Standard line element example 
In Euclidean three-space, we can define paraboloidal coordinates $(u,v,\phi)$ via
  \begin{align*}
x = uv\cos\phi,\quad y = uv\sin\phi,\quad z = \frac{1}{2}(u^2-v^2)
\end{align*}
  Find $ds^2 = dx^2 + dy^2 + dz^2$

So I just want people to check my working and answer for this question. I have that
\begin{align*}
dx &= v\cos\phi\,du + u\cos\phi\,dv - uv\sin\phi\,d\phi\\
dy &= v\sin\phi\,du + u\sin\phi\,dv + uv\cos\phi\,d\phi\\
dz &= u\,du - \frac{1}{2}v^2 + \frac{1}{2}u^2 - v\,dv\\
\\
dx^2 &= du^2v^2\cos^2\phi + 2du\,dv\,uv\cos^2\phi - 2du\,d\phi\,uv^2\sin\phi\cos\phi\\
&+ dv^2u^2\cos^2\phi - 2dv\,d\phi\,u^2v\sin\phi\cos\phi + d\phi^2u^2v^2\sin^2\phi.\\
\\
dy^2 &= du^2v^2\sin^2\phi + 2du\,dv\,uv\sin^2\phi + 2du\,d\phi\,uv^2\sin\phi\cos\phi\\
&+ dv^2u^2\sin^2\phi + 2dv\,d\phi\,u^2v\sin\phi\cos\phi + d\phi^2u^2v^2\cos^2\phi.\\
\\
dz^2 &= du\,u^3 - du\,uv^2 + \frac{(u^2-v^2)^2}{4}\\
&- dv\,u^2v + dv\,v^3 + du^2u^2 - 2du\,dv\,uv + dv^2v^2.\\
\\
\Rightarrow ds^2 &= du^2v^2\cos^2\phi + 2du^2uv\cos^2\phi + du^2u^2\cos^2\phi\\
&+ d\phi^2u^2v^2\sin^2\phi + du^2v^2\sin^2\phi + 2du^2uv\sin^2\phi\\
&+ du^2u^2\sin^2\phi + d\phi^2u^2v^2\cos^2\phi + du^2u^2 - du\,uv^2\\
&+ du\,u^3 - 2du\,dv\,uv + dv^2v^2 + dv\,v^3 - dv\,u^2v + \frac{(u^2-v^2)^2}{4}.\\
\\
\Rightarrow ds^2 &= (2u^2 + 2uv + v^2)du^2 + v^2dv^2 + u^2v^2d\phi^2\\
&- du\,uv^2 + du\,u^3 - 2du\,dv\,uv + dv\,v^3 - dv\,u^2v + \frac{(u^2-v^2)^2}{4}
\end{align*}
I don't understand why I have these extra terms!? I know this will take a while but just think that I have had to type set it into Latex as well :). So if anyone has the time to check this answer I would be very appreciative.

\begin{align*}
dz &= u\,du - v\,dv\\
\\
dz^2 &= u^2du^2 - 2uv\,du\,dv + v^2dv^2\\
\\
\Rightarrow ds^2 &= du^2v^2\cos^2\phi + 2du\,dv\,uv\cos^2\phi - 2du\,d\phi\,uv^2\cos^2\phi\\
&+ dv^2u^2\cos^2\phi - 2dv\,d\phi\,u^2v\sin\phi\cos\phi + d\phi^2u^2v^2\sin^2\phi\\
&+ du^2v^2\sin^2\phi + 2du\,dv\,uv\sin^2\phi + 2du\,d\phi\,uv^2\sin\phi\cos\phi\\
&+ dv^2y^2\sin^2\phi + 2dv\,d\phi\,u^2v\sin\phi\cos\phi + d\phi^2u^2v^2\cos^2\phi\\
&+ du^2u^2 + dv^2v^2 - 2du\,dv\,uv\\
\\
\Rightarrow ds^2 &= (u^2 + v^2)du^2 + (u^2 + v^2)dv^2 + (u^2v^2)d\phi^2
\end{align*}
