Question

Find the area of the region common to the parabolas 4y² = 9x and 3x² = 16y.

Answer 1

$$\begin{gathered} 3x^2=16y...\left(1\right)\mathrm{is}\mathrm{a}\mathrm{parabola}\mathrm{with}\mathrm{vertex}\mathrm{at}(0,0)\mathrm{opening}\mathrm{upwards}\mathrm{and}\mathrm{symmetrical}\mathrm{about}+\mathrm{ve}y-\mathrm{axis} \\ 4y^2=9x...\left(2\right)\mathrm{is}\mathrm{a}\mathrm{parabola}\mathrm{with}\mathrm{vertex}\mathrm{at}(0,0)\mathrm{opening}\mathrm{sideways}\mathrm{and}\mathrm{symmetrical}\mathrm{about}+\mathrm{ve}x-\mathrm{axis} \\ \mathrm{Solving}\mathrm{the}\mathrm{equations}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}\mathrm{the}\mathrm{points}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{the}\mathrm{two}\mathrm{parabolas}\mathrm{O}(0,0)\mathrm{and}\mathrm{A}(4,3) \\ \mathrm{Consider}\mathrm{a}\mathrm{vertical}\mathrm{strip}\mathrm{of}\mathrm{length}=\left|y_2-y_1\right|\mathrm{and}\mathrm{width}=dx\mathrm{such}\mathrm{that}\mathrm{P}(x,y_1)\mathrm{lies}\mathrm{on}\left(1\right)\mathrm{and}\mathrm{Q}(x,y_2)\mathrm{lies}\mathrm{on}\left(2\right) \\ \Rightarrow \mathrm{area}\mathrm{of}\mathrm{approximating}\mathrm{rectangle}=\left|y_2-y_1\right|dx \\ A\mathrm{pproximating}\mathrm{rectangle}\mathrm{moves}\mathrm{from}x=0\mathrm{to}x=4 \\ \Rightarrow \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{shaded}\mathrm{region}={\int }_0^4\left|y_2-y_1\right|dx={\int }_0^4\left(y_2-y_1\right)dx\left[\mathrm{As},\left|y_2-y_1\right|=y_2-y_1\mathrm{for}{\mathrm{y}}_2-{\mathrm{y}}_1>0\right] \\ \Rightarrow A={\int }_0^4\left(\sqrt{\frac{9}{4}x}-\frac{3}{16}{x}^2\right)dx \\ \Rightarrow A=\frac{3}{2}{\int }_0^4\sqrt{x}dx-\frac{3}{16}{\int }_0^4{x}^{2}dx \\ \Rightarrow A=\frac{3}{2}{\left[\frac{x^{{\displaystyle \frac{3}{2}}}}{{\displaystyle \frac{3}{2}}}\right]}_0^4-\frac{3}{16}{\left[\frac{x^3}{3}\right]}_0^4 \\ \Rightarrow A=4^{\frac{3}{2}}-\left(\frac{1}{16}\times 4^3\right) \\ \Rightarrow A=8-4=4\mathrm{sq}.\mathrm{units} \\ \therefore \mathrm{Area}\mathrm{bound}\mathrm{by}\mathrm{the}\mathrm{two}\mathrm{curves}=4\mathrm{sq}.\mathrm{units} \end{gathered}$$