# Surface Area in $R^n$

Let's say we have a closed domain $D \subset R^m$. We then define a function over $D$ explicitly $f:D\rightarrow \mathbb{R}^n$.

We can also state that is our function $f$ is differentiable $k$ times.

Is there a way to describe the surface area covered by $f$ in $\mathbb{R}^n$?

I had an intuition that we can use the integral of the gradient over the space of $D$ in some way to calculate this object. But I'm not sure exactly how to express this.

Any ideas on how we can construct the surface area object?

• By "surface area covered by $f$" do you mean the $m$-dimensional volume of the image $f(D)$? If so, the answer is $\int_D \sqrt{\det G}$ where $G = Df^T Df$ is the Gram matrix of the Jacobian $Df.$ – Anthony Carapetis Feb 1 '18 at 2:14
• In what way is volume different than surface area in what you just described? – Francois Wassert Feb 1 '18 at 2:16
• Area is 2-dimensional volume. When $m \ne 2$ I'm not sure what you mean by "surface area" - for example, if $m=1$ then the image of $f$ probably looks like a curve, so it makes more sense to talk about its 1-volume (length). Perhaps give an example of the dimensions $m,n$ you're really picturing when you talk about surface area? Maybe you actually mean $m=3$ and you're looking at the 2D "surface" of the 3D "blob" $f(D)$? – Anthony Carapetis Feb 1 '18 at 3:01
• Yes. I'm referring to surface in the context of minimal surface: en.wikipedia.org/wiki/Minimal_surface More generally surface area in the sense the infinite sum of points defined by the respective function if that makes sense. From the point of geometric measure theory, if $f$ maps onto a set with a well defined measure, surface area would be the haufsdorff measure. Please let me know if that doesn't make sense. – Francois Wassert Feb 1 '18 at 6:10