The area of the region: $$R = \{(x, y) : 5x^2 \leq y \leq 2x^2 + 9\}$$ is

We need to find the area of the region $$R = \{(x, y) : 5x^2 \leq y \leq 2x^2 + 9\}$$.

First, we find the points of intersection of $$y = 5x^2$$ and $$y = 2x^2 + 9$$. Setting $$5x^2 = 2x^2 + 9$$ gives $$3x^2 = 9$$, so $$x^2 = 3$$ and $$x = \pm\sqrt{3}$$.

The area is $$\int_{-\sqrt{3}}^{\sqrt{3}} [(2x^2 + 9) - 5x^2]\,dx = \int_{-\sqrt{3}}^{\sqrt{3}} (9 - 3x^2)\,dx$$.

Since the integrand is an even function, this equals $$2\int_0^{\sqrt{3}} (9 - 3x^2)\,dx = 2\left[9x - x^3\right]_0^{\sqrt{3}}$$.

Evaluating, we get $$2\left[9\sqrt{3} - (\sqrt{3})^3\right] = 2\left[9\sqrt{3} - 3\sqrt{3}\right] = 2 \cdot 6\sqrt{3} = 12\sqrt{3}$$.

Therefore, the area of the region is $$12\sqrt{3}$$ square units.