Question

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

Answer 1

Given line $y=x$

and the circle $x^2+y^2=32$

$\Rightarrow x^2+y^2=16\times 2=4^2\times 2$

$\Rightarrow x^2+y^2=(4\sqrt{2}{)}^2$

Centre $=(0,0)$

Radius $=4\sqrt{2}$

Drawing the diagram
$\Rightarrow x^2+x^2=32$

$\Rightarrow 2x^2=32$

$\Rightarrow x^2=16$

$\therefore x=\pm 14$

When $x=4$

$y=x=4$

When $x=-4$

$y=x=-4$

So point is $(4,4)\text{and}(-4,-4)$

As point $M$ is in $1^{st}$ Quadrant

$\therefore M=(4,4)$

$\therefore \text{Required Area = Area of}OMNO+\text{Area}NMAN$

$={\int }_0^4{y}_{1}dx+{\int }_0^{4\sqrt{2}}{y}_{2}dx$

Where $y_1=x\text{and}{y}_2=\sqrt{(4\sqrt{2}{)}^2-x^2}$

$={\int }_0^{4}xdx+{\int }_0^{4\sqrt{2}}\sqrt{(4\sqrt{2}{)}^2-x^2}dx$

$={\left[\frac{x^2}{2}\right]}_0^4+{\left[\begin{array}{c}\frac{x^2}{2}\sqrt{(4\sqrt{2}{)}^2-x^2}\\ +\frac{(4\sqrt{2}{)}^2}{2}{\sin }^{-1}\frac{x}{4\sqrt{2}}\end{array}\right]}_4^{4\sqrt{2}}$

$=\frac{16}{2}+[\frac{32}{2}{\sin }^{-1}1-(2\sqrt{32-16}+\frac{32}{2}{\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right))]$

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

$=8+8\pi -8-4\pi$

$=4\pi$

Method 2:

Let $\angle AOM=\theta$

We know the slope of $y=x$ is

$m=1\Rightarrow \tan \theta =1$

$\Rightarrow \tan \theta =1$

$\Rightarrow \theta =\frac{\pi }{4}$

Area of a sector of a circle

$=\frac{\theta }{2\pi }\times \pi r^2=\frac{\theta r^2}{2}$

$=\frac{\frac{\pi }{4}\times (4\sqrt{2}{)}^2}{2}=\frac{\pi \times 32}{8}$

$=4\pi$

Final answer:
Therefore, required area $=4\pi \text{square units}$.