Question

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

Answer 1

Given
$y=x$ --- (1)
$x^2+y^2=32$ ----(2)
Solving (1) and (2), we find that the line and the circle meet at B(4, 4) in the first quadrant as shown in the figure.
Draw perpendicular BM to the x-axis.
Therefore, the required area = area of the region OBMO + area of the region BMAB.
Now, the area of the region $OBMO={\int }_0^{4}ydx={\int }_0^{4}xdx$
$=\frac{1}{2}{\left[x^2\right]}_0^4=8$--- (3)
Again, the area of the region BMAB,
$={\int }_4^{4\sqrt{2}}ydx={\int }_4^{4\sqrt{2}}\sqrt{32-x^2}dx$
$={[\frac{1}{2}x\sqrt{32-x^2}+\frac{1}{2}\times 32\times si{n}^{-1}\frac{x}{4\sqrt{2}}]}_4^4$
$-=[\frac{1}{2}4\sqrt{2}\times 0+\frac{1}{2}\times 32\times si{n}^{-1}1]-[\frac{4}{2}\sqrt{32-16}+\frac{1}{2}\times 32\times si{n}^{-1}\frac{1}{\sqrt{2}}]$
$=8\pi -(8+4\pi )=4\pi -8$ --- (4)
Adding (3) and (4), we get the required area = $4\pi$ sq.units.