Quarter-circle and circle inscribed in square A circle is inscribed in square ABCD, and a quarter-circle with A as its centre is inscribed in the square. What is the area of the crescent like shape inside the full circle and created by the quarter circle. I'm terrible at explaining so here is a diagram. Is there any way to find the area of the crescent-like shape in relation to the smaller circle without using integration?

 A: Recall for any circular arc with radius $r$, if the half-angle subtended by the arc is $\theta$, then the area of corresponding circular segment (the region bounded by the arc and chord with same endpoints ) is given by the formula
$$\verb/Area/ = r^2(\theta - \sin\theta\cos\theta)
= r^2\sin^{-1}\frac{\ell}{r} - \ell\sqrt{r^2-\ell^2}
$$
where $\ell = r\sin\theta$ is half of the length of the chord.
For the problem at hand, label vertices/lengths/angles according to the figure below:
$\hspace 1in$ 
When the side of the square $ABCD$ is $1$, then the crescent shape (the orange area) is the difference of two circular segments. Both circular segments have $FH$ as the supporting chord. One has radius $r_1 = \frac12$ and half-angle $\theta_1 = \angle HED$. The other one has radius $r_2 = 1$ and half-angle $\theta_2 = \angle HBD$. By cosine rule, we have
$$\cos\theta_1 = -\cos(\angle BEH) = \frac{BH^2 - BE^2 - HE^2}{2(BE)(HE)} = \frac{1 - \frac12 - \frac14}{2(\frac12)(\frac{1}{\sqrt{2}})} = \frac{\sqrt{2}}{4}$$
This implies
$$\sin\theta_1 = \frac{\sqrt{14}}{4}
\quad\implies\quad \ell = \frac12 FH = r_1\sin\theta_1 = \frac{\sqrt{14}}{8}$$
The area of the crescent shape is
$$\begin{align} &
 \left( r_1^2 \sin^{-1}\frac{\ell}{r_1} - \ell\sqrt{r_1^2 - \ell^2}\right)
-\left( r_2^2 \sin^{-1}\frac{\ell}{r_2} - \ell\sqrt{r_2^2 - \ell^2}\right)\\
=  &\; \frac14\sin^{-1}\frac{\sqrt{14}}{4}- \sin^{-1}\frac{\sqrt{14}}{8} + \frac{\sqrt{7}}{8}\\
\approx &\; 0.1463812595303479
\end{align}
$$
A: The intersections of the two circles can be found by plugging the parametric equation of the small one into the implicit of the second.
$$(\cos\alpha+1)^2+(\sin\theta+1)^2=4,$$
giving
$$\alpha=\frac\pi4\pm\arccos\frac1{\sqrt8}.$$
As the amplitude of the arc of the small circle is $\theta=2\arccos\dfrac1{\sqrt8}$, the chord length is $c=\sqrt{\dfrac72}$.
The requested area is the difference between two circular segments, which can be computed by
$$a=\frac{r^2}2(\theta-\sin\theta).$$ 
For the large circle, the amplitude is obtained from the chord by
$$\theta=2\arcsin\frac c{2r}.$$
