Question

This is the questions I have trouble with :

Set up (do not evaluate) the integral that gives the surface area of the surface generated by rotating the curve y=tanhx on the interval (0, 1) around the x-axis.

Anyone who can help? Not really sure how to even begin!
Thanks

Answer 1

Surface area will be y dx dTheta

from x 0 to 1, and theta from 0 to 2PI


Area= Int (Int) tanhx dx dtheta

Answer 2

Area=2Pi(int from 0 to 1) y*sqrt(1+(y')^2)dx

Answer 3

Thanks so much! :)