Question

Draw a rough sketch to indicate the region bounded between the curve y² = 4x and the line x = 3. Also, find the area of this region.

Answer 1

$$\begin{gathered} y^2=4x\mathrm{represents}\mathrm{a}\mathrm{parabola}\mathrm{with}\mathrm{vertex}\mathrm{at}(0,0)\mathrm{and}\mathrm{axis}\mathrm{of}\mathrm{symmetry}\mathrm{along}\mathrm{the}+\mathrm{ve}\mathrm{direction}\mathrm{of}\mathit{}x\mathit{}\mathrm{axis} \\ x=3\mathrm{is}\mathrm{a}\mathrm{line}\mathrm{parallel}\mathrm{to}\mathrm{y}\mathrm{axis}\mathrm{and}\mathrm{cutting}x\mathrm{axis}\mathrm{at}(3,0) \\ \mathrm{Since}{y}^2=4x\mathrm{is}\mathrm{symmetrical}\mathrm{about}\mathrm{x}\mathrm{axis}, \\ \therefore \mathrm{Required}\mathrm{area}A=\mathrm{OA}\hspace{0.17em}\mathrm{CO}=2\times \mathrm{area}\mathrm{OABO} \\ \mathrm{Slicing}\mathrm{the}\mathrm{area}\mathrm{above}\mathit{}x\mathit{}\mathrm{axis}\mathrm{into}\mathrm{vertical}\mathrm{strips}\mathrm{of}\mathrm{length}=\left|y\right|\mathrm{and}\mathrm{width}=dx \\ \mathrm{Area}\mathrm{of}\mathrm{corresponding}\mathrm{rectangle}=\left|y\right|dx \\ \mathrm{The}\mathrm{corresponding}\mathrm{rectangle}\mathrm{moves}\mathrm{from}x=0\mathrm{to}x=3 \\ A=2\times \mathrm{area}\mathrm{OABO} \\ \Rightarrow A=2{\int }_0^3\left|y\right|dx=2{\int }_0^3\left|y\right|dx\left[\mathrm{As},\left|y\right|=y,y>0\right] \\ \Rightarrow A=2{\int }_0^3\sqrt{\left(4x\right)}dx \\ \Rightarrow A=4{\int }_0^3\sqrt{x}dx \\ \Rightarrow A=4{\left[\frac{x^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\right]}_0^3=\frac{8}{3}{\left[x^{\frac{3}{2}}\right]}_0^3=\frac{8}{3}\times 3\sqrt{3}=8\sqrt{3}\mathrm{sq}.\mathrm{units} \\ \therefore \mathrm{Area}\mathrm{of}\mathrm{region}\mathrm{bound}\mathrm{by}\mathrm{curve}{y}^2=4x\mathrm{and}x=3\mathrm{is}8\sqrt{3}\mathrm{s}\mathrm{q}.\mathrm{u}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{s} \end{gathered}$$