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:
$\{(x,y,z)\mid\frac{1}{2}(3-2y^2+4y-3x^2)=z,0<z<5/2\}$
 A: The first thing I would do is "complete the square" (that's the second time today I have written that!).  $2y^2- 4y= 2(y^2- 2y)= 2(y^2- 2y+ 1- 1)= 2(y^2- 2y+ 1)- 2= 2(y- 1)^2- 2$.  So we can write the formula as $3x^2+ 2(y- 1)^2- 2+ 2z- 3= 0$, $z= 5/2- (3/2)x^2- 2(y- 1)^2$.  Now let u= y- 1.  We have $z= 5/2- (3/2)x^2- u^2$.  In the Xu-plane that is the ellipse, $(3/2)x^2+ u^2= 5/2$.  When u= 0, $(3/2)x^2= 5/2$ so $x^2= 5/3$.  x goes from $-\sqrt{5/3}$ to $\sqrt{5/3}$.  And $u^2= 5/2- (3/2)x^2$ so for each x, u goes from $-\sqrt{5/2- (3/2)x^2}$ to $\sqrt{5/2- (3/3)x^2}$.  
The volume is given by $\int_{-\sqrt{5}{3}}^{\sqrt{5/3}}\int_{-\sqrt{5/2- (3/2)x^2}}^{\sqrt{5/2- (3/2)x^2}} \left(5/2- (3/2)x^2- u^2\right) dudx$.
A: 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
$$I= \pi\sqrt{\frac 23} \int_0^{5/2} \left( \frac 52 -z\right)dz
= \frac{25\pi}{8}\sqrt{\frac 23} $$
A: Transform it so you can use cylindrical coordinates, then. One can find without much work that the vertex of this paraboloid is at $(0,1,5/2)$, so let $x=u/\sqrt3$ and $y=v/\sqrt2+1$. The transformed equation is $u^2+v^2+2z-5=0$, which has the rotational symmetry about the $z$-axis that you’re looking for. At this stage you don’t really need to convert Yao cylindrical coordinates since you’re now dealing with a surface of revolution for the boundary. Remember to multiply by the Jacobian of the transformation (constant volume scale factor) after you’ve computed the volume of the transformed shape.
