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The two formulas I have for surface integrals are $\int_S \mathbf v \cdot \mathbf n dS$ and $\int_{\gamma} \mathbf v(\mathbf r(u,v)) \cdot \left(\frac {\partial \mathbf r}{\partial u} \times \frac {\partial \mathbf r}{\partial v}\right) dudv$. If these are the same thing then doesn't that mean that $\left(\frac {\partial \mathbf r}{\partial u} \times \frac {\partial \mathbf r}{\partial v}\right)$ has to be equal to $\mathbf n$, which is a unit vector? I don't see how for any parametrization we make, that cross product will always be of unit length.

Can someone explain to me what I'm missing here?

user196766
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    You are correct in that the cross product usually is not a unit vector. But that $dS$ compensates for this because we have $$dS=||\frac{\partial{\bf r}}{\partial u}\times\frac{\partial{\bf r}}{\partial v}||,du,dv.$$ Together with ${\bf n}$ we get the claimed formula. – Jyrki Lahtonen Dec 01 '14 at 16:13
  • @JyrkiLahtonen I seem to recall something like $dxdy = J\ dsdt$, where $J$ is the Jacobian. So is $| \frac{\partial{\bf r}}{\partial u}\times\frac{\partial{\bf r}}{\partial v}|$ equal to the Jacobian of the transformation $dxdy \mapsto dudv$? And if so, how could I prove that? – user196766 Dec 01 '14 at 16:17
  • More or less. I'm a bit hesitant to call it Jacobian, because $\bf r$ is (presumably) a 3D-vector. Anyway, the idea is that in 3D (as well as 2D) the parallelogram with sides $\vec{a}$ and $\vec{b}$ has area $||\vec{a}\times\vec{b}||$. Here we apply this to the infinitesimal vectors $\vec{a}=(\partial r/\partial u),du$ and $\vec{b}=(\partial r/\partial b),dv$. – Jyrki Lahtonen Dec 01 '14 at 16:21
  • So we start with a unit square with sides $dx$ and $dy$ (where the "units" in this unit square are $dx$ and $dy$). Then we transform $dx\ \hat x \mapsto (\partial \mathbf r/ \partial u)du$ and $dy\ \hat y \mapsto (\partial \mathbf r/ \partial v)dv$ which maps that unit square to some parallelogram. The norm of that parallelogram is then $|(\partial \mathbf r/ \partial u)du \times (\partial \mathbf r/ \partial v)dv| = |(\partial \mathbf r/ \partial u)\times (\partial \mathbf r/ \partial v)|dudv$? I'm not entirely sure where that mapping comes from but it does seem to make sense. Thanks. – user196766 Dec 01 '14 at 16:28
  • That's the idea. I do think you start with a tile with dimensions $du \cdot dv$. Those $u$ and $v$ are coordinates parametrizing the surface $S$. No $x$s and $y$s in that parameter spaces. More like $$(x,y,z)={\bf r}(u,v),$$ so all of $x,y,z$ are functions of both $u,v$. – Jyrki Lahtonen Dec 01 '14 at 16:33

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