The area of the region bounded by $$y^2 = 8x$$ and $$y^2 = 16(3 - x)$$ is equal to

We need to find the area bounded by the parabolas $$y^2 = 8x$$ and $$y^2 = 16(3-x)$$.

To find their points of intersection, set $$8x = 16(3-x)$$, which gives $$8x = 48 - 16x$$, so $$24x = 48$$ and $$x = 2$$. Substituting back into either equation yields $$y^2 = 16$$, hence $$y = \pm 4$$.

Expressing $$x$$ in terms of $$y$$, the first parabola opening to the right can be written as $$x_1 = \frac{y^2}{8}$$, and the second parabola opening to the left is $$x_2 = 3 - \frac{y^2}{16}$$.

By symmetry about the x-axis, the total area is twice the area for $$0 \le y \le 4$$. Therefore,

$$A = 2\int_0^4 \Bigl[\Bigl(3 - \frac{y^2}{16}\Bigr) - \frac{y^2}{8}\Bigr] dy = 2\int_0^4 \Bigl(3 - \frac{y^2}{16} - \frac{y^2}{8}\Bigr) dy = 2\int_0^4 \Bigl(3 - \frac{3y^2}{16}\Bigr) dy\,. $$

Evaluating this integral,

$$2\left[3y - \frac{3y^3}{48}\right]_0^4 = 2\left[3y - \frac{y^3}{16}\right]_0^4 = 2\bigl(12 - \frac{64}{16}\bigr) = 2(12 - 4) = 16\,. $$

Hence, the area bounded by the two parabolas is $$16$$.