Arc Length and Surface question about hyperbolic function

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Homework Help Overview

The discussion revolves around a homework problem involving the area of a surface obtained by rotating the curve y=cosh(x) between x=a and x=-a around the x-axis. The circumference of the region bounded by this curve and the lines y=0, x=a, and x=-a is given as 2a+4, where a>0.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to find the arc length and surface area, expressing concerns about the lack of numerical results in their answer. They initially derive an equation involving sinh(2a) and question its relation to the given circumference.
  • Some participants question the correctness of the original poster's interpretation of the arc length and its connection to the circumference, suggesting a potential misunderstanding of the problem statement.
  • There is a clarification regarding the arc length formula, with one participant correcting the original poster's expression from sinh(2a) to 2sinh(a).

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem and clarifying the relationships between the arc length, circumference, and surface area. There is no explicit consensus yet, but participants are actively engaging with the problem and correcting each other's statements.

Contextual Notes

Participants are navigating through the implications of the problem statement and the relationships between the various mathematical expressions involved. The original poster expresses uncertainty about their results and the correctness of their approach, indicating a need for further clarification.

e179285
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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...
 
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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.
 

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