The area of smaller part between the circle $x^2+y^2=4$ and the line $x=1$ is _______ sq.units.

The area of the smaller segment bounded by the circle $x^2 + y^2 = 4$ (radius $r = 2$) and the vertical line $x = 1$ can be determined using integration or geometry.

Using integration, the area is given by $A = 2 \int_{1}^{2} \sqrt{4 - x^2} \, dx$. The standard antiderivative yields:

$A = 2 \left[ \frac{x}{2} \sqrt{4 - x^2} + 2 \sin^{-1} \left( \frac{x}{2} \right) \right]_{1}^{2}$

Evaluating from $x = 1$ to $x = 2$ gives:

$A = 2 \left[ \left( 0 + 2 \cdot \frac{\pi}{2} \right) - \left( \frac{1}{2} \sqrt{3} + 2 \cdot \frac{\pi}{6} \right) \right] = 2 \left( \pi - \frac{\pi}{3} - \frac{\sqrt{3}}{2} \right)$

Simplifying results in $A = \frac{4\pi}{3} - \sqrt{3}$.

Geometrically, the intersections at $x = 1$ occur at $y = \pm \sqrt{3}$, forming a central angle of $\frac{2\pi}{3}$ radians. The sector area is $\frac{1}{2} \cdot 4 \cdot \frac{2\pi}{3} = \frac{4\pi}{3}$, and the triangle area is $\frac{1}{2} \cdot 2\sqrt{3} \cdot 1 = \sqrt{3}$. Subtracting gives the same result: $\frac{4\pi}{3} - \sqrt{3}$.