Question

Find the area bounded by the curves 4y = | 4 − x² |, x² + y² = 25 and x = 0 above the x-axis.

Answer 1

$$\begin{gathered} x^2+y^2=25\mathrm{represents}\mathrm{a}\mathrm{circle}\mathrm{with}\mathrm{centre}\mathrm{O}\left((0,0\right)\mathrm{radius}5,\mathrm{Cutting}\mathrm{the}+\mathrm{ve}x\mathit{-}\mathrm{axis}\mathrm{at}\mathrm{C}\left(5,0\right) \\ 4y=\left|4-x^2\right| \\ \Rightarrow 4y=\left\{\begin{array}{l}4-x^2-2\le x\le 2\\ x^2-4x\ge 2\end{array}\right. \\ \Rightarrow 4y=4-x^2,-2\le x\le 2\mathrm{represents}\mathrm{an}\mathrm{arc}\mathrm{of}\mathrm{the}\mathrm{parabola}\mathrm{opening}\mathrm{down}\mathrm{wards},\mathrm{vertex}\mathrm{at}V\left(0,1\right),\mathrm{cutting}\mathrm{the}x\mathrm{axis}\mathrm{at}A\left(2,0\right)\mathrm{and}\mathrm{A}\text{'}\left(-2,0\right). \\ \mathrm{and}4y=x^2-4,x\ge 2\mathrm{represents}\mathrm{an}\mathrm{arc}\mathrm{of}\mathrm{the}\mathrm{parabola}\mathrm{opening}\mathrm{up}\mathrm{wards},\mathrm{vertex}\mathrm{at}\left(0,-1\right),\mathrm{cutting}\mathrm{the}x\mathrm{axis}\mathrm{at}A\left(2,0\right)\mathrm{and}\mathrm{A}\text{'}\left(-2,0\right). \\ \mathrm{Clearly}A\left(2,0\right)\mathrm{and}\mathrm{A}\text{'}\left(-2,0\right)\mathrm{are}\mathrm{points}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{arcs}\mathrm{of}\mathrm{the}\mathrm{two}\mathrm{parabolas} \\ 4y=x^2-4\mathrm{cuts}\mathrm{the}\mathrm{circle}\mathrm{at}\mathrm{B}\left(4,3\right)\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{quadrant} \\ \mathrm{Thus}\mathrm{area}\mathrm{of}\mathrm{shaded}\mathrm{region} \\ =\left(\mathrm{area}\mathrm{of}\mathrm{circle}\mathrm{between}x=0,x=4\right)-\left(\mathrm{area}\mathrm{of}4y=4-x^2\mathrm{between}x=0,x=2\right)-\left(\mathrm{area}\mathrm{of}4y=x^2-4\mathrm{between}x=2,x=4\right) \\ ={\int }_0^4\sqrt{25-x^2}dx-{\int }_0^2\left(\frac{4-x^2}{4}\right)dx-{\int }_2^4\frac{\left(x^2-4\right)}{4}dx \\ ={\left[\frac{1}{2}x\sqrt{25-x^2}+\frac{1}{2}\times 25{\sin }^{-1}\left(\frac{x}{5}\right)\right]}_0^4-{\left[\frac{4x}{4}-\frac{x^3}{4\times 3}\right]}_0^2-{\left[\frac{x^3}{4\times 3}-\frac{4x}{4}\right]}_2^4 \\ =\left\{\frac{1}{2}\times 4\times 3+\frac{1}{2}\times 25{\sin }^{-1}\left(\frac{4}{5}\right)\right\}-\left\{2-\frac{2}{3}\right\}-\left\{\frac{16}{3}-4-\frac{2}{3}+2\right\} \\ =6+\frac{1}{2}\times 25{\sin }^{-1}\left(\frac{4}{5}\right)-\frac{4}{3}-\frac{8}{3} \\ =\frac{25}{2}{\sin }^{-1}\left(\frac{4}{5}\right)+2\mathrm{sq}\mathrm{units} \end{gathered}$$