Squaring the circle

suppose you have a circle with area pi (therefore r = 1)
also, then suppose you have a square with area pi too (side = sqrt(pi))

now suppose you put both one on another
(looking like this: http://upload.wikimedia.org/wikipedia/commons/a/a7/Squaring_the_circle.svg )
whats the area of the intersection?
I managed to come up with an answer using 2 integrals
is there a way to do it with just one? or even better, is there a way to do not use riemann integrals at all?!


the answer I thought:
suppose you have them both on cardinal positions
so the circle equation would be y² + x² = 1
so what I did was integrate sqrt(1-x²) from 0 till the place that the circle meets the side of the square (so the x = sqrt(pi)/2) so the 1st riemann integral is integrate sqrt(1 - x²) from 0 till sqrt(pi)/2

then all we have to do is take away the "top" extra part (of the circle) the part that is atop the square
so all we have to do is take away the riemann integral of 1 - x² - sqrt(pi)/2 from 0 to the point that it meets the "ground" (or square top) which is when y = sqrt(pi)/2 on x² + y² = 1, giving an x = sqrt(1 - pi/4)

and that's it
4 * (Riemann Integral of (sqrt(1 - x²) dx), from 0 till sqrt(pi)/2, - Riemann Integral of (sqrt(1 - x² - sqrt(pi)/2) dx), from 0 to sqrt(1 - pi/4))

its complicated
but that's the only thing I though
any alternative answers?

ps: to illustrate what I did:
http://img177.imageshack.us/img177/9190/squaringcirclekh8.png

You can sort of do it with simple geometry. Except you'll need to know (or treat as constants) the values arcsin(sqrt(pi)/2) and arccos(sqrt(pi)/2).

>>2
explain please

The area of the shaded region in the leftmost picture in your illustration is:

Let p = sqrt(pi)/2.

1/2*p*cos(arcsin(p)) + pi*(arcsin(p) - arccos(p)) + 1/2*p*sin(arccos(p))

= 1/2*p*(sqrt(1-p^2)) + pi*(arcsin(p) - arccos(p)) + 1/2*p*(sqrt(1-p^2))

= p*(sqrt(1 - p^2)) + (arcsin(p) - arccos(p))/2

The left term might be a bit simplifiable, but afaik there is no nice form for arcsin(p) - arccos(p).

(All I did to get the initial formula was break the region up into two triangles and a circular section. Triangle area is bh/2, circular section area is (upperangle-lowerangle)/2 * r^2)

>>4
thanks heh
now I get it

its perfect :P

Calculus, is there anything you can't solve?

>>4
Actually, I put this into Mathematica and it returned

(1/4)(pi + sqrt((4-pi)pi) - 4arcsec(2/sqrt(pi)))

So there is your answer slightly simplified.