# Find the area of the region lying in the first quadrant and enclosed by the x – axis, the line y = x and the circle

Circle: $x ^{2} + y^{2} = 32$....(i)

Line: y = x

Intersection point

Put value of y in equation (i)

$x ^{2} + x^{2} = 32$

$\\ x^{2} = 16$

$\\ x= \pm 4$

If $\\ x=4$ then $\\ y=4$

Intersection point A(4,4)

For point B

Put y= 0 in equation (i)

$x= {4 \sqrt{2}$

$B(4 \sqrt{2},0)$

$\\ Required Area = Area of arc OAB$

$\\ Area of arc OAB =\int_{0}^{4} x d x+\int_{4}^{4 \sqrt{2}} \sqrt{32-x^{2}} d x \\\\ =\left[\frac{x^{2}}{2}\right]_{0}^{4}+\left[\frac{x}{2} \sqrt{32-x^{2}}+\frac{32}{2} \sin ^{-1}\left(\frac{x}{4 \sqrt{2}}\right)\right]_{4}^{4 \sqrt{2}} \\\\ =8+(0+8 \pi-8-4\pi) \\\\=4 \pi$

## Related Chapters

### Preparation Products

##### Knockout JEE Main April 2021 (One Month)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 14000/- ₹ 4999/-
##### Knockout NEET Aug 2021 (One Month)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 14000/- ₹ 4999/-
##### Knockout JEE Main May 2021

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 22999/- ₹ 9999/-
##### Knockout NEET Aug 2021

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Weekend Live Classes, Mentorship from our Experts, Unlimited Mock Tests and Personalized Analysis Reports, 24x7 Doubt Chat Support,.

₹ 22999/- ₹ 9999/-