# Derivation of formula for surface integral

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?

• 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. – John Douma Nov 19 '18 at 13:15

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.
• 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? – John Doe Nov 19 '18 at 12:43
• 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. – John Doe Nov 19 '18 at 17:01