jmc

Substituting $y^2=3x$ into the equation $x^2+y^2=4,$ we get $x^2+3x=4,$ so $$x^2+3x-4=0.$$Factoring, we get $(x-1)(x+4)=0,$ so $x=1$ or $x=-4$. The $x$-coordinate of a point on the circle $x^2+y^2=4$ must be between $-2$ and 2, so $x=1.$ Then $y^2=4-x^2=3,$ so $y=\pm \sqrt{3}.$ This gives us the points of intersection $(1,\sqrt{3})$ and $(1,-\sqrt{3}).$



By symmetry, the parabola $y^2=-3x$ and circle $x^2+y^2=4$ intersect at $(-1,\sqrt{3})$ and $(-1,-\sqrt{3}).$ Thus, the four points form a rectangle whose dimensions are 2 and $2\sqrt{3},$ so its area is $\boxed{4\sqrt{3}}.$