Find a circle–parabola area and half-turn volume

$dx=-\sqrt{12}\sin\theta\,d\theta$. The bounds $x=0,\sqrt3$ correspond to $\theta=\pi/2,\pi/3$. Since sine is positive here, the integral becomes $12\int_{\pi/3}^{\pi/2}\sin^2\theta\,d\theta=6[\theta-\tfrac12\sin2\theta]_{\pi/3}^{\pi/2}=\pi+3\sqrt3/2$.

$x^2+x^4=12$ gives $(x^2+4)(x^2-3)=0$, so the real intersections are $(\pm\sqrt3,3)$. By symmetry, the area is $2[\int_0^{\sqrt3}\sqrt{12-x^2}dx-\int_0^{\sqrt3}x^2dx]=2\pi+\sqrt3$ square units.

The region straddles the y-axis. Rotating its positive and negative x halves through $\pi$ sweeps complementary semicircles, so each horizontal segment produces a full disc. For $0\le y\le3$, radius squared is $y$; for $3\le y\le\sqrt{12}$, it is $12-y^2$. Thus $V=\pi\int_0^3y\,dy+\pi\int_3^{\sqrt{12}}(12-y^2)dy=\pi(16\sqrt3-45/2)$ cubic units.