# Total length of ellipse

#### ktpr2

I'm asked to show that the total length of the ellipse
x = a sin x
y = b cos x, a>b>0 is

$$L = 4a \int_{1}^{pi/2} \sqrt{1-e^2sin^2x} dx$$
where e is the eccentricity of the ellipse (e = c/a, where c = $$\sqrt{a^2-b^2}$$

I've tried a whole bunch of algebraic and trignometric manipulation but I'm getting the feeling I'm overlooking something. How would you approach this problem? I've tried working backwards and eventually got something of the form:

$$L = \int_{1}^{pi/2} \sqrt{2b^2cos^2x+c^2cos^2x-b^2(cos 2x)} dx$$

but i figured I should post here for ideas as well.

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#### dextercioby

Homework Helper
HINT:The total length is 4 times the arc in the first quadrant.Choose the wise parametrization and use the definition of the eccentricity (the modulus of the elliptic integral).

Daniel.

#### BobG

Homework Helper
This can be a challenging problem. Checking out how http://home.att.net/~numericana/answer/ellipse.htm#elliptic [Broken] have solved this might give you some ideas on how to approach it.

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#### dextercioby

Homework Helper
There's no big deal,Bob

$$L_{\mbox{ellipse}}=4L_{\mbox{arc in the first quadrant}}$$ (1)

Parametrization (polar elliptical)

$$\left\{\begin{array}{c}x(\phi)=a\cos\phi \\ y(\phi)=b\sin\phi \end{array} \right$$ (2)

$$\left\{\begin{array}{c} (dx)^{2}(\phi)=a^{2}\sin^{2}\phi \ (d\phi)^{2}\\ (dy)^{2}(\phi)=b^{2}\cos^{2}\phi \ (d\phi) ^{2}\end{array} \right$$ (3)

$$L_{\mbox{arc in the first quadrant}} =\int \sqrt{(dx)^{2}+(dy)^{2}} =\int_{0}^{\frac{\pi}{2}} \sqrt{a^{2}\sin^{2}\phi+b^{2}\cos^{2}\phi} \ d\phi$$ (4)

Now i'll leave it to the OP to finish it.

Daniel.

#### ktpr2

wow. that was down and funky. I figured it out in four lines after an embarrassing number of scrap sheets before. Thanks people.

I'll give several hints to those with the same question in the future:

The integral is from 0 to 2pi originally, but you can cut it down to 0 to pi/2 and multiply by four. recall that c^2is really just 1 - b^2/a^2 and when at the end of a deserted road, question if you can multiply by 1 expressed in a twisty way.

#### peeyush_ali

the length of any geometrical arc is given by [ intg of ( (dx .dx +dy.dy)^1/2 ) ] within proper limits... that is according to "pythogorean theorem"..

on simplifying the above thing we get... intg [(1+(dy/dx)^2)^1/2) ].dx

Now, comming to the problem i.e., length of an ellipse, calculate (dy/dx) from the standard equation in terms of "x" and put it in the above integral and solve thus obtained integral within proper limits....thas alll........... :)

if u do tht u will get approx L= pi[ 3(a+b)- {(3a+b)(a+3b)}^1/2 ]

#### peeyush_ali

By "within limits" i mean the limits of corresponding variable by which the integrand is multiplied i.e. the factor by which the integrand is multiplied which is a differential magnitude.... ( eg., limits of "x" if the integrand is multiplied by "dx" and limits of "theta" if the integrand is multiplied by "d(theta)" )

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