1
$\begingroup$

The surface integral is given by $$\int_{S} f \,dS = \iint_{D} f(\mathbf{\sigma}(u, v)) \left\|{\partial \mathbf{\sigma} \over \partial u}\times {\partial \mathbf{\sigma} \over \partial v}\right\| \,du\,dv$$

Why is $dS =\left\|{\partial \mathbf{\sigma} \over \partial u}\times {\partial \mathbf{\sigma} \over \partial v}\right\|\,du\,dv$ ? I know that $\left\|{\partial \mathbf{\sigma} \over \partial u}\times {\partial \mathbf{\sigma} \over \partial v}\right\|$ is the magnitude of the normal vector to the differential surface area element but I can't see why this formula is true.

$\endgroup$
2
  • 1
    $\begingroup$ Hint: The area of the parallelagram formed by the vectors $a$ and $b$ is $\|a\times b\|$ $\endgroup$
    – whpowell96
    Apr 3 at 19:35
  • $\begingroup$ @whpowell96 Got it, Thanks! $\endgroup$
    – Sasikuttan
    Apr 3 at 19:42

1 Answer 1

2
$\begingroup$

So, your surface $S$ is presumably parameterized by a function $\sigma(u,v)$ for $(u,v) \in R$. Note that, in particular, the values of $f$ on $S$ are given by $f(\sigma(u,v))$, akin to line integrals that you may have discussed in the past.

So ultimately the discussion becomes: what is the differential surface area element, i.e. what is $\mathrm{d}S$?

To figure this out, we essentially perform the same process you might have seen to justify definitions of Riemann integrals, double integrals, line integrals, etc.

  • Break up the surface at a bunch of points. I assume we're working in $\mathbb{R}^3$, so those points will be $(x_k,y_k,z_k)$.

  • These points will correspond to some $(u_k,v_k) \in R$.

  • Let $\Delta S_k$ denote the area of the $k$th surface "patch", and $\Delta u_k, \Delta v_k$ the dimensions of the rectangle in $R$ it corresponds to.

  • Naturally, for a selection of very very many points, very close together, then $\Delta S_k$ can be approximated by a parallelogram above its surface.


To use a diagram from my university's main calculus textbook of choice - University Calculus, Early Transcendentals (4th edition):

Here, $\mathbf{r}$ is the parameterization of the surface, and $\Delta \sigma_{uv}$ the area of the patch of the surface $S$.

enter image description here


  • Well, we can find the area of this parallelogram, via the cross product of its sides. (Recall: the area of the parallelogram between $\vec{x}$ and $\vec{y}$ is $\|\vec{x} \times \vec{y}\|$.)

  • Naturally, since the partial derivatives of our parameterization give the vectors tangent to the surface, just find $\sigma_u$ and $\sigma_v$ at some corner of the patch. Then $$ \Delta S_k \approx \left\| \frac{\partial \sigma}{\partial u} \times \frac{\partial \sigma}{\partial v} \right\| \, \Delta u_k \, \Delta v_k $$ which becomes, in the limit, $$ \mathrm{d} S = \left\| \frac{\partial \sigma}{\partial u} \times \frac{\partial \sigma}{\partial v} \right\| \, \mathrm{d} u \, \mathrm{d} v $$

Ultimately this is just a rough-and-fast derivation though, with some details and niceness assumptions (e.g. $S$ is "smooth") skipped over.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .