Without using calculus, how can I find the formula for the area of the shaded region in the attached figure in terms of a and b? Note that the center of the larger circle is on the edge of the smaller. 
 A: To understand I recommend you open the picture here.
Lets denote :


*

*$A$ the center of the circle of radius $b$.

*$B$ the center of the circle of radius $a$.

*$C$ and $D$ the ontersection points betwin the two circles.

*$E$ the intersection point between the segments [CD] and [AB]

*$A_{shaded}$ the area of the shaded region.

*$A_{circle_A}$ the area of the circle of center A.

*$A_{circle_B}$ the area of the circle of center B.

*$\alpha$ the angle $\angle CAD$.

*$\beta$ the angle $\angle ABC$.

*$A_{Red}$ the area delimited by the arc (AC) and the segment [AC].

*$A_{Blue}$ the area delimited by the arc (AD) and the segment [AD].

*$A_{ACD}$ the area delimited by the arc (CD) and the segments [AC] and [AD]. 
We have two cases :
First case : $2a\leq b$ then $A_{shaded}=A_{circle_A} - A_{circle_B}= \pi b^2-\pi a^2 $
Second case $b<2a$ :
you can see that $A_{shaded}= A_{circle_A}-A_{ACD}-A_{Blue}-A_{Red}$ 
but $A_{Blue}=A_{Red}$ and $A_{circle_A}=\pi b^2$
So we get $A_{shaded}= \pi b^2-A_{ACD}-2A_{Blue}$
$A_{ACD}= \frac{b^2\alpha}{2}$ 
$A_{Blue} = \frac{a^2}{2}(\beta-sin(\beta))$ the difference between the area delimited by ( the segment [BC], the arc (CA) and the segment [AB] ) and (the area of the triangle BCA).
so finally we get $A_{shaded}= \pi b^2-\frac{b^2\alpha}{2}-a^2(\beta-sin(\beta))$. Now lets compute the angles $\alpha$ and $\beta$ :
applying the law of cosines on the triangle AED you get that
$cos(\frac{\alpha}{2}) = \frac{a^2+b^2-a^2}{2ab}=\frac{b}{2a}$ (notice the condition $b<2a$) and thus $\alpha=2 cos^{-1}(\frac{b}{2a})$
Now for $\beta$ notice that the triangle BCA is isosceles (BC=BA=a) so $\beta = \pi-\alpha$
Finally you get :
$A_{shaded}= \pi b^2-b^2 cos^{-1}(\frac{b}{2a})-a^2(\pi-2 cos^{-1}(\frac{b}{2a})-sin(\pi-2 cos^{-1}(\frac{b}{2a})))$
A: 
If we consider the centers of both the circles on y axis, origin as the center of the bigger circle, then let the two circles intersect at $A=(x_1,y_1), B=(x_2,y_2)$ 
Equation of the bigger circle is $x^2+y^2=b^2...(I)$
Equation of the smaller circle $(x-h)^2+(y-k)^2=a^2$, here $h=0, k=-a$
$\therefore x^2+(y+a)^2=a^2...(II)$
$(II)-(I)$ gives $y=\frac{-b^2}{2a}$ and $x=\pm \frac{-b\sqrt{4a^2-b^2}}{2a}$
$\therefore x_1=-\frac{b\sqrt{4a^2-b^2}}{2a}$ and $x_2=\frac{b\sqrt{4a^2-b^2}}{2a}$
It can be proved that $AB=2CB$ and $\angle OCB=90$ by Pythagoras theorem.
$\ell(AB)=\frac{b}{a}\sqrt{4a^2-b^2}$
angle subtended by $AB$ at the center of the bigger circle$=\phi=2\cdot \sin^{-1}\left(\frac{\sqrt{4a^2-b^2}}{2a}\right)$
angle subtended by $AB$ at the center of the smaller circle$=\phi'=2\cdot \sin^{-1}\left(\frac{b\sqrt{4a^2-b^2}}{2a^2}\right)$
The isosceles triangles $AO'O$ and $BO'O$, both with equal sides length$=a$ and base$=b$ and area$=\frac{b}{4}\sqrt{4a^2-b^2}$
Area of the crescent shape region or segment = area of sector of the smaller circle formed by $AO$ or $BO$ - combined area of isosceles triangles $AO'O$ and $BO'O$
$=\frac{a^2}{2}\cdot 2 \sin^{-1}\left( \frac{b\sqrt{4a^2-b^2}}{2a^2} \right)-\left( \frac{b}{2}\sqrt{4a^2-b^2} \right)$
area of the bigger circle sector$=\frac{b^2}{2}\cdot 2 \sin^{-1} \left( \frac{\sqrt{4a^2-b^2}}{2a} \right)$
$\therefore$ area of the required region $\color{red}{= \pi b^2 - b^2\cdot \sin^{-1} \left( \frac{\sqrt{4a^2-b^2}}{2a} \right)-\left(  a^2\cdot  \sin^{-1}\left( \frac{b\sqrt{4a^2-b^2}}{2a^2} \right)-\left( \frac{b}{2}\sqrt{4a^2-b^2} \right)\right)}$
To check if the formula is correct, put $b=\sqrt2 a$, so that points $O'$ and $C$ coincide and the $\angle AOB=90$ and the required area is (area of bigger circle) - (area of sector $AOB$) - (area of two segments or crescent shaped regions), which is 
$\frac{3\pi b^2}{4}-\frac{\pi a^2}{2}+\frac{b}{2}\sqrt{4a^2-b^2}=a^2(\pi+1)$
if we substitute $b=\sqrt2 a $ in the $\color{red}{red}$ formula, we get the same result
Check using calculus: $(b=\sqrt2 a)$

In the diagram consider only the region marked I, to the right of the dashed line
area I=$\int_0^a (\sqrt{b^2-y^2}-a)dy = \dfrac{\pi a^2}{4}-\dfrac{a^2}{2}$
In the diagram consider only the region marked II, to the left of the dashed line (note that axis are shifted vertically down by $a$ units)
area II=$\int_0^a(a-\sqrt{a^2-y^2})dy = a^2-\dfrac{\pi a^2}{4}$
area I + area II = $\dfrac{a^2}{2}$, two such areas = $a^2$, area of upper half circle with radius $ b = \dfrac{\pi b^2}{2}=\pi a^2$
adding everything together we get the required area = $\pi a^2+a^2=a^2(\pi+1)$, $\color{red}{voila!}$
