Arc Length and Surface question about hyperbolic function

e179285
Messages
24
Reaction score
0
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...
 
Physics news on Phys.org
e179285 said:
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

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.
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...
 
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
 
e179285 said:
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,

e179285 said:
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

Read the statement of the problem above.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top