Using integration, find the area of the region in the first quadrant enclosed by the x-axis, the line and the circle <img alt="x^{2}+y^{2}= 32\cdot" src="https://entra

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

Answers (1)

given curve is : $x^{2}+y^{2}= 32---(i)$
                   $x^{2}+y^{2}= \left ( \sqrt{32} \right )^{2}= \left ( 4\sqrt{2} \right )^{2}$
It is a circle with centre (0,0) and radius $4\sqrt{2}$
given line is $y= x---(ii)$
solving equations (i) and (ii) for points of intersection
$x^{2}+x^{2}= 32 \left [ using (ii) and (i) \right ]$
$\Rightarrow x^{2}= 16$
$\Rightarrow x= \pm 4$
$x= 4\, \, \left [ first\, quadrant \right ]$
$\therefore$ point of intersection is $\left ( 4,4 \right )$

Required area = Area OMA
                    = Area OMP + Area MPA.
$= \int_{0}^{4}x\, dx+\int_{4}^{4\sqrt{2}}\sqrt{\left ( 4\sqrt{2} \right )^{2}-x^{2}\, }dx$
$= \left [ \frac{x^{2}}{2} \right ]^{4}_{0}+\left [ \frac{x}{2}\sqrt{\left ( 4\sqrt{2} \right )^{2}-x^{2}} +\frac{\left ( 4\sqrt{2} \right )^{2}}{2}\sin^{-1}\left ( \frac{x}{4\sqrt{2}} \right )\right ]^{4\sqrt{2}}_{4}$
$= \frac{16}{2}+\left [ \left \{ \frac{4\sqrt{2}}{2}\sqrt{\left ( 4\sqrt{2} \right )^{2}-\left ( 4\sqrt{2} \right ) ^{2}}+\frac{32}{2}\sin^{-1}1\right \} -\left \{ \frac{4}{2}\sqrt{\left ( 4\sqrt{2} \right )^{2}-\left ( 4 \right )^{2}}+\frac{32}{2} \sin^{-1}\frac{1}{\sqrt{2}}\right \}\right ]$
$= 8+\left ( 2\sqrt{2}\left ( 0 \right )+16\times \frac{\pi }{2} \right )-\left ( 2\times 4+16\times \frac{\pi }{4} \right )$
$= 8+8\pi -8-4\pi$
$=4\pi \, sq.units$