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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Aicharya · Accepted Answer

$y^2=8(x+2)$ $y^2=4(1-x)$

The area of the required region is $A=2\left[\int_{-2}^{-1} y_1 d x+\int_{-1}^0 y_2 d x\right]$ where $y_1=\sqrt{8(x+2)}=2 \sqrt{2}(x+2)^{1 / 2}$ and $y_2=\sqrt{4(1-x)}=2(1-x)^{1 / 2}$ $\therefore \quad A=2\left[\int_{-2}^{-1} 2 \sqrt{2}(x+2)^{1 / 2} d x+2 \int_{-1}^0(1-x)^{1 / 2} d x\right]$ $=2\left[2 \sqrt{2}\left[\frac{(x+2)^{3 / 2}}{3 / 2}\right]_{-2}^{-1}+2\left\{\frac{-2}{3}(1-x)^{3 / 2}\right\}_{-1}^0\right]$ $=2\left[2 \sqrt{2} \times \frac{2}{3}+\frac{2 \times 2}{3}(2 \sqrt{2}-1)\right]$ $=\frac{8}{3}[\sqrt{2}+2 \sqrt{2}-1]=\frac{8}{3}[3 \sqrt{2}-1]$

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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