Area of the region (in square units) enclosed | vedclass

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(C) The given curves are $y^2=8(x+2)$ and $y^2=4(1-x)$.First,find the intersection point of the two parabolas:$8(x+2) = 4(1-x)$$2(x+2) = 1-x$$2x+4 = 1

Vedclass · Accepted Answer

$\frac{8}{3}(3-\sqrt{2})$

Vedclass · Answer

(C) The given curves are $y^2=8(x+2)$ and $y^2=4(1-x)$.First,find the intersection point of the two parabolas:$8(x+2) = 4(1-x)$$2(x+2) = 1-x$$2x+4 = 1-x$$3x = -3 \implies x = -1$.At $x=-1$,$y^2 = 8(-1+2) = 8$,so $y = \pm 2\sqrt{2}$.The region is bounded by the two parabolas and the $Y$-axis $(x=0)$.The area $A$ is given by $2 \int_{-2}^{0} |y| dx$. Specifically,from $x=-2$ to $x=-1$,the boundary is $y^2=8(x+2)$,and from $x=-1$ to $x=0$,the boundary is $y^2=4(1-x)$.$A = 2 \left[ \int_{-2}^{-1} \sqrt{8(x+2)} dx + \int_{-1}^{0} \sqrt{4(1-x)} dx \right]$$A = 2 \left[ 2\sqrt{2} \int_{-2}^{-1} (x+2)^{1/2} dx + 2 \int_{-1}^{0} (1-x)^{1/2} dx \right]$$A = 2 \left[ 2\sqrt{2} \left[ \frac{2}{3}(x+2)^{3/2} \right]_{-2}^{-1} + 2 \left[ -\frac{2}{3}(1-x)^{3/2} \right]_{-1}^{0} \right]$$A = 2 \left[ 2\sqrt{2} \cdot \frac{2}{3}(1)^{3/2} - \frac{4}{3} \left( (1-0)^{3/2} - (1-(-1))^{3/2} \right) \right]$$A = 2 \left[ \frac{4\sqrt{2}}{3} - \frac{4}{3} (1 - 2\sqrt{2}) \right]$$A = 2 \left[ \frac{4\sqrt{2}}{3} - \frac{4}{3} + \frac{8\sqrt{2}}{3} \right] = 2 \left[ \frac{12\sqrt{2}-4}{3} \right] = \frac{8}{3}(3\sqrt{2}-1)$.

Area of the region (in square units) enclosed by the curves $y^2=8(x+2)$,$y^2=4(1-x)$ and the $Y$-axis is

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