# Thread: Area of the surface

1. ## Area of the surface

Hey all,

How would you solve the following problem:

The arc of the catenary y=5coshx/5 between x=0 and x=5 rotates about OX. Find the area of the surface so generated.

Thank you

2. Here.

3. ## re;

Thanks for that TPH,

does the integral workout to 70.34π ? as i have tried and do not seem to get the answer.

Thanks again

Thanks for that TPH,

does the integral workout to 70.34π ? as i have tried and do not seem to get the answer.

Thanks again
No it is more like 175*pi

I'm afraid that ThePerfectHacker found the volume.

The arc of the catenary, y = 5·cosh(x/5) from x = 0 to x = 5

is rotated about the x-axis. .Find the area of the surface so generated.

The Area formula for a surface of revolution is given by:
. . . . . . . . . . . . . .___________
. . . S . = . 2π ∫ y √1 + (dy/dx)² dx

We have: .y .= .5·cosh(x/5)

. . .Then: .dy/dx .= .sinh(x/5)

. . . And: .1 + (dy/dx)² .= .1 + sinh²(x/5) .= .cosh²(x/5)
. . . . . . . . .___________ . . - . _________
. . Hence: .√1 + (dy/dx)² . = . √cosh²(x/5) . = . cosh(x/5)

Substitute: . S . = .
5·cosh(x/5)·cosh(x/5) dx . = . 10π cos²(x/5) dx

Using a double-angle identity, we have: . S . = .
[1 + cos(2x/5)] dx
. . with limits [0, 5]

Can you finish it now?