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*}