I have a circle of radius $1$ inside a square of side length $2$. Say I have an arbitrary line passing through the bottom left corner of the square. Find the area bounded by the line, square, and circle. I don't have a picture but if you imagine the concave triangle created by a circle being placed inside of a square, I am trying to find the area of that but with a line going through it. So I am trying to set up an integral to get the area. The equation of the bottom of a circle that has been shifted up and to the right by one is $$y=-\sqrt{1-(x-1)^2}+1$$ and the equation of the line is $$y=Ax$$
They intersect when $$-\sqrt{1-(x-1)^2}+1 = Ax \implies -(x-1)^2 = A^2x^2-2Ax$$ $$0 = (A^2+1)x^2-2(A+1)x +1$$ I will use the quadratic equation to solve, and I will throw away the bigger value since that hits the circle later. $$x=\frac{(A+1)- \sqrt{2A }}{(A^2+1)}$$
My final integral is $$\int _0^\frac{(A+1)- \sqrt{2A }}{(A^2+1)} Ax dx + \int _\frac{(A+1)- \sqrt{2A }}{(A^2+1)}^1 -\sqrt{1-(x-1)^2}+1 dx$$
Which is adding the portion of area under the line, and then the portion of area under the circle. For $A=1$ this should give me $\frac{4-\pi}{8}$, since it would cut that bottom left area in half. But this integral is not giving me the correct answer, does anyone know where I went wrong?
I don't think you've made a mistake anywhere. Your answer for $A=1$ does give me the correct answer. Essentially my solution is pretty much the same as yours.
First we need the intersection of $y=Ax$ and the circle, $(y-1)^2+(x-1)^2=1$. This we can solve by substitution:
$$(Ax-1)^2+(x-1)^2=1$$
which we can simplify and solve as a quadratic equation, after which we obtain:
$$x_0=\tfrac{A+1}{A^2+1}\pm\tfrac1{A^2+1}\sqrt{2A}$$
And since we have the smaller root, we can pick $-$ for the $\pm$, obtaining
$$x_0=\tfrac{A-\sqrt{2A}+1}{A^2+1}$$
Now we have a triangle in the bottom left with height $Ax_0$ and width $x_0$; this gives us an area of $\tfrac12Ax_0^2=\tfrac A2(\tfrac{A-\sqrt{2A}+1}{A^2+1})^2$. Now for the other part, we have the bottom piece of the circle from $x_0$ to $1$ (because it touches the square at $x=1$), and the equation of that piece of the circle is $y=1-\sqrt{1-(x-1)^2}$ (mind the $-$-sign in front of the root, since we want the lower part of the circle). So the area of that is
$$\int_{x_0}^1(1-\sqrt{1-(x-1)^2})\text{d}x$$
Split that into two to get
$$(1-x_0)-\int_{x_0}^1(\sqrt{1-(x-1)^2})\text{d}x$$
rewrite that a little bit to obtain
$$(1-x_0)-\int_0^{1-x_0}\sqrt{1-x^2}\text{d}x$$
So the end result is adding this together with the little triangle:
$$\tfrac A2(\tfrac{A-\sqrt{2A}+1}{A^2+1})^2+\tfrac{A^2-A+\sqrt{2A}}{A^2+1}-\int_0^{\tfrac{A^2-A+\sqrt{2A}}{A^2+1}}\sqrt{1-x^2}\text{d}x$$
which, for $A=1$, gives $\tfrac{4-\pi}{8}$.