Any help with the following problem would be greatly appreciated.

Set up, but do not evaluate, an integral for the length of the curve.

x^2/a^2 + y^2/b^2 = 1

Thanks.

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- Apr 4th 2010, 09:17 PMvlodgeIntegral for the length of a curve.
Any help with the following problem would be greatly appreciated.

Set up, but do not evaluate, an integral for the length of the curve.

x^2/a^2 + y^2/b^2 = 1

Thanks. - Apr 4th 2010, 10:47 PMsimplependulum
There are many methods , the choice depends on what you know about it .

we have $\displaystyle L = \int_a^b \sqrt{ 1 + y'^2 }~dx~,~ y = f(x) $

or parametric form :

$\displaystyle L = \int_{t_1}^{t_2} \sqrt{ x'^2(t) + y'^2(t) }~dt $

Make $\displaystyle y $ as the subject , we have

$\displaystyle y = (b/a) \sqrt{ a^2 - x^2 }$ and

$\displaystyle y' = (b/a) \cdot \frac{ -2x}{2 \sqrt{a^2-x^2}} = \frac{ -bx}{a \sqrt{ a^2 - x^2 } }$

Therefore , $\displaystyle L = \int_{-a}^a \sqrt{ 1 + \left ( \frac{ -bx}{a \sqrt{ a^2 - x^2} } \right )^2 } ~dx $

But $\displaystyle L $ indicates the length of the semi-ellipse but not the whole ellipse , for the whole ellipse you need to multiply $\displaystyle L $ by $\displaystyle 2 $. - Apr 5th 2010, 04:55 AMHallsofIvy
As simplependulum said, you can also do this with parametric equations.

Standard parametric equations for the ellipse $\displaystyle \frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1$ are x= a cos(t) and y= b sin(t). Then x'= -a sin(t) and y'= b cos(t) so the arclength is $\displaystyle \int_{t= 0}^{2\pi}\sqrt{a^2 sin^2 t+ b^2 cos^2 t} dt$

By the way, the reason the problem said "Set up, but do not evaluate" the integral is that "ellipitic integrals" [b]cannot[b] be integrated analytically.