2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

$\iint_S f \operatorname d S$ 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.

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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