Does every shape have zero volume?

Question

Consider the digram below: the red line ($c$) enclosing an area on the XY plane lies in the yz plane and the blue line is a surface with this line as its boundry curve. Let's say we are trying to work out the volume of this shape. Using the divergence theorem we have: $$\int_v dv=\frac{1}{3}\oint_s \begin{pmatrix} x \\ y\\z\end{pmatrix}\bullet d\vec s$$ Splitting this up into the the blue surface and the surface enclosed by the red line on the zy plane we have: $$\int_v dv=\frac{1}{3}\int_{s_1} \begin{pmatrix} x \\ y\\z\end{pmatrix}\bullet (\hat {\vec n})ds+\frac{1}{3}\int_{s_2} \begin{pmatrix} x \\ y\\z\end{pmatrix}\bullet (-\vec e_x)ds$$ On the yz plane $x=0$ so the second term on the right hand side vanishs and we are left with: $$\int_v dv=\frac{1}{3}\int_{s_1} \begin{pmatrix} x \\ y\\z\end{pmatrix}\bullet (\hat {\vec n})ds=\frac{1}{3}\int_{s_1} \begin{pmatrix} x \\ y\\z\end{pmatrix}\bullet d\vec s_1$$ Where $d \vec s_1=\hat{\vec n} ds$. We can, however, use Stoke's thoerem on this intergral using the fact that: $$\oint_c \vec F \bullet d\vec r=\frac{1}{3}\int_{s_1} \begin{pmatrix} x \\ y\\z\end{pmatrix}\bullet d\vec s_1=-\frac{1}{3}\int_{s_2} \begin{pmatrix} x \\ y\\z\end{pmatrix}\bullet (-\vec e_x)ds=0$$ The minus sign comes due to the orintation of the field? So in the end we therefore get: $$\int_v dv=0$$ i.e. The volume of this shape is 0. Obviously I have done something that is not mathematiclly correct, but I can't see it? Please can someone explain?

Answer 1

In the starred step of your purported application of Stokes' theorem $$ \oint_c \vec F \bullet d\vec r = \frac{1}{3}\int_{s_1} \begin{pmatrix} x \\ y\\z\end{pmatrix}\bullet d\vec s_1 \stackrel{*}{=} -\frac{1}{3}\int_{s_2} \begin{pmatrix} x \\ y\\z\end{pmatrix}\bullet (-\vec e_x)ds = 0, $$ ("trading" an integral over $s_{1}$ with an integral over $s_{2}$), you're implicitly assuming the radial field is the curl of some other field, so that the surface integral is determined by a line integral over the common boundary of $s_{1}$ and $s_{2}$.

But the radial field is not the curl of another field (since, for example, its divergence does not vanish).