Area of Ellipse Proof

[Tarmac27]

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;

[math]y=\frac{b}{a}(\sqrt{a^2-x^2})[/math]
Giving the area of the top half as;

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

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

[math](a^2\int(sin^2\theta)d\theta[/math]
Then once integrated and subbing x back in I got

[math]\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}))))[/math]
But then when evaluating the inner bracket from -a to a I got

[math]A=-\frac{ab\pi}{2}[/math]
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?

 

[topsquark]

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 [imath]1/2 \pi a^2[/imath].

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

-Dan

 

[Tarmac27]

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.

 

[LCKurtz]

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 [imath][-\frac \pi 2, \frac \pi 2][/imath]. 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 [imath]\arcsin x[/imath]. You probably saw that done by implicit differentiation of the equation [imath]x = \sin y[/imath] giving [imath]1=\cos y ~y'[/imath] so you get [math]y' = \frac 1 {\cos y} = \frac 1 {\pm \sqrt {1-\sin^2 y}}=\frac 1 {\pm \sqrt {1-x^2}}[/math]. 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 [imath][-\frac \pi 2, \frac \pi 2][/imath] if you use the + sign in the formula for the derivative of arcsine.