Area and volume relation (multivariable calculus problem) Let $D \subset R^3$ a region over the plane $z=0$, if $C$ is the cone of base $D$ and vertex at $(0,0,1)$, show that $Vol(C)=\dfrac{1}{3}A(D)$, where $A(D)$ is the area of the region $D$.
First I thought of applying Gauss (divergence) theorem over the region $W$, where $W$ is the region enclosed by the cone and the function $F=(x,y,z)$, in that case, we would have $$\iiint_W div(F)dV=\iint_{\partial W} F.dS.$$
We have $$\iiint_W div(F)dV=3\iiint_W 1dV=3Vol(C).$$ So I've tried to show $$\iint_{\partial W} F.dS=A(D)$$ but I couldn't. 
I've parametrized the cone ($\partial W$) by $T(u,v)=(u\cos(v),u\sin(v),u)$, then $T_u \times T_v=(-u\cos(v),-u\sin(v),u)$, so $$\iint_{\partial W} F.dS=\iint_{[0,1]\times [0,2\pi]} (u\cos(v),u\sin(v),u).(-u\cos(v),-u\sin(v),u)dudv=-u^2+u^2=0$$
Clearly the area of$D$ is not $0$. I don't know what I am doing wrong, I would appreciate corrections and suggestions.
 A: There are a couple of problems with your parametrization. First, you don't know that $D$ is a disk. Your parametrization is also of the cone with vertex at the origin rather than $(0, 0, 1)$. (The former turns out to be easier for computation, though.) Most importantly, your parametrization misses the bottom surface of the cone (even in the case where $D$ is a disk). Also, if I'm reading your notation correctly, you want to normalize the surface normal vector in your computation of $\int F.dS$. (It's also possible to derive the result by noting that the $z = h$ plane of the cone is just $D$ shrunk by a factor of $(1 - h)$ in both directions, which means that the total volume is $\int_0^1 (1 - h)^2 \text{vol}(D)\,dh = \frac{1}{3}\text{vol}(D)$; it looks like you're specifically asking for a multivariable-calculus derivation, though.)
Let's take the vertex of the cone $C$ to be at the origin and its base to be in the $z = 1$ plane (an assumption that obviously doesn't change its volume). Consider the field $F = (x, y, z)$. The boundary of the cone $C$ consists of two parts: the "lower" part $P$ below the base, which is exactly as you described, and the surface $D$. (If this is unclear, think of a solid cylinder: The boundary comprises not only the "vertical edges" of the cylinder, but also the two ends. It's the same with the cone, except one end of the cylinder has been squashed to a point.) By the argument you gave, the integral $\int_P F.dS$ vanishes. On the other part $D$ of the boundary, the (outward) normal vector $\hat n\equiv (0, 0, 1)$. Thus $F.\hat n = z \equiv 1$ on $D$, and we have $\int_{D} F.dS = \int_{D} 1 = \text{Area}(D)$. Putting it all together,
\begin{align*}
\text{Vol}(C) 
 &= \int_C 1\; dV  \\
 &= \frac{1}{3}\int_C \text{div} F\; dV  \\
 &= \frac{1}{3}\int_{\partial C} F.dS  \\
 &= \frac{1}{3}\int_{P} F.dS + \frac{1}{3}\int_{D} F.dS \\
 &= \frac{1}{3}\text{Area}(D).
\end{align*}
