Find the area enclosed between the curves $x^2+y^2=4, y=\sqrt 3 x, x >0$.  

The first curve is a circle, while the second one is a line. I think that this area equals $\int\_0^1(\sqrt{4-x^2}-x\sqrt 3)dx$. Am I right?  

EDIT: The area from above is the area above the line $y=x\sqrt 3$, I forgot to mention that.