Area of Ellipse Proof

Tarmac27

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Jan 29, 2021
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Hello,

I'm trying to prove the area of ellipse formula via integration and I have gotten close but I've got a negative sign on my final answer which I know shouldn't be there.

I reaaranged the equation of ellipse to get the top half of the ellipse as;

y=ba(a2x2)y=\frac{b}{a}(\sqrt{a^2-x^2})
Giving the area of the top half as;

baaaa2x2dx\frac{b}{a}\int_{-a}^{a}\sqrt{a^2-x^2}dx
I didn't know how to approach this integral at first so I tried one round of IBP and got:

xa2x2+x2a2x2dxx\sqrt{a^2-x^2}+\int\frac{x^2}{\sqrt{a^2-x^2}}dx
I realised that a trig sub would be better got the remaining integral down to the following by letting x = asinu

(a2(sin2θ)dθ(a^2\int(sin^2\theta)d\theta
Then once integrated and subbing x back in I got

ba(xa2x2+a2(12arcsin(xa)14sin(2arcsin(xa))))\frac{b}{a}(x\sqrt{a^2-x^2}+a^2(\frac{1}{2}arcsin(\frac{x}{a})-\frac{1}{4}sin(2arcsin(\frac{x}{a}))))
But then when evaluating the inner bracket from -a to a I got

A=abπ2A=-\frac{ab\pi}{2}
However I have realised now that if I used arcsin(-1)=-pi/2 instead of arcsin(-1) = 3pi/2 I would have got a positive answer. But shouldn't these two angles give the same result?
 
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Don't do IBP, just sub x = a sin(u).

Anyway, the integral on the second line is simply the integral over a 1/2 circle so you know the area is 1/2πa21/2 \pi a^2.

I don't have time to go over the arcsine problem but I'd just convert my limits to u in the π/2π/2sin2(u) du\int_{-\pi / 2}^{\pi /2} sin^2(u) ~ du integral.

-Dan
 
but I'd just convert my limits to u in the ∫−π/2π/2sin2(u) du\int_{-\pi / 2}^{\pi /2} sin^2(u) ~ du∫−π/2π/2sin2(u) du integral.
Yea I realised how this is a much more effecient approach after I got to my answer. I just went through it again with your suggested method and got the correct answer, Thanks!. I know you don't have time but if anyone else could explain where I went wrong in my method I would appreciate it.
 
However I have realised now that if I used arcsin(-1)=-pi/2 instead of arcsin(-1) = 3pi/2 I would have got a positive answer. But shouldn't these two angles give the same result?
No. The reason goes back to why the principle values are chosen as they are. The arcsin is a multiple valued function and the range is chosen as [π2,π2][-\frac \pi 2, \frac \pi 2]. The reason this interval is chosen is not usually explained in a trig course because it involves calculus. Think back to where you wanted to calculate the derivative of arcsinx\arcsin x. You probably saw that done by implicit differentiation of the equation x=sinyx = \sin y giving 1=cosy y1=\cos y ~y' so you get y=1cosy=1±1sin2y=1±1x2y' = \frac 1 {\cos y} = \frac 1 {\pm \sqrt {1-\sin^2 y}}=\frac 1 {\pm \sqrt {1-x^2}}. If you want to use the + sign in that formula, you want to choose a range of the multiple valued arcsine function where it is increasing. That is why you are stuck with [π2,π2][-\frac \pi 2, \frac \pi 2] if you use the + sign in the formula for the derivative of arcsine.
 
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