Integrating an Ellipse: Calculating the Width of a Melding Pillar and Roof

[Mamed]

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


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

I tried http://integrals.wolfram.com/index.jsp?expr=1/sqrt(1-sin^2(x)/b^2)&random=false" but i don't understand the result or how to integrate the result.

Thanks for any help

 

[dextercioby]

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.

 

[Mamed]

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.

 

[D H]

The special function you want is already implemented in Matlab. It's called ellipke in Matlab.

 

[dextercioby]

The circle is ok. It's a standard integral solvable by a substitution.

 

[Mamed]

I think i might have made a mistake when substituting.

I have
<br /> <br /> \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1<br /> <br />

And i substitute by

<br /> y = sin(\theta)<br /> x = cos(\theta)<br />


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

and then integrate as x = \int_0^b\frac{dy}{f(y)} or am i doing some big mistakes?

 

[Mamed]

I think i might have made a mistake when substituting.

I have
<br /> <br /> \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1<br /> <br />

And i substitute by

<br /> y = sin(\theta)<br /> x = cos(\theta)<br />


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

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

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


What i want to 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.