Practicing Success
The area enclosed by the circle $x^2 + y^2 = 2$ excluding the area bounded by $y = x$ and $y^2 =x$, is |
$\frac{1}{12}(24π-1)$ $\frac{1}{6}(12π-1)$ $\frac{1}{12}(6π-1)$ $\frac{1}{12}(12π-1)$ |
$\frac{1}{6}(12π-1)$ |
Clearly, required area A is given by A = Area of the circle - Area of the shaded region. $⇒A=2π-\int\limits_0^1(y_2-y_1)dx$ $⇒A=2π-\int\limits_0^1(\sqrt{x}-x)dx=2π-\left[\frac{2x^{3/2}}{3}-\frac{x^2}{2}\right]_0^1$ $=2π-\frac{1}{6}=\frac{1}{6}(12π-1)$ |