How do I solve this?
>>8060064
If you wanted to brute force it you can use double integration, but there are better approaches.
I would try integration in polar coordinates.
The ratio of the area of the top right square within the larger circle to the area outside is probably the same as the ratio of the smaller circle bounded by the larger circle to the area that isn't.
I say probably cause I cant be fucked to check.
>>8060067
I tried taking the top right square and putting an x-axis 1/2 way up and find the cartesian equation of the two circles in the form y=f(x) and then find the point of intersection and integrating them both between 0 and the point. I got to finding the point of intersection and gave up as I couldn't think of an easy way of doing it. Wondered if there was a simpler way (like A+B+C = X method) or something.
Fun fact, the area of the smaller circle is exactly the same as the area of the box bounded by the unit circle.
>>8060097
>>8060075
Alright I tried it out with geometry and algebra.
Anyone who did it with integrals wanna compare, see if I'm retarded?
>>8060064
lots of areas, proportions and subtracting.
>>8060064
have yuo tried turning it on and off again?
>>8060064
http://mathworld.wolfram.com/Circle-CircleIntersection.html
>>8060064
I posted this in the other thread, kek
It's a cute problem. Here is how you actually do it.
Frame the circles in terms of cartesian coordinates. Each circle has an equation. Find those equations (look them up if you need to), and then find the two points where both equations are satisfied. Those are the intercepts.
Then you can figure out two circular sectors/segments (I forget which, again, look up the formulas). Subtract one from the other, and you're done.
OR, a more elegant way: on a certain critical diagonal line, there are about 5 or 6 critical points; moreover, this arrangement can be rigidly translated so that you instead consider the shape whose area you wish to find, as the difference of two functions (the lower halves of each circle can be thrown out for these purposes).
You therefore consider the definite integral, that is, a /single integral/, of a function being itself simply the difference of the two involved functions.
I seem to remember that its expression is a somewhat hairy thing with some trig terms, but I'm not certain.
>>8060164
Cheers I'll give this a go tomorrow
This is as simplified as it gets
>>8060064
isn't it (pi^2)/16 ??
Substrate the curve subtract the curve
>>8060064
Oldie but goodie. Saved a loooong time ago.
