Integration of an ellips

  • Thread starter Mamed
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  • #1
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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
 
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Answers and Replies

  • #2
dextercioby
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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.
 
  • #3
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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.
 
  • #4
D H
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The special function you want is already implemented in Matlab. It's called ellipke in Matlab.
 
  • #5
dextercioby
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The circle is ok. It's a standard integral solvable by a substitution.
 
  • #6
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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?
 
  • #7
17
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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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