(Multivariable) Deriving a Formula for Volume of Cone The problem:
Let $P = (a, b, h), h > 0$, be a point in $\mathbf{R}^3$. If $D$ is a region in the $xy-$plane, the solid cone with base $D$ and apex $P$ is obtained by joining all points of $D$ to $P$ by line segments. If $A$ is the area of the base $D$, derive a formula of the cone in terms of $a, b, h$ and $A$. 
I'm confused where to even start! (I need to use Multivariable Calculus) 
 A: Let $M$ be a circle of radius $R$ with center $(a,b,0)$ i.e $(x-a)^2+(y-b)^2 = R^2$. In view of this equation, we know that our region of integration is:
$$\left\{(x,y):-\sqrt{R^2-(x-a)^2}+b \leq y \leq \sqrt{R^2-(x-a)^2}+b,  -R \leq x \leq R\right\}$$
You can use polar coordinates to simplify the region. Let $x = R \cos \theta, y = R \sin \theta$ then expanding the above we have:
$$x^2-2ax+a^2+ y^2 - 2yb+b^2 = R^2 \iff R^2-2(aR\cos\theta + b R \sin \theta)+(a^2+b^2) = R^2$$
Knowing that $(a,b)$ lies on some circle of radius $\lambda$ (i.e $\lambda^2 = a^2+b^2$) and cancelling some terms above, we have:
$$2R(a \cos\theta + b \sin \theta) = \lambda^2 \iff R = \frac{\lambda^2}{2(a \cos \theta + b \sin \theta)}$$
Therefore the volume is given by:
$$\int_{0}^{2\pi}\int_{0}^{\frac{\lambda^2}{2(a \cos \theta + b \sin \theta)}} r\int_0^h dz \ dr \ d\theta$$
A: We all know that the volume of this cone $C$ computes to
$${\rm vol}(C)={1\over3}{\rm area}(D)\cdot h\ .\tag{1}$$
A calculus proof of $(1)$ can be given as follows: Let
$C_z:=\{(x,y)\>|\>(x,y,z)\in C\}$ be the section of $C$ cut out by the horizontal plane at level $z$. One then has 
$${\rm area}(C_z)=\left({h-z\over h}\right)^2\cdot{\rm area}(D)\ .$$
Fubini's theorem now gives
$$\eqalign{{\rm vol}(C)&=\int_C 1\>{\rm d}(x,y,z)=\int_0^h\int_{C_z}1\>{\rm d}(x,y)\>dz\cr &=\int_0^h{\rm area}(C_z)\>dz={\rm area}(D)\int_0^h\left({h-z\over h}\right)^2\>dz\cr
&={1\over3}{\rm area}(D)\cdot h\ .\cr}$$
A: There's a really slick proof of this using the divergence theorem that I saw years ago, I think in "Grad, Div, Curl, and All That."  Position the apex of the cone at the origin and the base in a plane parallel to the $xy-$plane.  Let $F(x,y,z) = (x,y,z)$ and apply the divergence theorem.  On the left-hand side, you get 3 times the volume.  On the right hand side, the flux through the lateral surface is $0$, since the normal is orthogonal to the flow, and on the base, it's equal to the height times the area of the base.  QED.  I apologize for not putting this in formulas, but my MathJax skills just aren't up to it, and I like the proof too much not to share it.     
