Question

Find the area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x² + y² = 32.

Answer 1

We have, $x^2+y^2=32$ and $y=x$

The point of intersection of the circle and parabola is obtained by solving the two equations

$$\begin{gathered} \therefore x^2+x^2=32 \\ \Rightarrow 2x^2=32 \\ \Rightarrow x^2=16 \\ \Rightarrow x=\pm 4 \\ \therefore y=\pm 4 \\ \mathrm{Thus}C\left(4,4\right)\mathrm{and}C\text{'}\left(-4,-4\right)\mathrm{are}\mathrm{points}\mathrm{of}\mathrm{intersection}\mathrm{of}\mathrm{the}\mathrm{circle}\mathrm{and}\mathrm{straight}\mathrm{line}. \\ \mathrm{Required}\mathrm{shaded}\mathrm{area}\left(OCAPO\right)=\mathrm{area}\left(OCPO\right)+\mathrm{area}\left(PCAP\right) \\ ={\int }_0^4\left|y_1\right|dx+{\int }_4^{\sqrt{32}}\left|y_2\right|dx \\ ={\int }_0^4{y}_{1}dx+{\int }_4^{\sqrt{32}}{y}_{2}dx\left\{\because y_1>0\Rightarrow \left|y_1\right|=y_1\mathrm{and}{y}_2>0\Rightarrow \left|y_2\right|=y_2\right\} \\ ={\int }_0^{4}xdx+{\int }_4^{\sqrt{32}}\sqrt{32-x^2}dx \\ ={\left[\frac{x^2}{2}\right]}_0^4+{\left[\frac{1}{2}x\sqrt{32-x^2}+\frac{1}{2}\times 32{\sin }^{-1}\left(\frac{x}{\sqrt{32}}\right)\right]}_4^{\sqrt{32}} \\ =8+8\mathrm{\pi }-8-4\mathrm{\pi } \\ =4\mathrm{\pi }\mathrm{sq}\mathrm{units}. \end{gathered}$$