# Surface Integral for the plane

The formula for surface integral is given by:

$$\iint_S \overrightarrow F \cdot \hat n \space dS$$

I want to know about the $dS$ part.

For example: If the plane lies in the first octant then it is:

$$dS = \frac{dx \space dy}{|\hat k \cdot \hat n|}$$

There are also other case such as:

$$dS = \frac{dz \space dy}{|\hat i \cdot \hat n|}$$

and $$dS = \frac{dx \space dz}{|\hat j \cdot \hat n|}$$

How do I exactly decide which $dS$ formula to choose i.e. if the plane lies in fourth or second octant. How will I choose?

• Do you speak about surfaces immersed in $\mathbb R^3$? And if yes, what is $\hat i$, $\hat j$ and $\hat k$? – frog Nov 18 '15 at 8:53
• The question says to evaluate surface integral for $\overrightarrow F = 18z \hat i -12 \hat j + 3y \hat k$ where plane $2x + 3y + 6z = 12$ lies in first octant. – user963241 Nov 18 '15 at 9:00
• $dS$ depends on the parametrization of the surface, not on where it lies per se. The latter will instead determine the bounds of integration. – amd Nov 18 '15 at 9:12
• I suggest you use the natural parametrization of the plane and then compute the surface element using it. – Mark Fantini Nov 18 '15 at 9:22
• According to the given solution, I simply need to pick which $dS$ value to use and for first octant I was told to use $dS = \frac{dx \space dy}{|\hat k \cdot \hat n|}$ but I don't quite understand why. – user963241 Nov 18 '15 at 9:38

To calculate a surface integral over a vector field you need parametrization of that oriented surface. Without too much of rigour:

\begin{align}\iint_S {\mathbf v}\cdot\mathrm d{\mathbf {S}} &= \iint_S \left({\mathbf v}\cdot {\mathbf n}\right)\,\mathrm dS\\&{}= \iint_T \left({\mathbf v}(\mathbf{x}(s, t)) \cdot {\left({\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right) \over \left\|\left({\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right)\right\|}\right) \left\|\left({\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right)\right\| \mathrm ds\, \mathrm dt\\&{}=\iint_T {\mathbf v}(\mathbf{x}(s, t))\cdot \left({\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right) \mathrm ds\, \mathrm dt.\end{align}

$d{\mathbf {S}}$ means that you are integrating vector field over an oriented surface. When you "extract" a unit normal vector from it you are left with a scalar function and $dS$ which correspond to a "normal" (scalar) surface integral. In this notation $$dS= \left\|\left({\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right)\right\| \mathrm ds\, \mathrm dt$$

where $\mathbf{x}(s,t): T\rightarrow S$ is a parametrization of your surface.

Also please note how $\left\|\left({\partial \mathbf{x} \over \partial s}\times {\partial \mathbf{x} \over \partial t}\right)\right\|$ cancel each other and simplify calculations.

• Unfortunately, the book doesn't use this parametrization method. – user963241 Nov 18 '15 at 12:04
• I'm sure it does. It is the definition of surface integral. Notation might be different though. But idea is exactly the same. – luka5z Nov 18 '15 at 14:21
• It uses the $dS$ values as given in my question if it means the same like this as shown here – user963241 Nov 18 '15 at 14:39

The plane (which appears only in a comment of yours) is of course infinite. I guess you are told to integrate over the triangle $S$ cut out of this plane by the first octant. This triangle projects down to the $(x,y)$-plane to the triangle $$S':=\left\{(x,y)\>\biggm|\>0\leq x\leq 6, \ \>0\leq y\leq4-{2\over3}x\right\}\ .$$ Now use the parametrization $$\phi:\quad S'\to S,\qquad (x,y)\mapsto \left(x,y,\ 2-{1\over3}x-{1\over2}y\right)$$ to compute the requested flow.

In order to proceed further we need the exact description of the (supposedly nonconstant) vector field $\vec F$ and the intended orientation of $S$.

• Is the $z$ always $0$ in first octant? – user963241 Nov 18 '15 at 12:12
• In the first octant all coordinate variables $x$, $y$, $z$ are $\geq0$, resp. $>0$, if the open octant is meant. – Christian Blatter Nov 18 '15 at 15:50