Solving for $y$ produces:
$ y =\frac{1}{2} \sqrt{9x^2-36} $
What we need to find is the area of $y$ from $x=2$ to $x=3$.
Thus we put:
$A = \frac{1}{2}\int_2^3 \sqrt{9x^2-36} dx$
Make an "hiperbolic" substitution:
$3x = 6 \cosh u$
$3dx = 6 \sinh u du$
We get the new limits are $0$ and $b = \cosh^{-1}\frac{3}{2} = \log({\frac{3}{2} +\sqrt{\frac{5}{4}}})$
$A = 6\int_0^b {{{\sinh }^2}udu} $
This integral is very similar to the integral of $\sin^2 x$,
$A = 6\int_0^b {{{\sinh }^2}udu} =6 \left. {\frac{{\sinh \left( {2u} \right) - 2u}}{4}} \right|_0^b$
Then you have
$A = 3\frac{{\sinh 2b - 2b}}{2}$
After a myriad of algebraic steps I end up with
$A = \frac{9}{4}\sqrt 5 - 6\log \phi $
Where $\phi$ is the golden ratio number.
Note that
$\eqalign{ & \log 8 - 3\log \left( {3 + \sqrt 5 } \right) = 3\log 2 - 3\log \left( {3 + \sqrt 5 } \right) = \cr & - 3\log \left( {\frac{{3 + \sqrt 5 }}{2}} \right) = - 3\log \left( {1 + \frac{{1 + \sqrt 5 }}{2}} \right) = \cr & - 3\log \left( {1 + \phi } \right) = - 3\log {\phi ^2} = - 6\log \phi \cr} $
EDIT: Remeber the value is just half the value of the total area.