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here is my proof

Formula: [4b+a^2(π-2)/b+a(π-2)]

Let A minor axis Let B major axis

1.)Example: given (A= 0) (B= 1000)

Solving: [4*1000+0^2(π-2)/1000+0(π-2)] Answer= 4000 because A is 0 so the answer is 4000 or only value of B will remain 1000 and no more point something

2.)Example: given (A= 1000) (B= 1000)

Solving:[4*1000+1000^2(π-2)/1000+1000(π-2)] Answer= 6283.185308

because the A and B are the same so the answer is like full circle circumference

3.)Example: given (A=1) (B= 1000)

Solving:[4*1000+1^2(π-2)/1000+1(π-2)] Answer= 4001.142734247

The answer is close to 4000....

len jhan
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1 Answers1

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Let's first look at the circle for inspiration.

A circle with radius $r$ at the origin is defined up to translation by $\frac{x^2}{r^2} + \frac{y^2}{r^2} = 1$

so $x = \sqrt{r^2 - y^2}$ well-defines the circle in the first quadrant, and we calculate $$\dfrac{dx}{dy} = - \frac{y}{\sqrt{r^2 - y^2}}$$ We'll do an arclength integral in cartesian coordinates and then multiply by $4$ for symmetry.

The length of the arc (I hope you've seen this before) is $$\int\limits_{y=0}^{y=r} dy \sqrt{1 + \big(\frac{dx}{dy}\big)^2} = r\int\limits_{y=0}^{y=r} dy \sqrt{\frac{1}{r^2 - y^2}}$$

You can recognize the expression $\sqrt{\frac{1}{r^2 - y^2}}$ as the derivative of $\arcsin(y/r)$, and thus the integral evaluates to $$r (\arcsin(r/r) - \arcsin(0/r) = \frac{\pi r}{2} $$

As mentioned before, this was just the top right quarter of the circle, so multiplying by four gives us the actual perimeter of the circle, $2 \pi r$.

So what about the ellipse??

It's pretty much the same deal! Now our ellipse, up to translation, is defined by $(\frac{x}{a})^2 + (\frac{y}{b})^2 = 1$. By considerations of symmetry we can again focus on a single quadrant, in which $x = \frac{a}{b} \sqrt{b^2 - y^2}$, and $$\frac{dx}{dy} = -\frac{a}{b} \frac{y}{\sqrt{b^2 - y^2}}$$

Now the integral ranges from $y = 0$ to $y = b$, and you can set it up just like above, as

$$\text{Perimeter of Ellipse} = 4*\int\limits_{y=0}^{y=b} dy \sqrt{1 + \big(\frac{dx}{dy}\big)^2}$$

You'll see the integral is pretty messy!

About the best you can do here for closed form is to fiddle with a linear change of variable and define the eccentricity constant $e^2 = 1 - \frac{a^2}{b^2}$, to put this into the form

$$\text{Perimeter of Ellipse} = 4*b*\int\limits_{y=0}^{y=1} dy \sqrt{ \frac{1-e^2y^2}{1 - y^2}}$$ The integral expression here is known as a complete elliptic integral of the second kind.

Note that when the eccentricity is $0$, the minor and major axes are equal in length and $b$ becomes the radius $r$ of the circle, with the integral collapsing to $\pi / 2$, and we recover the circumference of a circle. When the eccentricity is $1$, $a$ is forced to be $0$ and the integral collapsis to $1$, so that the perimeter of the ellipse is two times the length of the line segment $[-b, b]$ (you alluded to this case in your question).

Badam Baplan
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  • i try to use other formulas of ellipse circumference but the problem is when the value of y=0 x=>0 the answers are not exact.. – len jhan May 18 '17 at 23:52
  • If you put the formula i stated here into a solver like wolframalpha, it will give you a very very precise answer – Badam Baplan May 19 '17 at 00:01
  • ah ok sir.. ive done it before.. they are all good..but all i want is exact number/answer..thanks for your comments.. – len jhan May 19 '17 at 00:08
  • unfortunately there's no pretty closed form for this, at least not one $2 \pi r$ or something rational in terms of powers of $a,b$ and constants. The problem is very famous for this, many great mathematicians have thought about it (e.g. Ramanujan and Bessel) and the best we've got is some highly accurate approximations and fast converging series. – Badam Baplan May 19 '17 at 00:21
  • thats why i always thought.. i found my own formula close to 0 value snd high value but i dont know in the middle value.. – len jhan May 19 '17 at 00:31