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