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 $