Troubleshooting a trigonometry/geometry question - spot the (silly) mistake! I'm back again!
Again, another error - not sure if it's them or me this time...
Here is the question

In order to calculate the shaded region, I added some lines to the diagram. I realise that there are other ways to solve this (and indeed the solution give uses a different breakdown and there's seems correct).

I then calculated the area of the sector ADB - the area of the triangle ADB and added this to the area of the sector BOD. I then multiplied that value by 2 to get the area of the shaded region.
Here is my working (Let a = alpha and t = theta)
$Area = 2[\frac12(3r)^2(\frac{a}2)-\frac12(3r)^2\sin(\frac{a}2)+\frac12(2r)^2(\frac{t}2)]$
Giving me a final answer of $r^2[(\frac{9a}2)-9\sin(\frac{a}2)+2t]$
Can you see where I went wrong...?
EDIT:
Here is the solution they provided:

 A: Another way to compute the area is to add the part on the right of chord $CD$
and the part on the left of chord $CD$.
The portion of the shaded region on one side of the chord is called a circular segment.
You can compute the area of the circular segment on the right by finding the area of the circular sector between $AC$ and $AC$ and subtracting the area of the triangle
$\triangle ACD$:
$$ \frac92 r^2 \alpha - \frac92 r^2 \sin\alpha. $$
Similarly the area of the circular segment on the left is the area of the sector between $BC$ and $BD$ minus the area of $\triangle BCD$:
$$ 2r^2 \theta - 2r^2 \sin\theta. $$
Add them together and you get:
$$ \frac92 r^2 \alpha - \frac92 r^2 \sin\alpha + 2r^2 \theta - 2r^2 \sin\theta. $$
Factor out $r^2$:
$$
r^2\left(\frac92 \alpha - \frac92 \sin\alpha + 2 \theta - 2 \sin\theta\right).\tag1
$$
Excluding the final term $- 2 \sin \theta,$ the part in the brackets looks a lot like your result, but it has a term $-\frac92 \sin\alpha$ where you have
$-9 \sin\frac\alpha2.$
Those are not the same terms.
Let $M$ be the point where chord $CD$ intersects the segment $AB$
and consider the two right triangles $\triangle AMD$ and $\triangle BMD.$
You can find that $DM = 3 \sin\frac\alpha2 = 2 \sin\frac\theta2.$
Moreover,  $AM = 3 \cos\frac\alpha2$ and $BM = 2 \cos\frac\theta2,$
so $3 \cos\frac\alpha2 + 2 \cos\frac\theta2 = AB = 3.$
You can also verify via the double-angle formula for the sine that
$\frac92 \sin\alpha = 9 \sin\frac\alpha2 \cos\frac\alpha2$
and $2 \sin\theta = 4 \sin\frac\theta2 \cos\frac\theta2.$
Then
\begin{align}
\frac92 \sin\alpha +  2 \sin \theta
&= 9 \sin\frac\alpha2 \cos\frac\alpha2 + 4 \sin\frac\theta2 \cos\frac\theta2 \\
&= \left(3 \sin\frac\alpha2\right)\left(3 \cos\frac\alpha2\right)
 + \left(2 \sin\frac\theta2\right)\left(2 \cos\frac\theta2\right) \\
&= \left(3 \sin\frac\alpha2\right)\left(3 \cos\frac\alpha2\right)
 + \left(3 \sin\frac\alpha2\right)\left(2 \cos\frac\theta2\right) \\
&= \left(3 \sin\frac\alpha2\right)
   \left(3 \cos\frac\alpha2 + 2 \cos\frac\theta2\right) \\
&= \left(3 \sin\frac\alpha2\right) \times 3\\
&= 9 \sin\frac\alpha2.
\end{align}
So if we now write Equation $(1)$ in the form
$$
r^2\left(\frac92 \alpha - \left(\frac92 \sin\alpha + 2 \sin\theta\right)
 + 2 \theta\right),
$$
by the substitution $\frac92 \sin\alpha + 2 \sin\theta = 9 \sin\frac\alpha2$
we can see this is equal to your result.
If you generalize the problem so that the circles can have any radius and $B$ is not necessarily on the circumference of the circle about $A,$
with the restriction only that the two circles intersect at two distinct points, then the provided solution can be adapted to a formula to solve the more general problem. That is one reason it is likely to have been chosen.
Your solution, which effectively simplifies
$\frac92 \sin\alpha + 2 \sin\theta$ to $9 \sin\frac\alpha2,$
is possible because $B$ is on the circumference of the other circle.
A: The two answers are equivalent.
You have: $r^2[(\frac{9 \alpha}2)-9\sin(\frac{\alpha}2)+2 \theta]$
They have: $ [\frac 9 2(\alpha -\sin \alpha) + 2(\theta - \sin \theta)]r^2$
The triangle $ABD$ is isosceles, so $\frac \alpha 2 + 2 \frac \theta 2=\pi$
$\frac \alpha 2 = \pi-\theta \Rightarrow \sin \frac \alpha 2=\sin(\pi-\theta)=\sin \theta$
That changes your answer to $r^2[9(\pi - \theta)-9\sin \theta+2 \theta]=r^2[9\pi - 9 \sin \theta -7 \theta]$
$\alpha +2\theta = 2\pi \Rightarrow \alpha = 2\pi - 2\theta$
$\sin \alpha = \sin (-2\theta)=-\sin (2\theta)=-2\sin \theta \cos \theta$
That changes their answer to $ [\frac 9 2(2\pi - 2\theta +2\sin \theta \cos \theta) + 2(\theta - \sin \theta)]r^2 = [9\pi - 9\theta +9\sin \theta \cos \theta + 2\theta - 2\sin \theta]r^2 $
$=r^2[9\pi +(9\cos \theta - 2)\sin \theta - 7\theta]$
These look more alike now they are both in terms of $\theta$ but there is still a difference.
We can apply the cosine rule to the triangle $ADB$ to get $\cos (\frac \theta 2)=\frac{2^2+3^2-3^2}{2.2.3}$
So $\cos  (\frac \theta 2)=\frac 1 3$
$\cos \theta = \cos (2 \frac \theta 2)=2 \cos^2 (\frac \theta 2) -1 = 2(\frac 1 3)^2-1=-\frac 7 9$
Then $9\cos \theta - 2= -7-2=-9$
which turns their answer into yours.
A: Cartesian reference.
Equation of center circumference $A(0,0)$ and radius $R=3r$:
$x^{2}+y^{2}=9r^{2}$.
Equation of center circumference $B(3r,0)$ and radius $R=2r$:
$(x-3r)^{2}+y^{2}=4r^{2}$.
The equation of the radical axis is given by the difference of the two equations of circumferences:
$x=\frac{7r}{3}$.
The area is equal to twice the areas of the two integrals:
$$\int_r^{\frac{7r}{3}}\sqrt{r(6x-5r)-x^{2}}dx=$$
$=4r^{2}tan^{-1}(\frac{\sqrt{2}}{2})-\frac{4\sqrt{2}r^{2}}{9}$,
$$\int_\frac{7r}{3}^{3r}\sqrt{9r^{2}-x^{2}}dx=$$
$=\frac{r^{2}(81\pi-28\sqrt{2})}{18}-18r^{2}tan^{1}(\frac{\sqrt{2}}{2})$.
Area is:
$r^{2}(9\pi-4\sqrt{2})-28r^{2} tan^{-1}(\frac{\sqrt{2}}{2})$.
