Let’s add up tiny rectangles of widths along the $x$-direction and infinitesimal height $dy$.
The two curves are defined as $x=y$ and $x=\frac14(12-y^2)$. Therefore, the width of each rectangle is $$\frac{12-y^2}{4}-y$$
That gives the infinitesimal area $$dA=\left(\frac{12-y^2}{4}-y\right)\,dy$$ We can see that we’re adding up rectangles from $y=-6$ to $y=2$, so we just have to evaluate
$$\begin{align}
A &=\int_{-6}^{2} \left(\frac{12-y^2}{4}-y\right) \, dy \\
&= \frac14 \int_{-6}^{2}\left(12-y^2\right)\, dy -\int_{-6}^{2} y\,dy \\
&= \frac14 \left(12y-\frac13y^3\middle)\right|_{-6}^{2} - \left( \frac12y^2\middle)\right|_{-6}^{2}
\end{align}$$
Can you handle it from there?
You were on track except for taking the square root.