The area of the region (in sq. units), enclosed by the circle $$x^2 + y^2 = 2$$ which is not common to the region bounded by the parabola $$y^2 = x$$ and the straight line $$y = x$$, is

We begin with the circle whose equation is $$x^{2}+y^{2}=2$$. The centre is at the origin and the radius is obtained from the standard form $$x^{2}+y^{2}=r^{2}$$, so here $$r^{2}=2$$ which gives $$r=\sqrt{2}$$.

Next we look at the region bounded by the parabola $$y^{2}=x$$ and the straight line $$y=x$$. To find their common points we equate the two expressions for $$x$$:

$$y^{2}=y \;\;\Longrightarrow\;\; y^{2}-y=0 \;\;\Longrightarrow\;\; y(y-1)=0.$$

Thus the two curves meet at $$y=0$$ and $$y=1$$. Substituting these $$y$$-values back gives the corresponding $$x$$-coordinates:

For $$y=0$$: $$x=0$$, for $$y=1$$: $$x=1$$.

Therefore the intersection points are $$(0,0)$$ and $$(1,1)$$.

Between $$y=0$$ and $$y=1$$ we compare the $$x$$-coordinates of the two curves. For any $$0<y<1$$ we have $$y>y^{2}$$, so the straight line $$x=y$$ lies to the right of the parabola $$x=y^{2}$$. Hence the region bounded by the two curves is the set of points with $$y$$ between 0 and 1, $$x$$ ranging from $$x=y^{2}$$ (left) to $$x=y$$ (right).

Let us confirm that this region actually sits inside the circle. Take an arbitrary point of the region: its coordinates satisfy $$y^{2}\le x\le y$$ with $$0\le y\le1$$. The maximum possible value of $$x^{2}+y^{2}$$ inside the strip occurs when $$x=y$$, so

$$x^{2}+y^{2}\le y^{2}+y^{2}=2y^{2}\le2\quad(\text{because }y\le1).$$

Thus every point of the parabola-line region satisfies $$x^{2}+y^{2}\le2$$ and is therefore inside (or on) the circle. In particular, the point $$(1,1)$$ lies exactly on the circle since $$1^{2}+1^{2}=2$$.

What we require is the area enclosed by the circle that is not common to the parabola-line region. In words:

Area wanted = (Area of the whole circle) - (Area common to circle & parabola-line region).

But we have just seen that the entire parabola-line region is contained in the circle, so “area common” is simply the area between the parabola and the line from $$y=0$$ to $$y=1$$.

Let us calculate these two areas one by one.

1. Area of the circle. The standard formula is

$$\text{Area}=\pi r^{2}.$$

Here $$r^{2}=2$$, hence

$$\text{Area of circle}= \pi(2)=2\pi.$$

2. Area bounded by $$x=y^{2}$$ and $$x=y$$ between $$y=0$$ and $$y=1$$. Using the strip (vertical) method in which $$y$$ is the variable of integration, the elementary area element is

$$dA = (x_{\text{right}}-x_{\text{left}})\,dy = (y - y^{2})\,dy.$$

Integrating from $$y=0$$ to $$y=1$$ gives

$$\text{Common area}= \int_{0}^{1}(y - y^{2})\,dy.$$

We integrate term by term:

$$\int y\,dy = \frac{y^{2}}{2}, \qquad \int y^{2}\,dy = \frac{y^{3}}{3}.$$

Hence

$$\text{Common area}= \left[\frac{y^{2}}{2} - \frac{y^{3}}{3}\right]_{0}^{1} = \left(\frac{1}{2}-\frac{1}{3}\right)-\left(0-0\right) = \frac{1}{2}-\frac{1}{3} = \frac{1}{6}.$$

Finally, subtracting this overlap from the whole circle area we get

$$\text{Required area}= 2\pi - \frac{1}{6}.$$

To match the given options we write everything with the common denominator 6:

$$2\pi= \frac{12\pi}{6},$$

so

$$2\pi - \frac{1}{6}= \frac{12\pi}{6}-\frac{1}{6}= \frac{1}{6}(12\pi - 1).$$

This expression exactly coincides with Option D.

Hence, the correct answer is Option D.