How do you find the surface area of the solid obtained by rotating about the #x#-axis the region bounded by #9x=y^2+18# on the interval #2<=x<=6# ?
The surface area can be found by
Let us look at some details.
By taking the derivative,
As x goes from 2 to 6, y goes from 0 to 6.
Since the surface area of revolution can be found by
As y goes from 0 to 6, u goes from 1 to 25/9.