Let $S$ be a surface given by the parametrization $\vec{r}(u,v)= x(u,v)\vec{i} + y(u,v)\vec{j} + z(u,v)\vec{k}$ for $(u,v)\in D$. If $f(x,y,z)$ is a function defined on $S$, then the surface integral can be found using the formula $$ \iint_S f(x,y,z) dS = \iint_D f(\vec{r}(u,v))\lvert \vec{r}_u\times\vec{r}_u\rvert dA $$

I have seen how to derive this by drawing some pictures, but I am wondering if there is a proof that makes use of the change of variable formula. I have tried a few things, but I can't find anything that gives a Jacobian of $\lvert \vec{r}_u\times\vec{r}_u\rvert$.

Is this possible?

  • $\begingroup$ Every derivation I have seen of this is geometric. The $|\vec r_u\times\vec r_v|dA$ term is the calculation of $dS$ so strictly speaking, we are not changing variables. We are given a parametrization of a surface and calculating the surface integral based on that. $\endgroup$ – John Douma Nov 19 '18 at 13:15

The formula you have written down is merely the definition of the surface integral(and as such doesn't need any proof). Ok, to be slightly more precise, if you define the integral as the sum of surface "area elements" times the value of the function at any point inside that surface, then you see that this gives you the above formula after noting that the vector product is the same as the area of the surface element.

The change of variable formula just shows that you may parametrize your surface S in any way you like and still get the same answer. The Jacobian comes up when we have an integral of the type $\int_{a}^{b} \int_{c}^{d}f(x,y) \,dx\,dy$ and some other params $u,v$ and $u=u(x,y)$ and $v=v(x,y)$ and then $\int_{a}^{b} \int_{c}^{d}f(x,y) \,dx\,dy = \int_{r}^{s} \int_{m}^{n}J(u,v)f(u,v) \,dx\,dy$.

In our case, assuming you have some other set of parameters $(p,q)$ such that $\vec{r}(p,q) = x(p,q)\vec{i}+y(p,q)\vec{j}+z(p,q)\vec{k}$ , then $|\vec{r}_u \wedge \vec{r}_v|dudv= J(u,v)|\vec{r}_p \wedge \vec{r}_q|dpdq$(write down this computation in full if you don't believe me!) and by the change of variable formula, the two integrals(one in which we parametrized using $p,q$ and the other with $u,v$) will be equal.

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    $\begingroup$ The above formula was presented to me as a theorem. The definition was the things with "area elements". So there is no way to choose some change of variables so that the $\lvert \vec{r}_u\times\vec{r}_u\rvert$ becomes the jacobian? $\endgroup$ – John Doe Nov 19 '18 at 12:43
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    $\begingroup$ @John Doe: I edited my answer to include a small discussion about the Jacobian, but I have to insist that the formula above is a consequence of the definition of the integral and has, strictly speaking, nothing to do with a Jacobian. However, given that when you define an integral you don't define it in terms of a certain parametrization, you may actually deduce the formula of change of variables(which does indeed involve the Jacobian) from this integral formula(but not the other way round). $\endgroup$ – Sorin Tirc Nov 19 '18 at 13:25
  • $\begingroup$ Ok, thanks for the answer. It just really looked to me like the $\lvert \vec{r}_u\times\vec{r}_v\rvert$ factor is some kind of Jacobian. $\endgroup$ – John Doe Nov 19 '18 at 17:01

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