The area (in sq. units) of the region A={(x, | vedclass

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(C) The region is defined by $|x|+|y| \leq 1$ and $2y^{2} \geq |x|$.Due to symmetry about both axes,we calculate the area in the first quadrant and mu

Vedclass · Accepted Answer

$\frac{7}{6}$

Vedclass · Answer

(C) The region is defined by $|x|+|y| \leq 1$ and $2y^{2} \geq |x|$.Due to symmetry about both axes,we calculate the area in the first quadrant and multiply by $4$.In the first quadrant $(x \geq 0, y \geq 0)$,the region is bounded by $x+y=1$ and $2y^{2}=x$.The intersection point is $2y^{2} = 1-y$,which gives $2y^{2}+y-1=0$.Solving $(2y-1)(y+1)=0$,we get $y=\frac{1}{2}$ (since $y \geq 0$).At $y=\frac{1}{2}$,$x=1-\frac{1}{2}=\frac{1}{2}$.The area in the first quadrant is the area under the line $x+y=1$ minus the area under the curve $x=2y^{2}$ from $y=0$ to $y=\frac{1}{2}$.Area $= \int_{0}^{1/2} (1-y) dy - \int_{0}^{1/2} 2y^{2} dy = [y - \frac{y^{2}}{2}]_{0}^{1/2} - [\frac{2y^{3}}{3}]_{0}^{1/2} = (\frac{1}{2} - \frac{1}{8}) - (\frac{2}{3} 	imes \frac{1}{8}) = \frac{3}{8} - \frac{1}{12} = \frac{9-2}{24} = \frac{7}{24}$.Total area $= 4 	imes \frac{7}{24} = \frac{7}{6}$ sq. units.

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

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