Question

The area bounded by the curve y² = 8x and x² = 8y is
(a) $\frac{16}{3}$sq. units

(b) $\frac{3}{16}$sq. units

(c) $\frac{14}{3}$sq. units

(d) $\frac{3}{14}$sq. units

Answer 1

Non of the given option is correct.



Point of intersection of both the parabolas y² = 8x and x² = 8y is obtained by solving the two equations

$$\begin{gathered} y^2=8x\mathrm{and}{x}^2=8y \\ \therefore \frac{y^4}{64}-8y=0 \\ \Rightarrow y\left(y^3-8^3\right)-0 \\ \Rightarrow y=0\mathrm{or}y=8 \\ \Rightarrow x=0\mathrm{or}x=8 \\ \therefore O\left(0,0\right)\mathrm{and}A\left(8,8\right)\mathrm{are}\mathrm{the}\mathrm{points}\mathrm{of}\mathrm{intersection}. \\ \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{shaded}\mathrm{region}={\int }_0^8\left|y_2-y_1\right|dx \\ ={\int }_0^8\left(y_2-y_1\right)dx \\ ={\int }_0^8\left(\sqrt{8x}-\frac{x^2}{8}\right)dx \\ ={\left[\frac{\sqrt{8}}{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{x}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{1}{8}\times \frac{x^3}{3}\right]}_0^8 \\ =\frac{2}{3}\times \sqrt{8}\times 8^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-\frac{1}{8}\times \frac{8^3}{3}-0 \\ =\frac{2}{3}\times \sqrt{8}\times 8\sqrt{8}-\frac{8^2}{3} \\ =\frac{2}{3}\times 8^2-\frac{8^2}{3} \\ =\frac{8^2}{3}\left(2-1\right) \\ =\frac{64}{3}\mathrm{sq}\mathrm{units} \end{gathered}$$