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

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$\frac{1}{6}$

$\frac{7}{6}$

$\frac{5}{6}$

$\frac{1}{3}$

answer is C.

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Detailed Solution

$\begin{array}{l} | x | + | y | \leq 1 represent region of inside a square \\ 2 y^{2} \geq | x | \Rightarrow \{\begin{cases} 2 y^{2} \geq x if x \geq 0 \\ \geq - x if x < 0 \end{cases} \end{array}$

$which represents region outside/on the two parabolas y^{2} = x / 2 and y^{2} = - x / 2$ By symmetry,

$\begin{matrix} Require area = 4 \times [A r e a b o u n d e d i n Q_{1}] \\ = 4 \times [A r e a b o u n d e d b y x + y = 1, y^{2} = \frac{x}{2} a n d y - a x i s] \\ = 4 \times [\int_{0}^{\frac{1}{2}} x d y + A r e a o f Δ^{l e} f o r m e d b y x + y = 1, y = \frac{1}{2}, y - a x i s] \\ = 4 \times [\int_{0}^{\frac{1}{2}} (2 y^{2}) d y + \frac{1}{2} (\frac{1}{2}) (\frac{1}{2})] \\ = 4 \times [{[\frac{2 y^{3}}{3}]}_{0}^{\frac{1}{2}} + \frac{1}{8}] \\ = 4 \times [\frac{2}{3} \times \frac{1}{8} + \frac{1}{8}] \\ = \frac{1}{3} + \frac{1}{2} \\ = \frac{5}{6} s q . u n i t s \end{matrix}$