When does a double integral represent a surface area, and when does it represent a volume? When does $\int_Af(x,y)dA$ represent a surface area geometrically, and when does it represent a volume?
In my lecture notes I'm told it represent the volume underneath the surface $z=f(x,y)$, but I've found examples online computing double integrals to find the surface area of a surface.
Thanks a bundle
 A: $\iint_A f(x,y)dA$ represents the volume underneath the surface $z=f(x,y)$. When $f(x,y)=1$ you get the area of $A$, which is the volume under the plane $z=1$ intersected with $A$. Note that $\iint_AdA$ represents a volume, and the value of this volume (the scalar part) is the area of $A$.
A: $\iint_S f \operatorname d S$ is the Surface Integral of scalar field $f$ over surface $S$.
When the curve can be described by a parameterised vector $S: \vec r(s,t)$, we have
$$\iint_S f(\vec r(s,t))\operatorname d S = \iint_S f(\vec r(s,t))\begin{Vmatrix} \frac{\partial \vec r(s,t)}{\partial s}\times \frac{\partial \vec r(s,t)}{\partial t}\end{Vmatrix} \operatorname d s \operatorname d t$$
Note: the scalar field takes a three dimensional vector as its argument, not simply the curvilinear coordinates.

The surface area of the curve $\;\vec r=\begin{bmatrix}x \\ y\\g(x,y)\end{bmatrix}$ projected above some section $T$ of the x,y plane we use:
$$\begin{align}
\iint_S \operatorname dS
 & = \iint_T \begin{Vmatrix}\dfrac{\partial}{\partial x}\begin{bmatrix}x \\ y\\g(x,y)\end{bmatrix}\times\dfrac{\partial\;}{\partial y}\begin{bmatrix}x \\ y\\g(x,y)\end{bmatrix}\end{Vmatrix}\operatorname d x \operatorname d y
\\[1ex]
& = \iint_T \begin{Vmatrix}\begin{bmatrix}1 \\ 0\\ g_x(x,y)\end{bmatrix}\times\begin{bmatrix}0 \\ 1\\g_y(x,y)\end{bmatrix}\end{Vmatrix}\operatorname d x \operatorname d y
\\[1ex]
& = \iint_T \sqrt{g_x(x,y)^2 +g_y(x,y)^2 + 1}\operatorname d x \operatorname d y
\end{align}$$
Note: here we are using a unit scalar field $f = 1$.

To find the volume between a surface $z=g(x,y)$ and a section $T$ of the x,y plane, we just use:
$$\iint_T g(x,y) \operatorname d x \operatorname d y$$
Which is a totally different beast altogether.
