# Compute the volume bounded by a parametric surface

Given a smooth (i.e. $C^1$ continuous) closed parametric surface $(x,y,z) = S(u,v)$, how do I compute the volume $\mathcal{V}$ bounded by it?

After browsing through a couple of books, I'm sure it is related to the divergence theorem (Gauss theorem), or perhaps Fubini's theorem. But I couldn't a proper proof for the following steps:

$$\mathcal{V} = \iiint_V \, dV = \iiint_V \, dx \, dy \, dz = \iiint_V \, dz \, dx \, dy = \iiint_V \frac{\partial}{\partial z} \left( z \right) \, dz \, dx \, dy = \iint_A \left( \int \frac{\partial}{\partial z} \left( z \right) \, dz \right) \, dx \, dy = \iint_A z \, dx \, dy.$$

In addition, I stumbled upon the expression

$$\iint_S z \, \cos(\gamma) \, dS = \iint_A z \, dx \,dy,$$

where $\cos(\gamma)$ — referred to as a direction cosine — is the $z$-component of the normal vector $n$ to the surface $S$. I don't quite understand why the left- and right-hand side are equivalent?

Anyway, provided that the above expressions are correct, the last step would be to rewrite

$$\iint_A z \, dx \, dy = \iint_\Omega z \left( \frac{\partial x}{\partial u}\frac{\partial y}{\partial v} - \frac{\partial y}{\partial u}\frac{\partial x}{ \partial v} \right) \, du \, dv,$$

which I can compute.

We are given a "body" $B\subset{\mathbb R}^3$ with boundary surface $\partial B=:S$, and it so happens that we have a $C^1$-parametrization $${\bf f}:\quad A\to{\mathbb R}^3,\quad (u,v)\mapsto{\bf f}(u,v)=\bigl(f_1(u,v),f_2(u,v), f_3(u,v)\bigr)$$ of $S$, where the parameter domain $A$ is a nice subset of the $(u,v)$-plane. It is understood that ${\bf f}$ is "essentially" (i.e., apart from a set of two-dimensional measure $0$) injective and that the (unnormalized) normal $${\bf n}(u,v):={\bf f}_u(u,v)\times {\bf f}_v(u,v)$$ is $\ne{\bf 0}$ almost everywhere and points to the outside of $B$. (An example is the well known parametrization of the unit sphere $S^2$ by means of geographical coordinates.)
Given any $C^1$-vector field ${\bf x}\mapsto{\bf p}({\bf x})$ defined in a neighborhood of $B$ Gauss' theorem says the following: $$\int\nolimits_B{\rm div}\,{\bf p}\ {\rm d}({\bf x})=\int\nolimits_{\partial B} {\bf p}\cdot d\vec\omega=\int\nolimits_A {\bf p}\bigl({\bf f}(u,v)\bigr)\cdot\bigl({\bf f}_u(u,v)\times {\bf f}_v(u,v)\bigr)\ {\rm d}(u,v)\ .\tag{1}$$ Here $d\vec\omega$ is the "vectorial surface element"; it "unpacks" in the way displayed on the right.
Now we choose ${\bf p}$ in such a way that ${\rm div}\,{\bf p}({\bf x})\equiv1$. In this way the left side of $(1)$ is just ${\rm vol}(B)$. A possible choice for ${\bf p}$ is $${\bf p}({\bf x}):=(0,0,x_3)\ .$$ The integrand on the right side then becomes $$q(u,v):=f_3(u,v)\bigl(f_{1.u}f_{2.v}-f_{1.v}f_{2.u}\bigr)_{(u,v)}$$ (this is the right side of your last formula). In this way we finally obtain $${\rm vol}(B)=\int\nolimits_A q(u,v)\ {\rm d}(u,v)\ .$$
• Thanks for your answer Christian. The first part is clear, but why do we need a vector field $\mathbf{x}$, I only mentioned scalar fields in my question? Also, how is $d\vec{\omega}$ defined? In general, I just wondered which steps in my original question are straightforward and which steps require theorems such as Gauss' or Fubini's theorem. Also, references to books are most welcome. Mar 13, 2014 at 16:17
• @Ailurus: See my update. Gauss' theorem is about vector fields. Your steps don't make much sense, among others because nobody knows what your $A$ is. Mar 13, 2014 at 16:31
• Yes I know, but I'm considering a scalar field and wondered whether there would be another way to turn the triple integral into a double integral. In my steps, the $A$ in the double integral is just the domain $V$ minus one of the axes, in this case $z$. Then the next step is to use change of variables to get to a double integral that I can actually compute. It's these two steps that are not completely evident to me. Perhaps I should have mentioned this a bit clearer in my original question, my apologies. Mar 14, 2014 at 12:06