Question

# Calculate the area of the region bounded by the parabolas y2 = x and x2 = y.

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Solution

## ${y}^{2}=x\mathrm{is}a\mathrm{parabola},\mathrm{opening}\mathrm{sideways},\mathrm{with}\mathrm{vertex}\mathrm{at}\mathrm{O}\left(0,0\right)\mathrm{and}+\mathrm{ve}x-\mathrm{axis}\mathrm{as}\mathrm{axis}\mathrm{of}\mathrm{symmetry}\phantom{\rule{0ex}{0ex}}{x}^{2}=y\mathrm{is}\mathrm{a}\mathrm{parabola},\mathrm{opening}\mathrm{upwards},\mathrm{with}\mathrm{vertex}\mathrm{at}\mathrm{O}\left(0,0\right)\mathrm{and}+\mathrm{ve}y-\mathrm{axis}\mathrm{as}\mathrm{axis}\mathrm{of}\mathrm{symmetry}\phantom{\rule{0ex}{0ex}}\mathrm{Soving}\mathrm{the}\mathrm{above}\mathrm{two}\mathrm{equations},\phantom{\rule{0ex}{0ex}}{x}^{2}={y}^{4}=y\phantom{\rule{0ex}{0ex}}⇒{y}^{4}-y=0\phantom{\rule{0ex}{0ex}}⇒y=0\mathrm{or}y=1.\phantom{\rule{0ex}{0ex}}\mathrm{So},x=0\mathrm{or}x=1\phantom{\rule{0ex}{0ex}}⇒\mathrm{O}\left(0,0\right)\mathrm{and}\mathrm{A}\left(1,1\right)\mathrm{are}\mathrm{points}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{two}\mathrm{curves}\phantom{\rule{0ex}{0ex}}\mathrm{Consider}\mathrm{a}\mathrm{vertical}\mathrm{strip}\mathrm{of}\mathrm{length}=\left|{y}_{2}-{y}_{1}\right|\mathrm{and}\mathrm{width}=dx\phantom{\rule{0ex}{0ex}}⇒\mathrm{Area}\mathrm{of}\mathrm{approximating}\mathrm{rectangle}=\left|{\mathrm{y}}_{2}-{\mathrm{y}}_{1}\right|dx\phantom{\rule{0ex}{0ex}}\mathrm{Approximating}\mathrm{rectangle}\mathrm{moves}\mathrm{from}x=0\mathrm{to}x=1\phantom{\rule{0ex}{0ex}}⇒\mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{shaded}\mathrm{region}={\int }_{0}^{1}\left|{y}_{2}-{y}_{1}\right|dx\phantom{\rule{0ex}{0ex}}⇒A={\int }_{0}^{1}\left({y}_{2}-{y}_{1}\right)dx\left[\mathrm{As},{y}_{2}-{y}_{1}>0⇒\left|{y}_{2}-{y}_{1}\right|={y}_{1}\right]\phantom{\rule{0ex}{0ex}}⇒A={\int }_{0}^{1}\left(\sqrt{x}-{x}^{2}\right)dx\phantom{\rule{0ex}{0ex}}⇒A={\left[\frac{{x}^{\frac{3}{2}}}{\frac{3}{2}}-\frac{{x}^{3}}{3}\right]}_{0}^{1}\phantom{\rule{0ex}{0ex}}⇒A=\left[\frac{2}{3}×{1}^{\frac{3}{2}}-\frac{{1}^{3}}{3}-0\right]\phantom{\rule{0ex}{0ex}}⇒A=\frac{2}{3}-\frac{1}{3}\phantom{\rule{0ex}{0ex}}⇒A=\frac{1}{3}\mathrm{sq}.\mathrm{units}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\mathrm{area}\mathrm{enclosed}\mathrm{by}\mathrm{the}\mathrm{curves}=\frac{1}{3}\mathrm{sq}.\mathrm{units}\phantom{\rule{0ex}{0ex}}$

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