So when I was taking Calculus a few years back, I remember a lesson where the teacher asked a question about cutting a pizza into three equal parts. But instead of normal cuts going from the center out, You decide to make two parallel cuts so that the area between them is the same. Obviously I was eating pizza recently with friends and seemed to remember this problem randomly. I think I can get the problem set up correctly but I'm having problems with coming up with a general formula for pizza radius length to 1st cut.
So here's what I have so far..
$$ A = \pi r^2 $$ I need a third of that for my one slice. So my integral has to equal
$$ = \frac{\pi r^2}{3} $$
Equation of a circle is $$ (y-h)^2 + (x-k)^2 = r^2 $$
If I set this on a plane so my circle moves along the x-axis such that the center (k,h) is always at the same distance as the radius, it gives us a better integral, where we can start at 0 for the x value instead of a negative number. That gives me something like this.
$$ (y-0)^2 + (x-r)^2 = r^2 $$ $$ (y)^2 + (x-r)^2 = r^2 $$ $$ (y)^2 = r^2 - (x-r)^2$$ $$ y = \sqrt {r^2-(x-r)^2} $$
So I'm not taking the negative part here even though I am supposed to when doing a radical. The reason is because it is a mirror copy of the positive side. When you graph this you can see that its just the bottom part of the circle. You don't need the bottom half to determine where to cut on the X-axis because these two are symmetrical. So If you plug all this into an indefinite integral I get this. (Where "C" is where I should make my 1st cut on the X-axis for my pizza)
$$ \frac{\pi r^2}{3}= \int_0^c \sqrt {r^2-(x-r)^2} dx $$
$$ \frac{\pi r^2}{3}= \int_0^c \sqrt {r^2-(x^2 -2rx + r^2)} dx $$
$$ \frac{\pi r^2}{3}= \int_0^c \sqrt {-x^2+2rx} dx $$
So I get it to here and them I'm stuck. I don't think integration by parts works here, and none of the trig substitutions look right.. Any help would be appreciated.