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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,the total area $A$ is twice the area

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) The curves are $x = y^2 - 1$ and $x = |y| \sqrt{1 - y^2}$.Since both curves are symmetric about the $x$-axis,the total area $A$ is twice the area in the first quadrant where $y \in [0, 1]$.In the first quadrant,$x = y \sqrt{1 - y^2}$ and $x = y^2 - 1$.The area $A$ is given by:$A = 2 \int_{0}^{1} [y \sqrt{1 - y^2} - (y^2 - 1)] \, dy$$A = 2 \left[ \int_{0}^{1} y \sqrt{1 - y^2} \, dy - \int_{0}^{1} (y^2 - 1) \, dy \right]$Let $u = 1 - y^2$,then $du = -2y \, dy$,so $y \, dy = -\frac{1}{2} du$.When $y=0, u=1$; when $y=1, u=0$.$\int_{0}^{1} y \sqrt{1 - y^2} \, dy = -\frac{1}{2} \int_{1}^{0} \sqrt{u} \, du = \frac{1}{2} \int_{0}^{1} u^{1/2} \, du = \frac{1}{2} \left[ \frac{u^{3/2}}{3/2} \right]_{0}^{1} = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}$.$\int_{0}^{1} (y^2 - 1) \, dy = \left[ \frac{y^3}{3} - y \right]_{0}^{1} = \frac{1}{3} - 1 = -\frac{2}{3}$.Therefore,$A = 2 \left[ \frac{1}{3} - (-\frac{2}{3}) \right] = 2 \left[ \frac{1}{3} + \frac{2}{3} \right] = 2(1) = 2$.

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

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