I know that the area of an ellipse is:

 

Pi x Semimajor x Semiminor

 

But after quite a long time of figuring, i thought, why can't we just treat the ellipse as a circle by hypothetically shortening the semimajor and lengthening the semiminor to the same length (averaging them) and then using Pi x Radius²

 

I know they give different answers, but i don't see why my way is wrong. I know that it is, but i would like someone to prove to me that it isn't, and prove to me that Pi x ab is

 

thanks once again

I won't do the work, but you can get the area of the ellipse via a direct integration. This is outlined all over the web. If you get stuck, ask for some help.

because [math]ab \neq \left(\frac{a+b}{2}\right)^2[/math] except in special cases.

Or put differently, [math]a^2 \neq (a-x)(a+x); \hspace{1mm} a, \hspace{1mm} x \in \mathbb{R} - \{0\}[/math].

Your general idea is sound, it just needs to use the correct "average".

 

Let's say that the two quantities, a and b, were along a *single* axis. On a straight road, a chauffeur drives a miles in the morning and b miles in the afternoon. The final distance is the same as saying the chauffeur drove (a+b)/2 miles in the morning and again in the afternoon.

 

But, we're talking about distances along two perpendicular axes whose distances are multiplied together to find an area, not added together to find a distance. Thus, their average will not be the "arithmetic average" (shown above), but will be the "geometric average". This means that r = √(a∙b), and the area = π r².

Edited February 13, 201114 yr by ewmon