To find the volume of a certain solid cone A solid cone is obtained by connecting (with a line segment in $3$ dimensional Euclidean space ) every point of a plane region $S$ with a vertex not in the plane $S$ . Let $A$ denote the area of $S$ and let $h$ denote the altitude of the cone . Then what is the volume of the cone ? Please help ; thanks in advance 
 A: Hint: Take a cross-section of the solid using a plane parallel to the $x$-$y$ plane, and $z$ units straight down from the vertex. The cross-section is similar to $S$, and has area equal to $\left(\dfrac{z}{h}\right)^2A$.
A: It depends on the polar equation of the base profile $ r =f(\theta)$.We can sum up for cones of variable radius as:
$$ A = \int \frac12 \sqrt{r^2+h^2} \cdot r\cdot d \theta,  \; V =  \int \frac12 \cdot \frac13 \cdot r^2 \cdot h\cdot d \theta. $$
In case of polar symmetric and certain other specific profiles a formula can be found,  but in general there is no general formula connecting integrated $A,V.$
A: The answer may vary and it strongly depends on the meaning of 'connecting'.
But if you mean connecting with a line segment in euclidean $3D$ space, then the volume is $\frac 13 S\cdot h$.
Edit
Err... of course I meant  $\frac 13 A\cdot h$.
Like others already said (or suggested), the volume can be obtained by integration. If you notice the sections of the cone parallel to the base plane are figures similar to $S$, and the similarity ratio is $x:h$, where $x$ and $h$ are distances of the section and the base, respectively, from the vertex; so the section area $A(x)=\left(\frac xh\right)^2A$; then the volume differential
$$dV = A(x)\,dx = \frac A{h^2}\,x^2\,dx$$
and the volume
$$
V = \int\limits_{x=0}^h dV 
  = \frac A{h^2}\,\int\limits_{x=0}^h x^2\,dx 
  = \frac A{h^2}\, \left[\frac{x^3}3\right]_{x=0}^h 
  = \frac A{h^2}\, \left(\frac{h^3}3-\frac{0^3}3\right) 
  = \frac 13 Ah
$$
