Expert: Abe Mantell - 6/8/2006

Question
I have been doing some study for an upcoming test and i have got stuck  on one to do with the area of surfaces of revolution. The question is:

Find the curved surface area of the frustum formed by rotating the segment of the line y=2x+3
between x=1 and x=3 about the x axis.
there is a hint saying: first show that ds/dx =root 5 and then evaluate the intergral of 2 pi y ds/dx dx frpm 1 to 3.

Your help would be greatly appreciated.
Thank you for your time

Answer
Hello Hayley,

Check these sites for some nice explanations and diagrams:
http://curvebank.calstatela.edu/arearev/arearev.htm
http://tutorial.math.lamar.edu/AllBrowsers/2414/SurfaceArea.asp

Thus, with y=2x+3, y'=2, so ds/dx=sqrt(1+2^2)=sqrt(5)
So the surface area is:
Integral(2pi(2x+3)*sqrt(5)*dx from x=1 to x=3)

I trust trust that you can take it from there...yes?

Let me know if you need more help....have fun!  ;-)

Abe