Question

Using integration, find the area of the region bounded between the line x = 2 and the parabola y² = 8x.

Answer 1

$$\begin{gathered} y^2=8x\mathrm{represents}\mathrm{a}\mathrm{parabola}\mathrm{with}\mathrm{vertex}\mathrm{at}\mathrm{origin}\mathrm{and}\mathrm{axis}\mathrm{of}\mathrm{symmetry}\mathrm{along}\mathrm{the}+\mathrm{ve}\mathrm{direction}\mathrm{of}x-\mathrm{axis} \\ x=2\mathrm{is}\mathrm{line}\mathrm{parallel}\mathrm{to}y-\mathrm{axis} \\ \mathrm{Let}(x,y)\mathrm{be}\mathrm{a}\mathrm{given}\mathrm{point}\mathrm{on}\mathrm{the}\mathrm{parabola},y^2=8x \\ \mathrm{Since}\mathrm{parabola}{y}^2=8x\mathrm{is}\mathrm{symmetric}\mathrm{about}x-\mathrm{axis}, \\ \therefore \mathrm{Required}\mathrm{area}=2\left(\mathrm{area}\mathrm{OCAO}\right) \\ \mathrm{On}\mathrm{slicing}\mathrm{the}\mathrm{area}\mathrm{above}x-\mathrm{axis}\mathrm{into}\mathrm{vertical}\mathrm{strips}\mathrm{of}\mathrm{length}=\left|y\right|\mathrm{and}\mathrm{width}=dx \\ \Rightarrow \mathrm{area}\mathrm{of}\mathrm{rectangular}\mathrm{strip}=\left|y\right|dx \\ \mathrm{The}\mathrm{approximating}\mathrm{rectangle}\mathrm{moves}\mathrm{between}x=0\mathrm{and}x=2. \\ \mathrm{So},\mathrm{area}A=2\left(\mathrm{area}\mathrm{OCAO}\right) \\ \Rightarrow A=2{\int }_0^2\left|y\right|dx=2{\int }_0^{2}ydx\mathrm{as}y>0 \\ \Rightarrow A=2{\int }_0^2\sqrt{8x}dx \\ \Rightarrow A=2\times 2{\int }_0^2\sqrt{2x}dx=4\sqrt{2}{\int }_0^2\sqrt{x}dx \\ \Rightarrow A=4\sqrt{2}{\left[\frac{x^{{\displaystyle \frac{\mathit{3}}{\mathit{2}}}}}{{\displaystyle \frac{3}{2}}}\right]}_0^2=\frac{8}{3}\sqrt{2}\left[2^{\frac{3}{2}}-0\right]=\frac{8}{3}\times 2^2=\frac{32}{3}\mathrm{sq}.\mathrm{units} \\ \therefore \mathrm{Area}A=\frac{32}{3}\mathrm{sq}.\mathrm{units} \end{gathered}$$