Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Q.

The area bounded by the circle $x^{2} + y^{2} = 16$ , and the ordinates x = 0, x = 2 is

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

a

$4 \sqrt{3} + \frac{8 π}{3}$

b

$2 \sqrt{3} - \frac{4 π}{3}$

c

$4 \sqrt{3} - \frac{8 π}{3}$

d

$2 \sqrt{3} + \frac{4 π}{3}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$A r e a = 2 \int_{0}^{2} \sqrt{16 - x^{2}} d x$

$= 2 {(\frac{x}{2} \sqrt{16 - x^{2}} + \frac{16}{2} \sin^{- 1} (\frac{x}{4}))}_{0}^{2} = 2 (\sqrt{12} + 8 \frac{π}{6}) = 4 \sqrt{3} + \frac{8 π}{3}$

Related Blogs

Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs

Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Not sure what to do in the future? Don’t worry! We have a FREE career guidance session just for you!