Question

How does a double integral give the area ? Is it the same as the surface area ?

Answer 1

Without going into the details of computation, a few pictures will probably get the point across the best.

Consider the one-dimensional analogy. The standard definite integral gives the area under a curve over some interval (incidentally a line segment). But we can abstract the notion of the interval to instead be an arbitrary curve. In the first image, I've drawn a visual representation of the definite integral of a function $f(x)$ over an interval $\mathcal I$. In the second, I take the same function $f(x)$, but this time the integration domain is the curve itself, denoted $\mathcal C$.

Imagine $\mathcal C$ was a piece of string that we pick up and tighten so that it takes the same shape as the interval $\mathcal I$. Then the length of $\mathcal C$ could be computed by imagining we integrate the function $f(x)=1$ over this new interval.

It's a similar story in two dimensions. In the third image, we integrate $f(x,y)$ over a region $\mathcal D$, and that gives the volume under the surface $f(x,y)$ over $\mathcal D$. On the other hand, we can also treat a subset of the surface itself as the integration region, denoted $\mathcal S$.

Suppose we cut a $\mathcal S$-shaped piece from the whole surface $f(x,y)$. Then we could try to flatten it like we did the curve $\mathcal C$ in one dimension so that it resembles a flat region like $\mathcal D$, and the area of $\mathcal S$ can be computed by integrating the function $f(x,y)=1$ over this new region.

We have special names for the latter two types of integrals - line integrals and surface integrals - which you might be interested in looking up. The W*k*pedia page on line integrals in particular has a very nice animation showing what it means to compute a line integral when $\mathcal C$ is some path traversing a surface.

PS: Not sure why the last two are rotated. Sorry for the inconvenience. Hope you find this helpful.

To answer your second question, the answer is yes in a certain sense. A surface integral can only be set up for a differentiable surface, which is to say the surface can't have sharp turns, cusps, holes, etc. But, for example, if you have a cube, you can consider each face of the cube as individual surfaces. Then the total surface area of the cube can be computed by integrating along each component surface, then taking the sum of all six integrals.