Volume of $\{(x,y,z)\mid 3x^2-4y+2y^2+2z-3=0, z>0\}$ Consider S=$\{(x,y,z)\mid| 3x^2-4y+2y^2+2z-3=0\}$

How can I determine the volume of $\{(x,y,z)\mid 3x^2-4y+2y^2+2z-3=0, z>0\}$? That is the part of $S$ over the x-y axis.
My wrong approach was to integrate $z^2\pi=(\frac{1}{2}(3-2y^2+4y-3x^2))^2\pi$ from $z=0$ to the top $z=5/2$ but as you can see this thing isn't  circular. Maybe using cylindrical coordinates?
Furthermore I am looking for equations to describe the bottom  and the lateral surface of it.
For the bottom I have
$\{(x,y,z)\mid 3x^2-4y+2y^2-3\le 0, z=0\}$
And the surface:
$\{(x,y,z)\mid\frac{1}{2}(3-2y^2+4y-3x^2)=z,0<z<5/2\}$

Rewrite the equation as 
$$3x^2+2(y-1)^2+2z-5=0$$
Again, rewrite it in a standard elliptical form
$$\frac{x^2}{\frac{2}{3}(\frac52-z)}+\frac{(y-1)^2}{\frac 52 -z}=1$$
and note that the volume can be viewed as a stack of elliptical   disks from 0 to 5/2 along the vertical direction. Use the area formula $\pi ab$ for an ellipse, where $a$ and $b$ are the major and minor axises, Then, the area of each disk  at $z$ is  
$$ \sqrt{\frac 23} \left(\frac52 -z\right) \pi$$
So, its volume integral can be simply  expressed as
$$I= \pi\sqrt{\frac 23} \int_0^{5/2} \left( \frac 52 -z\right)dz
= \frac{25\pi}{8}\sqrt{\frac 23} $$
