The area (in sq. units) of the region $$A = \{(x, y) : |x| + |y| \leq 1,\; 2y^2 \geq |x|\}$$

We have to find the area of the set of all points $$A=\{(x,y):|x|+|y|\le 1,\;2y^{2}\ge |x|\}.$$

The inequality $$|x|+|y|\le 1$$ represents a diamond-shaped region (a square of side $$\sqrt2$$) whose vertices are $$(\pm1,0)$$ and $$(0,\pm1).$$

The second inequality $$2y^{2}\ge |x|$$ is the same as $$|x|\le 2y^{2},$$ which describes the vertical “double parabola” opening both left and right from the $$y$$-axis.

Every point in the desired region must satisfy both conditions, so for any fixed $$y$$ the admissible $$x$$ must lie in the interval

$$-\,\min\!\bigl(1-|y|,\;2y^{2}\bigr)\le x\le \min\!\bigl(1-|y|,\;2y^{2}\bigr).$$

Thus the horizontal width for that $$y$$ is

$$\text{width}=2\,\min\!\bigl(1-|y|,\;2y^{2}\bigr).$$

Because both inequalities involve only $$|y|$$ and $$y^{2},$$ the figure is perfectly symmetric about the $$x$$-axis. Hence we may compute the area for $$y\ge 0$$ and then double it.

For $$y\ge 0$$ we can drop the absolute value $$|y|$$ and write

$$f_{1}(y)=1-y,\qquad f_{2}(y)=2y^{2}.$$

The smaller of these two functions controls the half-width. We first locate their point of intersection:

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

Using the quadratic‐formula $$y=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$ with $$a=2,\;b=1,\;c=-1,$$ we get

$$y=\dfrac{-1\pm\sqrt{1+8}}{4}=\dfrac{-1\pm3}{4}.$$

The positive root is $$y=\dfrac{1}{2}.$$

So,

• For $$0\le y\le\dfrac12,\;f_{2}(y)=2y^{2}\le f_{1}(y)=1-y,$$ • For $$\dfrac12\le y\le1,\;f_{1}(y)=1-y\le f_{2}(y)=2y^{2}.$$

Now we write the area integral. First for the upper half ( $$y\ge0$$ ):

$$ \text{Area}_{y\ge0} =\int_{0}^{1}2\,\min\bigl(1-y,\,2y^{2}\bigr)\,dy =\int_{0}^{1/2}2\cdot(2y^{2})\,dy+\int_{1/2}^{1}2\cdot(1-y)\,dy. $$

Compute the two integrals one by one.

For $$0\le y\le\dfrac12$$:

$$ \int_{0}^{1/2}2\cdot(2y^{2})\,dy =4\int_{0}^{1/2}y^{2}\,dy =4\left[\frac{y^{3}}{3}\right]_{0}^{1/2} =4\left(\frac{(1/2)^{3}}{3}\right) =4\left(\frac{1}{24}\right) =\frac{1}{6}. $$

For $$\dfrac12\le y\le1$$:

$$ \int_{1/2}^{1}2\cdot(1-y)\,dy =2\int_{1/2}^{1}(1-y)\,dy =2\left[\;y-\frac{y^{2}}{2}\right]_{1/2}^{1} =2\left[\left(1-\frac12\right)-\left(\frac12-\frac{1}{8}\right)\right] =2\left[\frac12-\frac38\right] =2\left(\frac18\right) =\frac14. $$

Therefore

$$ \text{Area}_{y\ge0} =\frac16+\frac14 =\frac{2}{12}+\frac{3}{12} =\frac{5}{12}. $$

Because the region is symmetric about the $$x$$-axis, the total area is twice this value:

$$ \text{Total Area}=2\times\frac{5}{12}=\frac{5}{6}. $$

Hence, the correct answer is Option D.