The area of the region enclosed by $$y \leq 4x^2$$, $$x^2 \leq 9y$$ and $$y \leq 4$$, is equal to

We need to find the area of the region enclosed by $$y \leq 4x^2$$, $$x^2 \leq 9y$$, and $$y \leq 4$$.

Curve 1: $$y = 4x^2$$ is a narrow upward parabola.

Curve 2: $$x^2 = 9y$$, i.e., $$y = \frac{x^2}{9}$$ is a wide upward parabola.

Line: $$y = 4$$ is a horizontal line.

Intersection of $$y = 4x^2$$ and $$y = 4$$: $$4x^2 = 4 \Rightarrow x = \pm 1$$.

Intersection of $$x^2 = 9y$$ and $$y = 4$$: $$x^2 = 36 \Rightarrow x = \pm 6$$.

Intersection of $$y = 4x^2$$ and $$x^2 = 9y$$: Substituting $$x^2 = 9y$$ into $$y = 4x^2$$ gives $$y = 36y$$, so $$y = 0$$ (origin).

The three inequalities are:

$$y \leq 4x^2$$: region below the narrow parabola $$y = 4x^2$$.

$$x^2 \leq 9y$$, i.e., $$y \geq \frac{x^2}{9}$$: region above the wide parabola.

$$y \leq 4$$: region below the horizontal line.

For $$0 \leq x \leq 1$$: $$4x^2 \geq \frac{x^2}{9}$$, so the region is $$\frac{x^2}{9} \leq y \leq 4x^2$$.

For $$1 \leq x \leq 6$$: $$4x^2 \geq 4$$, so the constraint $$y \leq 4$$ is tighter, giving $$\frac{x^2}{9} \leq y \leq 4$$.

By symmetry about the y-axis, Area = $$2 \times$$ (area for $$x \geq 0$$).

Part 1 ($$0 \leq x \leq 1$$):

$$\int_0^1 \left(4x^2 - \frac{x^2}{9}\right) dx = \int_0^1 \frac{35x^2}{9}\, dx = \frac{35}{9} \cdot \frac{x^3}{3}\Big|_0^1 = \frac{35}{27}$$

Part 2 ($$1 \leq x \leq 6$$):

$$\int_1^6 \left(4 - \frac{x^2}{9}\right) dx = \left[4x - \frac{x^3}{27}\right]_1^6 = \left(24 - \frac{216}{27}\right) - \left(4 - \frac{1}{27}\right) = (24 - 8) - \left(\frac{107}{27}\right) = 16 - \frac{107}{27} = \frac{432 - 107}{27} = \frac{325}{27}$$ $$\text{Area} = 2\left(\frac{35}{27} + \frac{325}{27}\right) = 2 \times \frac{360}{27} = 2 \times \frac{40}{3} = \frac{80}{3}$$

The correct answer is Option D: $$\frac{80}{3}$$.