Question

Could someone please give me derivation for volune of frustrum of a cone???

Answer 1

the

R = Radius of larger cone, H = height of larger cone

r = radius of smaller cone, h = height of frustum

So, height of smaller cone = H-h

Volume( Cone ABC) = 1/3 pi R² H………(1)

Volume ( Cone ADE) = 1/3 pi r² (H-h) ………..(2)

Volume( frustum DBCE) =

1/3 pi R² H — 1/3 pi r² (H-h) ………. (3)

Now, here we require only h but not H, as h is the height of frustum. So we try to eliminate H by similar triangles property.

Tri AFE ~ tri AGC ( 2 right triangles are similar by AAA similarity criterion) F & G are centres of bases of frustum

=> (H-h)/r = H/R ( corresponding sides of similar triangles)

=> HR - hR = Hr

=> HR - Hr = hR

=> H( R-r) = hR

=> H = hR/(R-r)

By putting the value of H in eq (3)

Volume of frustum DBCE =

= 1/3 pi R² hR/(R-r) — 1/3 pi r² {hR/(R-r) — h }

= 1/3 pi R^3* h/(R-r) —1/3 pi r²(hR-hR+hr)/(R-r)

= 1/3 pi R^3*h/(R-r) — 1/3 pi r^3*h/(R-r)

= 1/3 pi h/(R-r) * (R^3 - r^3)

=1/3 pi h/(R-r) * (R-r) ( R² + r² + R.r ) ( by applying identity

a^3 - b^3 = ( a-b) ( a² + ab + b²)

So, Volume of frustum =

V=1/3πH(R2+Rr+r2)