Surfaces contained in a ball

A possible generalization is an upper bound for the diameter of a surface $S \subset \mathbb{R}^3$ in terms of its area $A(S)$ and its gaussian and mean curvatures $K$, $H$. Let me give an answer in the case of positively curved surfaces.

We define an ovaloid as a compact connected surface $S \subset \mathbb{R}^3$ with $K(p) >0$ for all $p \in S$. If $S$ is an ovaloid then it is orientable, and by using Hadamard-Stoker theorem it follows that its inner domain is a convex subset of $\mathbb{R}^3$ and that its Gauss map $N \colon S \to \mathbb{S}^2$ is a diffeomorphism. In particular, $S$ is a topological sphere.

The following classical estimate for the diameter of an ovaloid is due to Bonnet. Modern proofs are based on Hopf-Rinow theorem, see for instance

S. Montel, A. Ros, Curves and Surfaces, Theorem 7.45.

Theorem (Bonnet). Let $S \subset \mathbb{R}^3$ be a closed connected surface whose Gaussian curvature $K$ satisfies $\inf_{p \in S} K(p) >0$. Then $S$ is compact, that is $S$ is an ovaloid, and its diameter satisfies $$\textrm{diam} \, S \leq \frac{\pi}{\sqrt{\inf_{p \in S} K(p)} }.$$

Now, by using the isoperimetric inequality we can show that for any ovaloid the inequality $$\bigg(\int_S H(p) \bigg)^2 \geq 4 \pi A(S)$$ holds, see Exercise 10 p. 196 in the book quoted above. Substituting into the inequality provided by Bonnet's theorem, we finally obtain $$\textrm{diam} \, S \leq \frac{\big(\int_S H(p) \big)^2}{4 \, A(S) \sqrt{\inf_{p \in S} K(p)} }.$$