Question

Find the area of the sector of a circle bounded by the circle $x^2+y^2=16$ and the line $y=x$ in the first quadrant.

Answer 1

Centre of circle is $(0,0)$
Radius of circle $=4$
$x^2+y^2=16$ .....(i)
$y=x$ ......(ii)
By equation (i)$ and (ii), we get
$x^2+x^2=16\Rightarrow$ $x^2=8$
$\therefore$ $x=2\sqrt{2},y=2\sqrt{2}$
Shaded area $=$ Area of $OMPO+$ Area of $PMQP$
$={\int }_0^{2\sqrt{2}}ydx+{\int }_{2\sqrt{2}}^{4}ydx$
by line by circle
$={\int }_0^{2\sqrt{2}}xdx+{\int }_{2\sqrt{2}}^4\sqrt{16-x^2}dx$
$={\left[\frac{x^2}{2}\right]}_0^{2\sqrt{2}}+{[\frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}{\sin }^{-1}\frac{x}{4}]}_{2\sqrt{2}}$
$=(\frac{8}{2}-0)+[0+8{\sin }^{-1}1]-[\frac{2\sqrt{2}}{2}\sqrt{16-8}+8{\sin }^{-1}\frac{1}{\sqrt{2}}]$
$=4+[8{\sin }^{-1}1-\sqrt{2}\times 2\sqrt{2}-8{\sin }^{-1}\frac{1}{\sqrt{2}}]$
$=4+(8\times \frac{\pi }{2}-4-8\times \frac{\pi }{4})=4+4\pi -4-2\pi =2\pi$ square unit