Question

The area (in $\text{sq.units}$) enclosed between the curve $|x|+|y|<3$ and $x^2-6x+y^2<0$ is

• A. $2\pi$
• B. $\frac{7\pi }{4}$
• C. $3\pi$
• D. $\frac{9\pi }{4}$

Answer 1

The correct option is D $\frac{9\pi }{4}$

Given curve : $|x|+|y|<3$ and
$x^2-6x+y^2<0\Rightarrow (x-3{)}^2+y^2<9$
The region bounded is shown as :

So, Required Area $=$ Area of Arc $\overline{\left(CAOB\right)}$
$=\frac{\theta }{{360}^{\circ }}\times \left(\pi R^2\right)$
Here, $\theta =\frac{\pi }{2}$ $($angle $ACB)$$,R=3$(radius of arc)
$=\frac{\pi }{2}\cdot \frac{\pi (3{)}^2}{2\pi }=\frac{9\pi }{4}\text{sq.units}$