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I am stuck on this question from a tutorial sheet I am going through.

Compute the mean and Gaussian curvature of a surface in $\mathbb{R}^3$ that is given by


for some good functions $f(x)$ and $g(y)$.

I tried calculating the first and second fundamental forms to then find $\kappa=\dfrac{det II}{det I}$ but it seems long and I feel like there should be an easier way. Also that doesn't tell me the mean curvature.

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It is a slightly tedious calculation, yes. The mean curvature is the trace of the 2nd fundamental form. – user20266 Apr 22 '12 at 15:08
@Tomas The mean curvature is the trace of the Weingarten map, to be more precise – Yuri Vyatkin Apr 22 '12 at 15:38
@YuriVyatkin yes, you are right. – user20266 Apr 22 '12 at 16:36

Let us first recall some definitions and useful formulas for a surface in $\mathbb{R}^3$ given by an immersion $F\colon U \rightarrow \mathbb{R}^3$, where $U \subset \mathbb{R}^2$ is an open set.

Denote $F_x = \frac{\partial F}{\partial{x}}$, $F_{x x} = \frac{\partial{F_x}}{\partial{x}}$, and so on.

The components of the first fundamental form are $I_{x x} = F_x \cdot F_x$, $I_{y y}=F_y \cdot F_y$, $I_{x y} = I_{y x} = F_x \cdot F_y$

For the second fundamental form we may use the expression for the unit normal vector $$ n = \frac{F_x \times F_y}{|F_x \times F_y|} $$ so that $II_{x x} = n \cdot F_{x x}$, $II_{y y} = n \cdot F_{y y}$, $II_{x y} = II_{y x} = n \cdot F_{x y}$

The Gaussian curvature $K(x,y)$ has the following expression $$ K(x,y) = \frac{II_{x x} II_{y y} - II_{x y}^2}{I_{x x} I_{y y} - I_{x y}^2} \tag{1} $$ and the mean curvature can be computed by $$ H(x,y) = \frac{I_{x x} II_{y y} - 2 I_{x y} II_{x y} + I_{y y} II_{x x}}{I_{x x} I_{y y } - I_{x y}^2} \tag{2} $$ In the proposed problem we a given a surface represented by a graph of function $z = f(x) + g(y)$, so our immersion has the following form: $$ F(x,y) = \begin{pmatrix} x \\ y \\ f(x) + g(y) \end{pmatrix} $$ and we calculate $F_x = \begin{pmatrix} 1 \\ 0 \\ f' \end{pmatrix}$, $F_y = \begin{pmatrix} 0 \\ 1 \\ g' \end{pmatrix}$, $F_{x x} = \begin{pmatrix} 0 \\ 0 \\ f'' \end{pmatrix}$, $F_{y y} = \begin{pmatrix} 0 \\ 0 \\ g'' \end{pmatrix}$, $F_{x y} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$.

This is enough to find the unit normal $$ n = \frac{(f', -g', 1)^T}{\sqrt{1 + (f')^2 + (g')^2}} $$ so we find the components of the second fundamental form $II_{x x} = \frac{f''}{\sqrt{1 + (f')^2 + (g')^2}}$, $II_{y y} = \frac{g''}{\sqrt{1 + (f')^2 + (g')^2}}$, $II_{x y} = 0$

The components of the first fundamental form are, of course, $I_{x x} = 1 + (f')^2$, $I_{y y} = 1 + (g')^2$, and $I_{x y} = f'g'$.

I would let you to finish the job by substituting these quantities into equations (1) and (2).

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