Question

Find the area of the region in the first quadrant enclosed by the x axis line $y=x$, and the circle $x^2+y^2=32$ ?

Answer 1

The given equation is; $y=x⟶\left(1\right)and,x^2+y^2=32⟶\left(2\right)$

By solving (1) and (2) we find that the line and circle meet at $(4,4)$ in the first quadrant

Draw perpendicular BC to the x-axis

therefore,

Required area$=$area of the region$BCOB$ $+$ area of the region $CBAC$

Now, area of region $BCOB$$={\int }_0^{4}ydx={\int }_0^{4}xdx={\left[\frac{x^2}{2}\right]}_0^4=8$

and, area of the region $CBAC$ $={\int }_4^{4\sqrt{2}}ydx={\int }_4^{4\sqrt{2}}\sqrt{32-x^2}dx={\int }_4^{4\sqrt{2}}\sqrt{{\left(4\sqrt{2}\right)}^2-x^2}dx={[\frac{x}{2}\sqrt{32-x^2}+\frac{32}{2}{\sin }^{-1}\frac{x}{4\sqrt{2}}]}_4^{4\sqrt{2}}=4\pi -8$

therefore required area $=8+4\pi -8=4\pi sq.unit$