Parametrisation of Surfaces!! An area A of $(x, y)$ plane bounded by $y$-axis and the parabola of the equation
$x=6-y^2$
Furthermore, a surface $F$ is given by the portion of the graph of the function $h (x, y) = 6-x-y^2$ which fulfils that $x\geq 0$ and $z\geq 0. $

*

*Determine a parametrization for $A$ and $F$.
Let $B$ denote the completed spatial regions related to those who are (vertically) between $A$ and $F$.


*Determine a parametrization for $B$.
I am stuck on this problem for quite a while now. Any help with the parametrization will be really helpful. Thank you :)

I did this by hand, so please verify these answers before trusting them. Notice that I maintained the same domains for all functions.
$A$ is defined by
$
\\x_A(t,u,v)=t
\\y_A(t,u,v)=u\sqrt{6-t}
\\z_A(t,u,v)=0
$
where 
$
\\0 \le t \le 6
\\-1 \le u \le 1
\\0 \le v \le 1
$
$F$ is defined by
$
\\x_F(t,u,v)=t
\\y_F(t,u,v)=u\sqrt{6-t}
\\z_F(t,u,v)=u^2(t-6)-t+6
$
where 
$
\\0 \le t \le 6
\\-1 \le u \le 1
\\0 \le v \le 1
$
And finally, $B$ is defined by
$
\\x_B(t,u,v)=t
\\y_B(t,u,v)=u\sqrt{6-t}
\\z_B(t,u,v)=v(u^2(t-6)-t+6)
$
where 
$
\\0 \le t \le 6
\\-1 \le u \le 1
\\0 \le v \le 1
$
