Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Doesn't a 2D shape have a surface? In that case, is it incorrect to call the area of it surface area?

share|cite|improve this question

closed as unclear what you're asking by Andrey Rekalo, Branimir Ćaćić, Dan Rust, azimut, Start wearing purple Aug 8 '13 at 16:31

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

Area for a plane and surface area for 2D curved surfaces embedded in 3-space. – Narasimham May 2 at 12:08

Yes, unless someone has some specific shade of meaning in mind, "surface area" and "area" are the same. I guess someone might ask, for example, "How much area does a tent cover?" in which case, he probably wants to know how much area on the ground is under the tent, and not the actual surface area of the tent itself. But usually, surface area is not different from area.

share|cite|improve this answer

The answer to your first question is Yes, because every single object that you can imagine on a $2D$ plane has a specific surface, which is essentially the shape itself!

Regarding your second assumption I will say Yes, it is incorrect to call "area" and "surface area" referring to $2D$ objects, because the moment you start talking about a surface you imply that you are considering a 2D object "inside" the third dimension.

To make an example, let $S$ be the area of a circle of radius $r$, then you know exactly it's area by: $$ S=\pi\cdot r^2 $$ But, when you take this object into the third dimension and "stretch" the vertexes, the now Surface area created is different because now it counts the movement inside a new dimension, even if this shape is still with height $=0$!

Numerically you can have the element $S'$ inside the third dimension, calculated by a particular integral, because your square is now considered as a function over the $xy$ plane ($z$): $$ S'=\iint z\:\mathrm{d}x\mathrm{d}y $$ So now the normal shape it is the same, but it does change by a correct reference inside the third dimension as the effective "Surface" of the new $3D$ object.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.