Practicing Success
Practicing Success
CUET Preparation Today
Find the area in the first quadrant, enclosed by the x-axis, the line $x=\sqrt{3}y$ and the circle $x^2+y^2=4$. |
$\frac{\pi}{2}$ $\frac{\pi}{3}$ $\frac{\pi}{6}$ $\frac{\pi}{4}$ |
$\frac{\pi}{3}$ |
Solve $y=\frac{x}{\sqrt{3}}$ and $x^2+y^2⇒A≡(\sqrt{3},1)$
Required Area = $\int\limits_0^1[\sqrt{4-y^2-3\sqrt{3}y}]dy=\frac{\pi}{3}$ |
