Original isoperimetric problem

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blueyellow
solve the original isoperimetric problem
a curve y=y(x) meets the x-axis at x=+a abd x=-a and has fixed length pi*a between these points. Show that the curve which encoses the maximum area between itself and the x-axis is the semicircle x^2+y^2=a^2 for y (greater than or equal to) 0



Homework Equations



euler lagrange equation?


The Attempt at a Solution


i tried reading thru the lect notes - little help
don't really know how to get started on this
pleas help. thanks in advance
 
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i'd start by writing the area in terms of the arbitrary function.. then consider to maximise it...

euler lagrange is probably way to go, though its best to set up the integral first, so all the terms are clear
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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