Question

Using integration, find the area of the region in the first quadrant enclosed by the $x-axis$, the line $y=x$ and the circle $x^2+y^2=32$.

Answer 1

Equation of the circle given is $x^2+y^2=32$

Radius of the given circle is $=\sqrt{32}=4\sqrt{2}$

$x^2+y^2=32$

$\Rightarrow y=\sqrt{32-x^2}$

$(4,4)$ is the point of Intersection of the circle and the line $y=x$

This can be calculated by solving the equations $y=x$ and

$x^2+y^2=32$ simultaneously

The required area is the area of region $OAB$.

$A=(4\sqrt{2},0),B=(4,4)$ and $O=(0,0)$

Area $={\int }_0^{4}xdx+{\int }_4^{4\sqrt{2}}\sqrt{32-x^2}dx$

$={\left(\frac{x^2}{2}\right)}_0^4+{(\frac{x\sqrt{32-x^2}}{2}+\frac{32}{2}{\sin }^{-1}\left(\frac{x}{4\sqrt{2}}\right))}_4^{4\sqrt{2}}$

$=8+0-8+16\frac{\pi }{2}-16\frac{\pi }{4}$

$=4\pi$

This is the required answer.