surface area of y=c+acosh(x/a)

M

monokill

New member
Joined
Mar 2, 2008
Messages
6
The Question is:

Find the area of the surface obtained by rotating the curve about the x-axis.

y=c+acosh(x/a), 0?x?a

------------------------------------------
I know that for y = cosh(x) I have to do:

1+(dy/dx)^2 = 1+sinh(x)^2 = cosh(x)^2
S=2pi x integral of coshxcoshx dx = 2pi x integral of 1/2(1+cosh2x) = pi{x+1/2sinh(2x)} with w/e limits.
------------------------------------------

But I don't know how to solve the question given. ><
 
You have the limits, [0,a]. A quick examination shows the function strictly increasing on the interval. That makes it simpler.

Seems to me there is a radius missing. You don't believe you will get the same result for c = 0 and c = 10, do you?
 
I tried it, can someone check my answer? My answer looks so complicated.

S = 2 pi integral:{ [c+acosh(x/a)]cosh(x/a) dx } from 0 to a

= 2 pi c integral:{cosh(x/a)} from 0 to a + 2 pi a integral: {cosh(x/a)^2} from 0 to a

then i subbed u=x/a du=1/a

= 2 pi ca integral:{cosh(u/a)} from 0 to a + 2 pi a^2 integral: {cosh(u/a)^2} from 0 to a

when x=a, a/a=1=u
when x=0, 0/a=0=u

=2 pi ca[sinhu]from 0 to 1 + 2 \pi a^2 integral:{1/2(1+cosh2u)} from 0 to a
(will use 0 to 1 as limits after integration)

=2 pi ca[sinhu] from 0 to 1 + pi a^2 integral:{1+cosh2u} from 0 to a

=2 pi ca[sinhu] from 0 to 1 + pi a^2[u+1/2sinh2u] from 0 to 1

=2 pi ca[sinh1-sinh0] + pi a^2[1+1/2sinh2-1/2sinh0]

=2 pi ca sinh1 + pi a^2(1+1/2sinh2)
=2 pi ca ((e-e^(-1))/2) +pi a^2(1+1/2 ((e^2-e^(-2))/2)
=pi ca(e-e^(-1)) + pi a^2(1+(e^2-e^(-2))/2)

I hope im right
 
You must log in or register to reply here.
Top