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:
$\{(x,y,z)\mid\frac{1}{2}(3-2y^2+4y-3x^2)=z,0<z<5/2\}$

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 $\int_{-\sqrt{5}{3}}^{\sqrt{5/3}}\int_{-\sqrt{5/2- (3/2)x^2}}^{\sqrt{5/2- (3/2)x^2}} \left(5/2- (3/2)x^2- u^2\right) dudx$.
