3] = pi.[(4/3)a^3] For volume of cone of height h and base radius a, consider the line y = (a/h)x. By rotating this about the x axis you will generate the cone. The volume of an elementary disk of thickness dx and radius y is pi.y^2.dx = pi.(a^2/h^2)x^2.dx Integrate between 0 and h: Volume = pi.INT[(a^2/h^2)x^3/3] between 0 and h = pi.(a^2/h^2)(h^3/3) = (1/3).pi.a^2.h In the case of a pyramid, you take slices, parallel to the base, of thickness dx and area calculated by ratio of similar rectangles or square with the shape of the base. You get the same formula as for the cone, namely(1/3).area of base times perpendicular height. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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