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(D) The curves are $x = y^2 - 1$ and $x = |y| \sqrt{1 - y^2}$.Since both curves are symmetric about the $x$-axis (as they involve $y^2$ or $|y|$),the

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$2$

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(D) The curves are $x = y^2 - 1$ and $x = |y| \sqrt{1 - y^2}$.Since both curves are symmetric about the $x$-axis (as they involve $y^2$ or $|y|$),the total area is twice the area in the first quadrant $(y \ge 0)$.In the first quadrant,the curves are $x = y^2 - 1$ and $x = y \sqrt{1 - y^2}$.The intersection points are found by setting $y^2 - 1 = y \sqrt{1 - y^2}$. Squaring both sides gives $(y^2 - 1)^2 = y^2(1 - y^2)$,which simplifies to $(y^2 - 1)^2 + y^2(y^2 - 1) = 0$,leading to $(y^2 - 1)(y^2 - 1 + y^2) = 0$. Thus $y^2 = 1$ or $2y^2 = 1$,so $y = 1$ or $y = 1/\sqrt{2}$.The area $A$ is given by $2 \int_{0}^{1} |y \sqrt{1 - y^2} - (y^2 - 1)| dy$.Evaluating the integral: $A = 2 \int_{0}^{1} (y \sqrt{1 - y^2} - y^2 + 1) dy$.$A = 2 \left[ -\frac{1}{3}(1 - y^2)^{3/2} - \frac{y^3}{3} + y \right]_{0}^{1}$.$A = 2 \left[ (0 - 1/3 + 1) - (-1/3 - 0 + 0) \right] = 2 \left[ 2/3 + 1/3 \right] = 2(1) = 2$.

Area of the region enclosed between the curves $x = y^2 - 1$ and $x = |y| \sqrt{1 - y^2}$ is

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