Find the area of the region bounded by the curves $y =\sqrt x$, $y=x-6$ and the x-axis by integral with respect to x 
Find the area (in green) of the region bounded by the curves $f(x) =\sqrt x$,  $g(x)=x-6$ and the x-axis  by integral with respect to x
  

Attempt:
since $x-6 =\sqrt x$
$x=9$
$A = \displaystyle\int _a^b [f(x)-g(x)] dx$
$A = \displaystyle\int _0^9 \sqrt x -\int _0^9x-6$
$=\frac{2}{3}(9)^{\frac{3}{2}}-\left(\frac{1}{2}(9)^2-6(9)\right)  $
$=31.5$ $unit^2$
I don't think this method is correct since I am not getting the same area when I solve the integral with respect to y which is also different when I take the whole area under f(x) then subtract the area of the triangle on the right (base=3, height=3)
Can anyone help?

Alternatively, you can use
\begin{align} 
f^{-1}(y)&=y^2
,\\
g^{-1}(y)&=y+6
\end{align}  

and find the area as
\begin{align} 
\int_0^3 \int_{f^{-1}(y)}^{g^{-1}(y)} \,dx \,dy
&=\int_0^3 6+y-y^2\,dy
\\
&= \left. 6y+\tfrac12\,y^2-\tfrac13\,y^3\right|_0^3
\\
&=\frac{27}2
.
\end{align}  
