Question

# Find the area bounded by the curves x = y2 and x = 3 − 2y2.

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Solution

## x = y2 is a parabola opening towards positive x-axis , having vertex at O (0,0) and symmetrical about x-axis x = 3 − 2y2 is a parabola opening negative x-axis, having vertex at A (3, 0) and symmetrical about x-axis, cutting y-axis at B and B' Solving the two equations for the point of intersection of two parabolas $x={y}^{2}\phantom{\rule{0ex}{0ex}}x=3-2{y}^{2}\phantom{\rule{0ex}{0ex}}⇒{y}^{2}=3-2{y}^{2}\phantom{\rule{0ex}{0ex}}⇒3{y}^{2}=3\phantom{\rule{0ex}{0ex}}⇒y=±1\phantom{\rule{0ex}{0ex}}y=1,⇒x=1\mathrm{and}y=-1⇒x=1\phantom{\rule{0ex}{0ex}}⇒E\left(1,1\right)\mathrm{and}F\left(1,-1\right)\mathrm{are}\mathrm{two}\mathrm{points}\mathrm{of}\mathrm{intersection}.\phantom{\rule{0ex}{0ex}}\mathrm{The}\mathrm{curve}\mathrm{character}\mathrm{changes}\mathrm{at}E\mathrm{and}F.\phantom{\rule{0ex}{0ex}}\mathrm{Draw}EF\mathit{}\mathrm{parallel}\mathrm{to}\mathit{}y\mathit{-}\mathrm{axis}.\phantom{\rule{0ex}{0ex}}C\left(1,0\right)\mathrm{is}\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}EF\mathrm{ith}\mathit{}x\mathit{-}\mathrm{axis}\phantom{\rule{0ex}{0ex}}\mathrm{Since}\mathrm{both}\mathrm{curves}\mathrm{are}\mathrm{symmetrical}\mathrm{about}x\mathit{-}\mathrm{axis},\phantom{\rule{0ex}{0ex}}\mathrm{Area}\mathrm{of}\mathrm{shaded}\mathrm{region}OEAFO=2\mathrm{Area}OEAO \phantom{\rule{0ex}{0ex}}=2\left(\mathrm{Area}OECO+\mathrm{area}CEAC\right).....\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{Area}OECO={\int }_{\mathrm{o}}^{1}\left|{y}_{1}\right|dx\left\{\mathrm{where}P\left(x,{y}_{1}\right)\mathrm{is}\mathrm{a}\mathrm{point}\mathrm{on}\mathit{}x={y}^{2}\right\}\phantom{\rule{0ex}{0ex}}={\int }_{\mathrm{o}}^{1}{y}_{1}dx\left\{\mathrm{as}{\mathrm{y}}_{1}>0\right\}\phantom{\rule{0ex}{0ex}}={\int }_{\mathrm{o}}^{1}\sqrt{x}dx\phantom{\rule{0ex}{0ex}}={\left[\frac{{\mathrm{x}}^{3}{2}}}{3}{2}}\right]}_{0}^{1}\phantom{\rule{0ex}{0ex}}=\frac{2}{3}\mathrm{sq}\mathrm{units}.....\left(2\right)\phantom{\rule{0ex}{0ex}}\mathrm{area}\mathrm{CEAC}={\int }_{1}^{3}\left|{y}_{2}\right|dx\left\{\mathrm{where}\mathrm{Q}\left(x,{y}_{2}\right)\mathrm{is}\mathrm{a}\mathrm{point}\mathrm{on}x=3-2{y}^{2}\right\}\phantom{\rule{0ex}{0ex}}={\int }_{1}^{3}{y}_{2}dx\left\{\mathrm{as}{\mathrm{y}}_{2}>0\right\}\phantom{\rule{0ex}{0ex}}={\int }_{1}^{3}\sqrt{\frac{3-x}{2}}dx\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{2}}{\int }_{1}^{3}\sqrt{3-x}dx\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{2}}{\left[-\frac{{\left(3-x\right)}^{3}{2}}}{3}{2}}\right]}_{1}^{3}\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{2}}×\frac{2}{3}\left[0+{2}^{3}{2}}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{2×2\sqrt{2}}{\sqrt{2}×3}=\frac{4}{3}\mathrm{sq}.\mathrm{units}.....\left(3\right)\phantom{\rule{0ex}{0ex}}\mathrm{From}\left(1\right),\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{area}\mathrm{of}\mathrm{Shaded}\mathrm{region}OEAFO\mathit{}=2\left[\frac{2}{3}+\frac{4}{3}\right]=2×2=4\mathrm{sq}\mathrm{units}$

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