In this [Paper](http://www.ams.org/journals/proc/1976-057-01/S0002-9939-1976-0402611-2/S0002-9939-1976-0402611-2.pdf) there is a proof that a closed plane curve of length
$L$ and curvature bounded by $K$ can be contained inside a circle of radius
$L/4 - (\pi - 2)/2K$. Are there similar results for smooth surfaces in $\mathbb{R}^3$?



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](http://bookstore.ams.org/gsm-69-r)*, Theorem 7.45. 



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> **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 $$\textrm{diam} \, S \leq \frac{\pi}{\sqrt{\inf\_{p \in S} K(p)} }.$$
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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
$$\textrm{diam} \, S \leq \frac{\big(\int\_S H(p) \big)^2}{4 \, A(S) \sqrt{\inf\_{p \in S} K(p)} }.$$


