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Integration of an ellips

  1. Nov 21, 2011 #1
    Hi

    Im trying to estimate a semicircle or ellips of a kind with an integral.
    And right now i'm trying to get the integral of a ellips.

    I need to integrate the equation


    [itex]x(\theta)[/itex] = [itex]\int_0^{\pi/2} \frac {d\theta}{a\sqrt{1-sin^2(\theta)/b^2}}[/itex]

    I tried http://integrals.wolfram.com/index.jsp?expr=1/sqrt(1-sin^2(x)/b^2)&random=false" but i dont understand the result or how to integrate the result.

    Thanks for any help
     
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Nov 21, 2011 #2

    dextercioby

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    Science Advisor
    Homework Helper

    That's one of the reasons elliptic integrals are called <special functions>, because they can't be written in terms of 'elementary' functions, such as polynomials, sin, cos, e^x, ln, sinh, cosh, ...The perimeter of an ellipse is an elliptic integral.
     
  4. Nov 21, 2011 #3
    so there is no way for me to just but in the boundaries? i want to implement this in a matlab function, later on does it mean that i have to use a nummerical method to solve it then?

    and what do you do if you if you have a circle is that also impossible to solve? because the only difference is the a and b constants.
     
  5. Nov 21, 2011 #4

    D H

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    Staff Emeritus
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    The special function you want is already implemented in Matlab. It's called ellipke in Matlab.
     
  6. Nov 21, 2011 #5

    dextercioby

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    The circle is ok. It's a standard integral solvable by a substitution.
     
  7. Nov 21, 2011 #6
    I think i might have made a mistake when substituting.

    I have
    [itex]

    \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1

    [/itex]

    And i substitute by

    [itex]
    y = sin(\theta)
    x = cos(\theta)
    [/itex]


    So when i do the substitution should i include that [itex]dy = cos(\theta)d\theta [/itex] or can i just substitute
    [itex]
    x = \sqrt{a^2 + \frac{a^2}{b^2}y^2} → \sqrt{a^2 + \frac{a^2}{b^2}sin(\theta)^2}
    [/itex]

    and then integrate as [itex] x = \int_0^b\frac{dy}{f(y)} [/itex] or am i doing some big mistakes?
     
  8. Nov 21, 2011 #7
    I think i might have made a mistake when substituting.

    I have
    [itex]

    \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1

    [/itex]

    And i substitute by

    [itex]
    y = sin(\theta)
    x = cos(\theta)
    [/itex]


    So when i do the substitution should i include that [itex]dy = cos(\theta)d\theta [/itex] or can i just substitute
    [itex]
    x = \sqrt{a^2 + \frac{a^2}{b^2}y^2} → \sqrt{a^2 + \frac{a^2}{b^2}sin(\theta)^2}
    [/itex]

    and then integrate as [itex] g = \int_0^b\frac{dy}{c-f(y)} [/itex]

    where c is a fixed length and f(y) is the function x. or am i doing some big mistakes?


    What i wanna do is calculate the width of a pillar at the top as it is melding together with the roof. I assume that the arc is in the form of an ellipse so i have

    c = the length of the pillar at the top.
    b = the length between when the pillar starts curving to the roof
    a = c - half of the pillar length

    g = c - x(y)

    and then integrate as 1/g.

    and then multiple by to get the other side.
     
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