# Derivation of the formula for parametrized surface area element

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.

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

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 $$S$$.

• 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.