I have come across the answer to a surface integral here:


And at one stage it says:

$$ds=\frac{dxdy}{|n\cdot k|}$$

I don't understand this step or where this formula comes from for surface integrals.

Can anyone explain why this is the case and what $k$ is?


  • $\begingroup$ Are you looking for a rigorous derivation or an intuitive explanation? $\endgroup$ – John Douma Nov 1 '15 at 0:06
  • $\begingroup$ Intuitive preferably although a rigorous definition would be okay if it is comprehensible to me. $\endgroup$ – Peter Nov 1 '15 at 0:11

The formula you see is the result of projecting $ds$ onto the plane. When we calculate any integral we are taking an infinite sum of small pieces. In this case, $ds$ represents an infinitesimal piece of surface area.

Imagine a surface in $3D$ space and a little square on the surface. This little square is $ds$ and it will have a unit normal vector pointing outward. That is the $\hat n$ in the formula. How does $ds$ relate to $dxdy$? If you draw vertical lines from the corners of $ds$ down to the plane, they will touch the corners of $dxdy$. It is important that you either draw this or visualize it so that you can understand it.

How does the area represented by $ds$ relate to the area represented by $dxdy$? If you have drawn a picture you can see that $ds\cos\theta=dxdy$ where $\theta$ is the angle between $ds$ and $dxdy$. That is the critical observation. The rest is calculation.

The angle $\theta$ between the two infinitesimal surfaces is the same as the angle between the unit normal vector $\hat n$ and the unit normal vector in the plane, which is $\hat k$. Using the formula for the dot product gives us $\hat n \cdot\hat k=|\hat n||\hat k|\cos\theta$. But $\hat n$ and $\hat k$ are unit vectors so this gives $\hat n\cdot\hat k=\cos\theta$.

Therefore, $ds\cos\theta=dxdy\implies ds=\frac{dxdy}{\cos\theta}\implies ds=\frac{dxdy}{\hat n\cdot\hat k}$.

Notice that this formula doesn't work if $\cos\theta = 0$. If that is the case then $ds$ is perpendicular to the plane so you would project it onto the $yz$ plane.


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