I am stuck on this problem. I need to calculate the curvatures but my shape matrix is not happening. Any help is appreciated. So far I have done this:
$X(u,v) = (cosv-(u+v)sinv, sinv+(u+v)cosv, u+2v)$
So I take the partial velocities
$X_u = (-sinv,cosv,1)$ and, $X_v = (-2sinv-(u+v)cosv, 2cosv-(u+v)sinv,2)$
Taking the cross product gives me:
$((u+v)sinv,-(u+v)cosv,u+v)$
Hence the unit normal is:
=$1/\sqrt{2} (sinv,-cosv,1)$
So the shape operators are
$S_p(X_u) = 0$ and, $S_p(X_v) = 1/\sqrt2 (cosv,sinv,0)$
This is where I get stuck. How do I calculate the shape matrix from here? If there is a more efficient way that would be enlightening.
Ah, so the matrix of the shape operator $S_p$ (with respect to the basis $\{X_u,X_v\}$) is $$\begin{bmatrix} a & c \\ b & d \end{bmatrix},$$ where $S_p(X_u) = aX_u + bX_v$ and $S_p(X_v) = cX_u + dX_v$. In your case, you find that $a=b=0$ and $c=-\dfrac{\sqrt2}{u+v}$ and $d=\dfrac1{\sqrt2(u+v)}$. (By the way, you dropped the negative sign in your computation.)
The customary way of computing this is to note that (using $n\cdot X_u = n\cdot X_v = 0$) \begin{align*} S_p(X_u)\cdot X_u &= -n_u\cdot X_u = n\cdot X_{uu} \\ S_p(X_u)\cdot X_v &= -n_u\cdot X_v = n\cdot X_{uv} = n\cdot X_{vu} = S_p(X_v)\cdot X_u \tag{$\star$}\\ S_p(X_v)\cdot X_v &= -n_v\cdot X_v = n\cdot X_{vv}. \end{align*} You then make the second fundamental form matrix $$II = \begin{bmatrix} X_{uu}\cdot n & X_{uv}\cdot n \\ X_{vu}\cdot n & X_{vv}\cdot n\end{bmatrix}$$ and the first fundamental form matrix $$I = \begin{bmatrix} \|X_u\|^2 & X_u\cdot X_v \\ X_v\cdot X_u & \|X_v\|^2\end{bmatrix},$$ and note that the equations ($\star$) can be written in matrix form as $$I\begin{bmatrix}a & c\\b& d\end{bmatrix} = II,$$ so $S_p = I^{-1}II$.
(You might want to download my differential geometry text for more examples and explanations.)