A right circular cone is generated by revolving the region bounded by `y=3x/4, y=3, x=0` about the y-axis. Find the lateral surface area of the cone.

Surface area (S) obtained by rotating the curve x=g(y), c `<=`  y `<=` d about y-axis is,

S=`int2pixds`

 where, ds=`sqrt(1+(dx/dy)^2)dy`

We are given `y=(3x)/4 , y=3 , x=0`

`y=(3x)/4`

`=>x=(4y)/3`

`dx/dy=4/3`

`S=int_0^3(2pi)xds`

`S=2piint_0^3((4y)/3)sqrt(1+(4/3)^2)dy`

`S=2piint_0^3(4y)/3sqrt(1+16/9)dy`

`S=2piint_0^3(4y)/3sqrt(25/9)dy`

`S=2pi(4/3)(5/3)int_0^3ydy`

`S=(40pi)/9[y^2/2]_0^3`

`S=(40pi)/9[3^2/2-0^2/2]`

`S=(40pi)/9(9/2)`

`S=20pi`

So the Lateral surface area of the cone is `20pi`

Surface area (S) obtained by rotating the curve x=g(y), c `<=`  y `<=` d about y-axis is,

S=`int2pixds`

 where, ds=`sqrt(1+(dx/dy)^2)dy`

We are given `y=(3x)/4 , y=3 , x=0`

`y=(3x)/4`

`=>x=(4y)/3`

`dx/dy=4/3`

`S=int_0^3(2pi)xds`

`S=2piint_0^3((4y)/3)sqrt(1+(4/3)^2)dy`

`S=2piint_0^3(4y)/3sqrt(1+16/9)dy`

`S=2piint_0^3(4y)/3sqrt(25/9)dy`

`S=2pi(4/3)(5/3)int_0^3ydy`

`S=(40pi)/9[y^2/2]_0^3`

`S=(40pi)/9[3^2/2-0^2/2]`

`S=(40pi)/9(9/2)`

`S=20pi`

So the Lateral surface area of the cone is `20pi`