Posted by **Nycw** on Friday, February 1, 2013 at 10:15pm.

The curve y=sinh(x),0<=x<=1, is revolved about the x-axis. Find the area of the resulting surface.

- calculus 2 -
**Eman**, Friday, February 1, 2013 at 10:32pm
Revolved about the x-axis~ 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)

- calculus 2 -
**Reiny**, Friday, February 1, 2013 at 10:35pm
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

- totally ignore my post - calculus 2 -
**Reiny**, Friday, February 1, 2013 at 10:37pm
scrap my reply,

I found the area, silly me

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