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.


1 Answer 1


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 $${\rm vol}(B)=\int\nolimits_A q(u,v)\ {\rm d}(u,v)\ .$$

  • $\begingroup$ 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. $\endgroup$
    – Ailurus
    Mar 13, 2014 at 16:17
  • 1
    $\begingroup$ @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. $\endgroup$ Mar 13, 2014 at 16:31
  • $\begingroup$ 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. $\endgroup$
    – Ailurus
    Mar 14, 2014 at 12:06

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