= Solution
The boundaries intersect where $x^2=2x+8$, giving $x=-2$ and $x=4$. Between these values, the straight line lies above the <parabola>. Therefore
$$
I=\int_{-2}^{4}\int_{x^2}^{2x+8}x\,dy\,dx=\int_{-2}^{4}(-x^3+2x^2+8x)\,dx.
$$
Evaluation gives
$$
\boxed{I=\left[-\frac{x^4}{4}+\frac{2x^3}{3}+4x^2\right]_{-2}^{4}=36}.
$$
A useful independent check uses the region's symmetry about $x=1$: its vertical width is $9-(x-1)^2$. Its area is $\int_{-3}^{3}(9-u^2)\,du=36$, and its <centroid> has $x$ coordinate one, so its first moment $\iint x\,dA$ is again $36$.
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