Question

Find the area of the region bounded by the parabola y² = 4ax and the line x = a.

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{equation}{y}^2=4ax\mathrm{represents}\mathrm{a}\mathrm{parabola},\mathrm{with}\mathrm{vertex}\mathrm{at}(0,0)\mathrm{and}\mathrm{symmetric}\mathrm{about}\mathrm{positive}\mathrm{side}\mathrm{of}x-\mathrm{axis} \\ x=a\mathrm{is}\mathrm{a}\mathrm{line}\mathrm{par}a\mathrm{llel}\mathrm{to}y-\mathrm{axis},\mathrm{and}\mathrm{cutting}x-\mathrm{axis}\mathrm{at}(a,0) \\ \mathrm{Making}\mathrm{vertical}\mathrm{strips}\mathrm{of}\mathrm{length}=\left|y\right|\mathrm{and}\mathrm{width}=dx\mathrm{in}\mathrm{the}\mathrm{quadrant}\mathrm{OLSO}. \\ \mathrm{Area}\mathrm{of}\mathrm{approximating}\mathrm{rectangle}=\left|y\right|dx \\ \mathrm{Since}\mathrm{the}\mathrm{approximating}\mathrm{rectangle}\mathrm{can}\mathrm{move}\mathrm{between}x=0\mathrm{and}x=a, \\ \mathrm{and}\mathrm{as}\mathrm{the}\mathrm{parabola}\mathrm{is}\mathrm{symmetric}\mathrm{about}x-\mathrm{axis}, \\ \mathrm{Required}\mathrm{shaded}\mathrm{area}\mathrm{OLSO}=\mathrm{A}=2\times \mathrm{Area}\mathrm{OLMO} \\ A=2{\int }_0^a\left|y\right|dx=2{\int }_{\mathit{0}}^{a}y\mathit{}dx\mathit{}\mathit{}\mathit{}\mathit{}\left[\mathrm{As},y>0\Rightarrow \left|y\right|=y\right] \\ \Rightarrow A=2{\int }_0^a\sqrt{4ax}dx \\ \Rightarrow A=4{\int }_0^a\sqrt{ax}dx \\ \Rightarrow A=4\sqrt{a}{\int }_0^a\sqrt{x}dx \\ \Rightarrow A=4\sqrt{a}{\left[\frac{x^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\right]}_0^a \\ \Rightarrow A=\frac{8}{3}\sqrt{a}\left[a^{\frac{3}{2}}-0\right] \\ \Rightarrow A=\frac{8}{3}{a}^2\mathrm{sq}.\mathrm{units} \end{gathered}$$