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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

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  • $\begingroup$ Could you add some of your own thoughts? $\endgroup$ – Rory Daulton Dec 19 '15 at 16:18
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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 $\left(\dfrac{z}{h}\right)^2A$.

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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

$$ V = \int\limits_{x=0}^h dV = \frac A{h^2}\,\int\limits_{x=0}^h x^2\,dx = \frac A{h^2}\, \left[\frac{x^3}3\right]_{x=0}^h = \frac A{h^2}\, \left(\frac{h^3}3-\frac{0^3}3\right) = \frac 13 Ah $$

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  • $\begingroup$ Yes ; I mean exactly that ; could you please elaborate how did you get the volume ? Did you mean $\dfrac1{3} Ah$ ? $\endgroup$ – user228168 Dec 18 '15 at 9:36
  • $\begingroup$ @SaunDev Of course, it should be $A$ instead of $S$, area, not the set. Please see the expanded answer. $\endgroup$ – CiaPan Dec 18 '15 at 9:58
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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 $A,V.$

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  • $\begingroup$ There is no suggestion the set $S$ in the cone's base is given by the $r=f(\theta)$ polar equation. It needn't even be connected, let alone simply connected, which is necessary (altough still incufficient) for such description (consider a base composed of three disjoint circles; or just a figure in a shape of letter 'C'). Anyway if you want to 'sum up for cones of variable radius' then you need to obtain a formula for the cone first... $\endgroup$ – CiaPan Dec 18 '15 at 11:30

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