Register to reply

Circumference of an ellipse

by Libertine
Tags: circumference, ellipse
Share this thread:
May23-06, 03:15 PM
P: 17
An ellipse has an equation which can be written parametrically as:
x = a cos(t)
y = b sin(t)

It can be proved that the circumference of this ellipse is given by the integral:
[tex]\int^{2\pi}_0 \sqrt{a^2 \sin^2 t + b^2 \cos^2 t} \ \ dt[/tex]

Prove that, if [tex]a=r(1+c)[/tex] and [tex]b=r(1-c)[/tex], where c is a positive number small enough for powers higher than [tex]c^2[/tex] to be neglected, then this circumference is approximately:
[tex]2 \pi r (1+\frac{1}{4}c^2)[/tex]

So I substituted in the expressions for a and b:
[tex]$ \int^{2\pi}_0 \sqrt{a^2 \sin^2 t + b^2 \cos^2 t} \ \ dt \\
=\int^{2\pi}_0 \sqrt{(r(1+c))^2 \sin^2 t + (r(1-c))^2 \cos^2 t} \ \ dt \\
=r \int^{2\pi}_0 \sqrt{\sin^2 t + 2c \sin^2 t + c^2 \sin^2 t + \cos^2 t - 2c \cos^2 t + c^2 \cos^2 t} \ \ dt \\
=r \int^{2\pi}_0 \sqrt{(\sin^2 t + \cos^2 t)+ c^2( \sin^2 t + \cos^2 t) + 2c(\sin^2 t - \cos^2 t)} \ \ dt \\
=r \int^{2\pi}_0 \sqrt{1 + c^2 + 2c(\sin^2 t - \cos^2 t)} \ \ dt \\ $
After this point, I seem to hit a brick wall and can't simplify it any further or factorise to get rid of that annoying square root. Any help appreciated.

(I'm assuming the tex won't come out all right first time, so I'll be trying to correct it for a little while)
Phys.Org News Partner Science news on
Scientists discover RNA modifications in some unexpected places
Scientists discover tropical tree microbiome in Panama
'Squid skin' metamaterials project yields vivid color display
May23-06, 08:51 PM
P: n/a
Although I can't explicitly tell you how to help, perhaps approximating the square root function in the integral will get you somewhere. I see one or two term Taylor expansions justify equations in classes all the time as "approximations" :) ( )

Edit: It was worth a shot. Came out to 2*pi*r(1+0.5*c^2) by my calculations.

Edit 2: Whoops I said Newton's Method instead of Taylor series.
May23-06, 10:00 PM
Sci Advisor
PF Gold
Hurkyl's Avatar
P: 16,091
where c is a positive number small enough for powers higher than [itex]c^2[/itex] to be neglected,
That sounds like a command to use Taylor series, and neglect all the terms with powers higher than .

You won't be able to compute this integral directly -- it doesn't have an expression in terms of "elementary" functions.

George Jones
May23-06, 11:09 PM
George Jones's Avatar
P: 6,248
Circumference of an ellipse

Considering the first 3 terms in the power series expansion of

[tex](1 + x)^{\frac{1}{2}},[/tex]


[tex]x = c^2 + 2c \left( \sin^2 t - \cos^2 t \right) = c^2 - 2c \cos 2t[/tex]

is small, seems to give the right answer.

Edit: I got the same result as vsage when I considered the first 2 terms of the power series expansion, but this does not include all terms of order c^2.

May25-06, 09:27 AM
P: 17
Ok, thanks guys. I'll have a go doing that (although my Taylor Series expansion knowledge is sketchy at best).
May25-06, 02:25 PM
HW Helper
P: 1,025
[tex](1 + x)^{\frac{1}{2}}= 1 + \frac{1}{2} x - \frac{1}{8}x^2 + \frac{1}{16} z^3+\cdots , |x|<1[/tex]
May25-06, 09:33 PM
Sci Advisor
PF Gold
Hurkyl's Avatar
P: 16,091
You don't actually have to know the Taylor expansion: you only need a couple terms, so it's easy enough to compute. You simply need to differentiate with respect to c twice to get all the terms up to order c^2.
May26-06, 07:06 AM
P: 17
Just for the curious: I did managed to get the answer in the end - it was necessary to use the third term of the expansion. Thanks.

Register to reply

Related Discussions
Circumference of cirlce Precalculus Mathematics Homework 6
Proove its a circumference Precalculus Mathematics Homework 13
Reduced Circumference Astronomy & Astrophysics 1
Circumference of an Ellipse Introductory Physics Homework 2
Circumference of an ellipse General Math 1