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Math Help - Area of the surface

  1. #1
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    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
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  2. #2
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    Here.
    Attached Thumbnails Attached Thumbnails Area of the surface-picture5.gif  
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  3. #3
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    Post 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
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  4. #4
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    Quote Originally Posted by dadon View Post
    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
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  5. #5
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    Hello, dadon!

    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?

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