DERIVATION OF AREA AND CIRCUMFERENCE OF CIRCLE USING CALCULUS
CIRCLE
DERIVATION OF AREA OF CIRCLE
LETS TAKE dπ‘ , A VERY SMALL ANGLE IN ABOVE FIGURE
SO; THE SMALL PART OF CIRCLE IN DIAGRAM
CAN BE CONSIDERED AS RIGHT ANGLED TRIANGLE
tan dπ‘ = ds/r
AS ANGLE IS VERY SMALL, tan dπ‘ = dπ‘
SO, dπ‘ =ds/r
SO, ds=r dπ‘ ..................(1)
AREA OF TRIANGLE= 1/2 * BASE *HEIGHT
= 1/2 * r * ds
D(A) = 1/2 * r *r dπ‘............. FROM (1)
IF WE INTEGRATE THESE SMALL TRIANGLES HAVING THIS AREA, WE WILL GET
AREA OF CIRCLE
SO AREA OF CIRCLE= ∫D(A)
2π
= ∫ 1/2* r * r dπ‘. [AS dπ‘ IS FROM 0 TO 2π ]
0
[AS 2π REPRESENTS ANGLE OF ENTIRE CIRCLE]
2π 2
= ∫ (1/2) r dπ‘.
0
2 2π
=(1/2 )* r * ∫ dπ‘
0
2
= [ r / 2 ] * 2π
2
= π r
SO, AREA OF CIRCLE IS π«R*R
CIRCUMFERENCE OF CIRCLE
AS WE KNOW, ds=r dπ‘ ............from 1
SO; CIRCUMFERENCE OF CIRCLE CAN BE OBTAINED BY ADDING ALL THESE LENGTHS
I.E INTEGRATING ds
SO C= ∫ ds
2π₯
C = ∫ r dπ‘
0
2π₯
C = r ∫ dπ‘
0
C= r * 2π₯
C= 2π₯r
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