Question

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

Answer 1

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

$$\begin{gathered} x=y^2 \\ x=3-2y^2 \\ \Rightarrow y^2=3-2y^2 \\ \Rightarrow 3y^2=3 \\ \Rightarrow y=\pm 1 \\ y=1,\Rightarrow x=1\mathrm{and}y=-1\Rightarrow x=1 \\ \Rightarrow E\left(1,1\right)\mathrm{and}F\left(1,-1\right)\mathrm{are}\mathrm{two}\mathrm{points}\mathrm{of}\mathrm{intersection}. \\ \mathrm{The}\mathrm{curve}\mathrm{character}\mathrm{changes}\mathrm{at}E\mathrm{and}F. \\ \mathrm{Draw}EF\mathit{}\mathrm{parallel}\mathrm{to}\mathit{}y\mathit{-}\mathrm{axis}. \\ C(1,0)\mathrm{is}\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{of}EF\mathrm{ith}\mathit{}x\mathit{-}\mathrm{axis} \\ \mathrm{Since}\mathrm{both}\mathrm{curves}\mathrm{are}\mathrm{symmetrical}\mathrm{about}x\mathit{-}\mathrm{axis}, \\ \mathrm{Area}\mathrm{of}\mathrm{shaded}\mathrm{region}OEAFO=2\mathrm{Area}OEAO\hspace{0.17em} \\ =2\left(\mathrm{Area}OECO+\mathrm{area}CEAC\right).....\left(1\right) \\ \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\} \\ ={\int }_{\mathrm{o}}^1{y}_{1}dx\left\{\mathrm{as}{\mathrm{y}}_1>0\right\} \\ ={\int }_{\mathrm{o}}^1\sqrt{x}dx \\ ={\left[\frac{{\mathrm{x}}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}_0^1 \\ =\frac{2}{3}\mathrm{sq}\mathrm{units}.....\left(2\right) \\ \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=3-2y^2\right\} \\ ={\int }_1^3{y}_{2}dx\left\{\mathrm{as}{\mathrm{y}}_2>0\right\} \\ ={\int }_1^3\sqrt{\frac{3-x}{2}}dx \\ =\frac{1}{\sqrt{2}}{\int }_1^3\sqrt{3-x}dx \\ =\frac{1}{\sqrt{2}}{\left[-\frac{{\left(3-x\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]}_1^3 \\ =\frac{1}{\sqrt{2}}\times \frac{2}{3}\left[0+2^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right] \\ =\frac{2\times 2\sqrt{2}}{\sqrt{2}\times 3}=\frac{4}{3}\mathrm{sq}.\mathrm{units}.....\left(3\right) \\ \mathrm{From}\left(1\right),\left(2\right)\mathrm{and}\left(3\right),\mathrm{we}\mathrm{get} \\ \mathrm{Therefore},\mathrm{area}\mathrm{of}\mathrm{Shaded}\mathrm{region}OEAFO\mathit{}=2\left[\frac{2}{3}+\frac{4}{3}\right]=2\times 2=4\mathrm{sq}\mathrm{units} \end{gathered}$$