Answer
The volume is $2\pi$.
Work Step by Step
Notice that:
$3x^2+3y^2\leq 4-x^2-y^2$
$4x^2+4y^2\leq 4$
$x^2+y^2\leq 1$
The last expression tells us that the paraboloid $z_1=3x^2+3y^2$ is under the paraboloid $z_2=4-x^2-y^2$ over the disk $x^2+y^2\leq 1$ (in polar coordinates $0\leq r\leq 1$.
Then, the volume of the solid bounded by the two paraboloids is given by $V=\int_0^1\int_0^{2\pi} (z_2-z_1)rd\theta dr$.
Evaluate $V$:
$\int_0^1\int_0^{2\pi}((4-r^2)-3r^2)rd\theta dr=\int_0^1\int_0^{2\pi}(4r^3-4r)d\theta dr$
$=\int_0^1(4r-4r^3)\theta]_0^{2\pi} dr$
$=\int_0^1 (8r-8r^3)\pi dr$
$=[(4r^2-2r^4)\pi ]_0^1$
$=2\pi $
