The area bounded by the curves $$y = |x^2 - 1|$$ and $$y = 1$$ is

We need to find the area bounded by $$y = |x^2 - 1|$$ and $$y = 1$$.

Setting $$|x^2 - 1| = 1$$:

Case 1: $$x^2 - 1 = 1 \implies x^2 = 2 \implies x = \pm\sqrt{2}$$

Case 2: $$x^2 - 1 = -1 \implies x^2 = 0 \implies x = 0$$

Also, $$x^2 - 1 = 0$$ at $$x = \pm 1$$, which is where the absolute value changes behavior.

For $$|x| \leq 1$$: $$|x^2 - 1| = 1 - x^2 \leq 1$$. The curve is below $$y = 1$$.

For $$1 \leq |x| \leq \sqrt{2}$$: $$|x^2 - 1| = x^2 - 1 \leq 1$$. The curve is below $$y = 1$$.

For $$|x| > \sqrt{2}$$: $$|x^2 - 1| = x^2 - 1 > 1$$. The curve is above $$y = 1$$.

By symmetry about the $$y$$-axis, the area equals:

$$A = 2\int_0^{\sqrt{2}} \left(1 - |x^2 - 1|\right) dx$$

$$= 2\left[\int_0^{1}\left(1 - (1-x^2)\right)dx + \int_1^{\sqrt{2}}\left(1 - (x^2-1)\right)dx\right]$$

$$= 2\left[\int_0^{1} x^2\,dx + \int_1^{\sqrt{2}} (2 - x^2)\,dx\right]$$

$$\int_0^{1} x^2\,dx = \dfrac{1}{3}$$

$$\int_1^{\sqrt{2}} (2-x^2)\,dx = \left[2x - \dfrac{x^3}{3}\right]_1^{\sqrt{2}} = \left(2\sqrt{2} - \dfrac{2\sqrt{2}}{3}\right) - \left(2 - \dfrac{1}{3}\right) = \dfrac{4\sqrt{2}}{3} - \dfrac{5}{3}$$

$$A = 2\left[\dfrac{1}{3} + \dfrac{4\sqrt{2}}{3} - \dfrac{5}{3}\right] = 2\left[\dfrac{4\sqrt{2} - 4}{3}\right] = \dfrac{8(\sqrt{2}-1)}{3}$$

The correct answer is Option D: $$\dfrac{8}{3}(\sqrt{2} - 1)$$.