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  3. 대수 및 대수기하 Algebra & Algebraic Geometry

Volume of a body made by a cube (regular hexadron) rotated along its diagonal

작성자 Uploader : 냉면중독 작성일 Upload Date: 2018-08-31변경일 Update Date: 2018-08-31조회수 View : 438

Volume of the body made by rotating a cube (regular hexadron) with a side length being "a", as shown in figure, along its diagonal is calculated as the following:

Letting the line segment AG as X axis, length of AG is 3^(1/2)*a.

The body of rotation is symmetrical with
respect to AG/2 and its volume is estimated by making double the integration to AG/2.

Up to AG/3, the cross section is an equilateral triangle, and that from AG/3 to AG/2 is an hexagon. Thus, integration is calculated by this segmentation.

Volume of the rotated body is obtained by integrating the cross section of π*r^2. Then it can be obtained by expressing the "r" as a function of x and integrating,


□ 0 ~ AG/3 segment (0 <= x <= (1/3)^(1/2)*a)  
where the cross section is an equilateral
triangle of which median line length multipled by 2/3 makes the largest circle.

For an equilateral triangle of its side length being "a", the distance from its center to its vertex is :
(3^(1/2))/2 * (2/3) * a = (1/3)^(1/2) * a

By x=(1/3)^(1/2)*a, r is equal to the distance between the center and vertex of a equilateral triangle of which side length is 2^(1/2)*a, then,

r= (1/3)^(1/2) * 2^(1/2)*a = (2/3)^(1/2) * a

Establishing an proportional relationship between x and r,

r/x = 2^(1/2),

Then,

r = 2^(1/2) * x

By integrating, we get,

V1 = π∫2*x^2 dx = 2π*(1/3)*x^3, (x=0 ~ x=(1/3)^(1/2)*a)

V1 = 2πa^3 / (9*3^(1/2))  


□ AG/3 ~ AG/2 segment ((1/3)^(1/2)*a <= x <= 3^(1/2)*a/2 )
where the cross section is an hexagon, the point B moves to the point I.

Length of the line segment IJ is constant at (1/2)^(1/2) * a. Expressing h as a function of x, we get r :

From x=(1/3)^(1/2)*a,
h=BJ/2 =(1/6)^(1/2)*a

At x=(3^(1/2))/2,
h=0,  

Then,
h = -(2)^(1/2)*x + (3/2)^(1/2)*a

r^2 = (1/2)*a^2 + 2*x^2 - 2*3^(1/2)*a*x + (3/2)*a^2 = 2*( x^2 - 3^(1/2)*a*x + a^2)

Integrating the above,

V2 = π∫2*( x^2 - 3^(1/2)*a*x + a^2) dx  
 = 2π*((1/3)*x^3 - (1/2)*3^(1/2)*a*x^2 + a^2*x), (x=(1/3)^(1/2)*a ~ x=3^(1/2)*a/2)

V2 = 2π*a^3 { ( 3^(1/2) / 8 - 3*3^(1/2) / 8 + 4 / 8 )
           - ( 1 / (9*3^(1/2)) - 3^(1/3) / 6 + 1 / 3^(1/2) ) }
 = 2π*a^3 { 3^(1/2) / 4 - (2-9+18) / (18*3^(1/2)) }
 = 2π*a^3 ( (9*3 - 22) / (36*3^(1/2)) )
 = 2π*a^3 * 5 / (36*3^(1/2))

□ 2*V1*V2

V = 2*V1*V2 = 4π*a^3*( 1/(9*3^(1/2)) + 5 / (36*3^(1/2)) )
= 4π*a^3*(4+5) / (36*3^(1/2))  
= π*a^3 / 3^(1/2)

*** 참고문헌[References] ***
V = π*a^3 / 3^(1/2)
변수명 Variable 변수값 Value 변 수 설 명 Description of the variable


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