calculus 2
posted by Nycw .
The curve y=sinh(x),0<=x<=1, is revolved about the xaxis. Find the area of the resulting surface.

Revolved about the xaxis~ must use: integral 2*Pi*y ds (Eq 1)
For ds use:
ds = sqrt(1+[F'(x)]^2) dx (Eq 2)
sinh(x) dx = cosh(x) = F'(x)
sub cosh(x) into Eq 2 and you get:
ds = sqrt(1+[cosh(x)]^2) dx
now sub ds into Eq 1
integral 2*Pi*y*sqrt(1+[cosh(x)]^2) dx
We are taking the derivative with respect to x so change y to it's equivalent.
integral 2*Pi*sinh(x)*sqrt(1+[cosh(x)]^2) dx, from 0 to 1
you now have your final equation to integrate over the interval, 0 to 1
and you get 5.52989 (final answer) 
by definition
sinh(x) = (e^x  e^x)/2
so the area
= ∫sinh(x) dx from 0 to 1
= ∫(e^x  e^x)/2 dx from 0 to 1
= [ e^x + e^x)/2] from 0 to 1
= (e^1 + e^1)/2  (e^0 + e^0)/2
= (e + 1/e)/2  (1+1)/2
= e/2 + 1/(2e)  1
= (e^2 + 1  2e)/(2e)
= (e  1)^2 / (2e)
you better check my arithmetic 
scrap my reply,
I found the area, silly me