The area (in sq. units) of the region enclosed by the curves $$y = x^2 - 1$$ and $$y = 1 - x^2$$ is equal to:

We begin by recalling the standard formula for the area enclosed between two curves that can be written as functions of $$x$$. If, on a closed interval $$[a,b]$$, the curve $$y = f(x)$$ always lies above the curve $$y = g(x)$$, then the required area is given by

$$\text{Area} \;=\; \int_{a}^{b} \bigl[f(x) - g(x)\bigr]\;dx.$$

In the present problem the two curves are

$$y = x^{2} - 1 \quad\text{and}\quad y = 1 - x^{2}.$$

First, we locate their points of intersection because those $$x$$-values will serve as the limits of integration. To find the intersection points, we equate the two expressions for $$y$$:

$$x^{2} - 1 \;=\; 1 - x^{2}.$$

Transposing all terms to one side gives

$$x^{2} - 1 - 1 + x^{2} \;=\; 0,$$

which simplifies to

$$2x^{2} - 2 \;=\; 0.$$

Dividing every term by $$2$$, we obtain

$$x^{2} - 1 \;=\; 0.$$

Adding $$1$$ to both sides yields

$$x^{2} \;=\; 1.$$

Taking the square root on both sides, we have the two intersection abscissae

$$x = +1 \quad\text{and}\quad x = -1.$$

Next, we must decide which curve is on top (has the larger $$y$$-value) between $$x=-1$$ and $$x=1$$. Taking any convenient point in this interval, say $$x = 0$$, we compute the corresponding $$y$$-values.

For $$y = x^{2}-1$$ at $$x = 0$$:

$$y = 0^{2} - 1 = -1.$$

For $$y = 1 - x^{2}$$ at $$x = 0$$:

$$y = 1 - 0^{2} = 1.$$

Clearly $$1 > -1$$, so the curve $$y = 1 - x^{2}$$ lies above the curve $$y = x^{2} - 1$$ throughout the closed interval $$[-1,1]$$. Therefore, in the formula for area we take $$f(x) = 1 - x^{2}$$ and $$g(x) = x^{2} - 1$$.

Substituting into the area formula, we get

$$\begin{aligned} \text{Area} &= \int_{-1}^{1} \Bigl[(1 - x^{2}) - (x^{2} - 1)\Bigr]\;dx \\ &= \int_{-1}^{1} \Bigl[1 - x^{2} - x^{2} + 1\Bigr]\;dx \\ &= \int_{-1}^{1} \bigl[2 - 2x^{2}\bigr]\;dx. \end{aligned}$$

It is convenient to factor out the common factor $$2$$:

$$ \text{Area} = 2 \int_{-1}^{1} \bigl[1 - x^{2}\bigr]\;dx. $$

We now evaluate the two simpler integrals separately. Using the basic antiderivatives $$\int 1\,dx = x$$ and $$\int x^{2}\,dx = \dfrac{x^{3}}{3}$$, we proceed:

$$\begin{aligned} \int_{-1}^{1} 1 \;dx &= \Bigl[\,x\,\Bigr]_{-1}^{1} = 1 - (-1) = 2, \\ \int_{-1}^{1} x^{2}\;dx &= \left[\frac{x^{3}}{3}\right]_{-1}^{1} = \frac{1^{3}}{3} - \frac{(-1)^{3}}{3} = \frac{1}{3} - \left(-\frac{1}{3}\right) = \frac{2}{3}. \end{aligned}$$

Substituting these numerical results back into the expression for the area, we obtain

$$\begin{aligned} \text{Area} &= 2 \Bigl[\,2 - \frac{2}{3}\Bigr] \\ &= 2 \Bigl[\frac{6}{3} - \frac{2}{3}\Bigr] \\ &= 2 \left[\frac{4}{3}\right] \\ &= \frac{8}{3}. \end{aligned}$$

Thus the enclosed region has an area of $$\dfrac{8}{3}$$ square units.

Hence, the correct answer is Option B.