# Homework Help: Arc Length and Surface question about hyperbolic function

1. May 2, 2012

### e179285

If the circumference of the region bounded by the curve y=cosh(x) and the lines y=0 x=a

and x=-a is 2a+4, where a>0 find the area of the surface obtained by rotating the part of

the curve y=cosh(x) between x=a x=-a and around the x axis.

This is my homework question.I tried to solve it.I get a result but ı'm not sure because

there is not number in my answer and this is area question.ı want to say my approach to

this question.

Firstly,ı found arc length interms of a. this is ea-e-a = 2a+4 ,i.e,
sinh(2a)=a+2

Secondly,ı found surface.∫2∏cosh(x)√1+sinh2(x) dx (from -a to a)

After calculations,S=sinh(2a)+a∏/2

Did ı do wrong something when solving the question,the answer is strange...

2. May 2, 2012

### LCKurtz

Are you sure you don't get $2\sinh(a)$ and not $\sinh(2a)$? And it isn't that that is equal to $2a+4$. It is the circumference of the rotated region that is equal to $2a+4$. That may change things.

3. May 2, 2012

### e179285

I'm sorry,ı made a mistake by writing value.It will be 2sinh(a),but ı don't understand why it is not equal to 2a+4

4. May 2, 2012

### LCKurtz

Read the statement of the problem above.