The hight from vertex of cone to upper surface of disk is equal to h₁ and the hight from vertex of cone to lower surface of disk is equal to h₂ .
Then we find hight of circular disk is equal to (h₂ - h₁ ).
By unitary method we find hight of this circular disk. Because radius is leneraly increasing with hight from vertex.

When radius R unit then hight will be H unit
radius hight
R H
1 H/R
r₁ (H/R)r₁ = h₁
Similarly r₂ (H/R)r₂ =h₂ then,
(h₂ - h₁ ) = (H/R)r₂ - (H/R)r₁
= H/R(r₂ - r₁ )

Let the disk be very very thin so that r₁ nearly equal to r₂ and h₁ nearly equal to h₂
As r₁ nearly equal to r₂ . This disk is similar to a right circular cylender.
It's volume is equal to average base area x hight. Where average base area = (22/7)r² unit.
Area of disk = (22/7)r²unit
= 3(22/7)r²/3
= (22/7)/3[3r²]
= (22/7)/3[r₁² + r₂² +r₁ x r₂]
Volume of disk = area of disk x hight of disk
= (22/7)/3[r₁² + r₂² + r₁ r₂] x (h₂ - h₁) unit
= (22/7)/3[r₁² + r₂² + r₁ r₂] (r₂ - r₁)H/R
= (22/7)/3 H/R (r₂ - r₁) (r₁² + r₂² + r₁ r₂)
= (22/7)/3 H/R (r₂³ - r₁³)

As like this disk we cut many more small circular disk. Sum of all these small disk we find volume of cone.
So now we write,
(22/7)/3 H/R (r₂³ - r₁³) + (r₃³ - r₂³) + (r₄³ - r₃³) +---------+(r_n³ - r_n-1³)
Where r_n = R
r₁=0 because r₁ is the radius of the vertex point of cone where radius is 0
=(22/7)/3 H/R(-r₁³ + r_n³)
=(22/7)/3 H/R(0+R³)
=(22/7)/3 H.R³/R = (22/7)R²H/3
Volume of cone = (22/7)R²H/3 unit
Volume of cone = [(22/7) x 2² x 1]/3 unit
Volume of cone = (12.571428571429 x 1)/3 unit
Volume of cone = 4.1904761904762 unit