Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Application of Integrals

Question:

The area of the region enclosed between the two circles $x^2+y^2 = 1 $ and $(x-1)^2 +y^2 =1$ is

Options:

$\left(\frac{2\pi }{3} -\frac{\sqrt{3}}{4}\right)$ units

$\left(\frac{\pi }{3} -\frac{2\sqrt{3}}{3}\right)$ units

$\left(\frac{2\pi }{3} -\frac{\sqrt{3}}{2}\right)$ units

$\left(\frac{\pi }{6} -\frac{\sqrt{3}}{2}\right)$ units

Correct Answer:

$\left(\frac{2\pi }{3} -\frac{\sqrt{3}}{2}\right)$ units