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


$\iint_S f \operatorname d S$ 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.


$\iint_A f(x,y)dA$ 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 $A$.

  • $\begingroup$ So what about the surface area? $\endgroup$ – ThanksABundle Jan 27 '15 at 23:53
  • $\begingroup$ @ThanksABundle That's different. Are you taking a calculus course or reading a book? Look for surface integrals. If you want I could show you the formula... $\endgroup$ – Vladimir Vargas Jan 27 '15 at 23:55
  • $\begingroup$ I would not use a $\int \int_A ... dA$ to indicate what is instead $\int \int ... dx~dy$ or $\int \int ... dr~d\theta$. $\endgroup$ – David G. Stork Jan 27 '15 at 23:55
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    $\begingroup$ @Dillon it does, but when the surface is in $\mathbb R^2$ (a flat surface). $\endgroup$ – Vladimir Vargas Jan 27 '15 at 23:55
  • $\begingroup$ @DavidG.Stork I prefer to use the double integral notation $\iint dA$, never $\iint dxdy$, but $\int\int dx dy$ which are iterated integrals from calculus in one-variable. $\endgroup$ – Vladimir Vargas Jan 27 '15 at 23:58

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