Surface Integral for the plane

Question

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?

Answer 1

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 $S$.

Answer 2

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.

Answer 3

Those dot products come about by projecting a small area $dS$ on a surface onto a small area $dA$ in the plane. Specifically, we get that $dS\cos{\theta}=dA$ where $dA=dxdy$, for example, and $\theta$ is the angle between $dA$ and $dS$. Note that the angle between two rectangles is the same as the angle between their normal vectors.

If $dS$ were being projected onto the $xz$ plane then we would have $dA=dxdz$.

We know how to integrate over $dxdy$ but not how to integrate over $dS$ so we want to convert $\int dS$ into an integral over $dA=dxdy$. Now, $$dS\cos{\theta}=dxdy\implies dS=\frac{dxdy}{\cos{\theta}}$$

We use the dot product to compute $\cos{\theta}$.

Let $\hat n$ be the normal to the surface at the point around which we consider $dS$. We get $$\hat n\cdot\hat k=\lvert\hat n\rvert\lvert\hat k\rvert\cos{\theta}=\cos{\theta}$$

Therefore, we get that $$dS=\frac{dxdy}{\hat n\cdot\hat k}$$

For the case when we project upon another plane like $xz$, for example, we get that $$dS=\frac{dxdz}{\cos{\theta}}=\frac{dxdz}{\hat n\cdot\hat j}$$ because $\hat j$ is the normal vector to the $xz$ plane.