The area (in sq. unit) of the region described by $$A = \{(x, y) : x^2 + y^2 \leq 1$$ and $$y^2 \leq 1 - x\}$$ is:

We have to find the area of the set

$$A=\{(x,y):\;x^2+y^2\le 1\;\text{ and }\;y^2\le 1-x\}.$$

The first inequality $$x^2+y^2\le 1$$ represents the closed unit disk, that is, every point inside or on the circle of radius 1 centred at the origin.

The second inequality $$y^2\le 1-x$$ can be rewritten as $$x\le 1-y^2,$$ which is the region lying to the left of the parabola $$x=1-y^2.$$

Hence the required region is the part of the unit circle that also lies to the left of the parabola.

First we locate the points where the circle and the parabola meet. Putting $$y^2=1-x$$ into $$x^2+y^2=1,$$ we get

$$x^2+(1-x)=1,$$

which simplifies step by step as follows:

$$x^2 - x + 1 = 1,$$

$$x^2 - x = 0,$$

$$x(x-1)=0.$$

Thus $$x=0 \quad\text{or}\quad x=1.$$

For $$x=0$$ we have $$y^2=1,$$ so $$y=\pm1.$$ For $$x=1$$ we have $$y^2=0,$$ so $$y=0.$$

Therefore the curves intersect at the three points $$(0,1),\;(0,-1),\;(1,0).$$

Now we set up a vertical strip (integration w.r.t. $$y$$). For a fixed $$y$$ satisfying $$-1\le y\le 1,$$ the circle allows $$x$$ from the left boundary $$x=-\sqrt{1-y^2}$$ to the right boundary $$x=\sqrt{1-y^2}.$$ The parabola, however, restricts us further to $$x\le 1-y^2.$$

We have to take the smaller of the two right boundaries:

$$x_{\text{right}}=\min\!\Bigl(\sqrt{1-y^2},\;1-y^2\Bigr).$$

Between $$y=-1$$ and $$y=1$$ it is easy to check (substituting any number like $$y=0.5$$) that

$$1-y^2 \;\lt \;\sqrt{1-y^2},$$

so the parabolic bound is the effective right‐hand edge. The left‐hand edge remains the circle’s $$x_{\text{left}}=-\sqrt{1-y^2}.$$

Hence, for each $$y\in[-1,1],$$ the horizontal width contained in the desired region is

$$\bigl(1-y^2\bigr)-\bigl(-\sqrt{1-y^2}\bigr)=1-y^2+\sqrt{1-y^2}.$$

The total area is therefore

$$ \begin{aligned} \text{Area}&=\int_{y=-1}^{1}\Bigl[1-y^2+\sqrt{1-y^2}\Bigr]\;dy. \end{aligned} $$

The integrand depends only on $$y^2,$$ so it is an even function. Consequently

$$ \text{Area}=2\int_{0}^{1}\Bigl[1-y^2+\sqrt{1-y^2}\Bigr]\;dy. $$

We now evaluate the two parts separately.

1. Integrating $$1-y^2$$:

Using $$\displaystyle\int y^n\,dy=\frac{y^{n+1}}{n+1},$$ we obtain

$$ \int_{0}^{1}(1-y^2)\,dy=\Bigl[y-\frac{y^3}{3}\Bigr]_{0}^{1}=1-\frac13=\frac23. $$

2. Integrating $$\sqrt{1-y^2}$$ from 0 to 1:

We recall the standard result that

$$\int_{-1}^{1}\sqrt{1-y^2}\;dy=\frac{\pi}{2},$$

which is the area of a semicircle of radius 1. Therefore the integral from 0 to 1 (half of that semicircle) is

$$\int_{0}^{1}\sqrt{1-y^2}\;dy=\frac{1}{2}\cdot\frac{\pi}{2}=\frac{\pi}{4}.$$

Putting the two results back into the area expression we get

$$ \begin{aligned} \text{Area}&=2\left[\;\frac23+\frac{\pi}{4}\;\right] \\ &=2\cdot\frac23+2\cdot\frac{\pi}{4} \\ &=\frac43+\frac{\pi}{2}. \end{aligned} $$

Thus, expressed in the same order as the options, the area is

$$\displaystyle\frac{\pi}{2}+\frac{4}{3}.$$

Hence, the correct answer is Option C.