Fun with multiple integrals

The bounding curves of the region are $y = x^2$ and $y = 2x + 8$ . They intersect where $x^2 - 2x - 8 = 0$ so $x=-2,4$ . In this interval, $2x + 8 \geq x^2$ so the integral is:

$\int_{-2}^{4} \int_{x^2}^{2x + 8} x dy dx = \int_{-2}^4 yx \bigg|_{x^2}^{2x + 8} dx = \int_{-2}^4 2x^2 + 8x - x^3 dx = 36$ .