Volume of a Cone

**bilalsaeedkhan** · July 22nd 2010, 08:23 PM

Can someone help me with calculating the volume of a Cone by taking a vertical cross-sectional area of the cone and integrating it? I know how to solve it for the horizontal cross-section, which gives an area of circle. But if we take a vertical cross-section then we get a Hyperbola, am I correct?

**earboth** · July 22nd 2010, 10:14 PM

Originally Posted by bilalsaeedkhan

Can someone help me with calculating the volume of a Cone by taking a vertical cross-sectional area of the cone and integrating it? I know how to solve it for the horizontal cross-section, which gives an area of circle. But if we take a vertical cross-section then we get a Hyperbola, am I correct? <=== No

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If the cut is parallel to the axis of the cone then you'll get parabolas as cross-sections.

Have a look here: Conic section - Wikipedia, the free encyclopedia

**bilalsaeedkhan** · July 22nd 2010, 10:36 PM

Originally Posted by earboth

If the cut is parallel to the axis of the cone then you'll get parabolas as cross-sections.

Have a look here: Conic section - Wikipedia, the free encyclopedia

My case is the third figure in your link, which is a hyperbola. Is the cross-section that I have taken in my original attachment a parabola or a hyperbola?

**fobos3** · July 22nd 2010, 10:59 PM

It is a lot easier to do it the other way but here it goes.

The equation of a cone is $x^2+y^2=R^2z^2$ .

Suppose the radius of the base at $z=h$ is $r$ . Then we have $Rh=r$ or $R=\dfrac{r}{h}$

Now we let $z\in\left[0,h\right]$

Next we have the equation of a general plane $Ax+By+Cz=D$ . We want to construct a plane $\mathcal{P}||(\hat{\textbf{j}},\hat{\textbf{k}})$

This would mean that $x=const$ or $\mathcal{P}\rightarrow x=x_0$ .

Now lets intersect this plane with our cone. This gives us the curve

$R^2z^2-y^2=x_0^2$ or $y=\pm\sqrt{R^2z^2-x_0^2}$

Now at $z=h$ $y=\pm\sqrt{r^2-x_0^2}$ which will be our limit of integration. Keep in mind that the vertex of the cone is at $z=0$ . We know that $z>0$ therefore $z=\dfrac{\sqrt{y^2+x_0^2}}{R}$

Next we find the area $A(x_0)=2hr-\int_{-\sqrt{r^2-x_0^2}}^{\sqrt{r^2-x_0^2}}\dfrac{\sqrt{y^2+x_0^2}}{R}\,dy=$ and I'm going to stop here as it is getting quite messy. I guess that helps to a point but I don't see any point in continuing.

**bilalsaeedkhan** · July 22nd 2010, 11:07 PM

Thank you! It is complicated beyond my scope but thanks nonetheless.

**earboth** · July 23rd 2010, 06:46 AM

Originally Posted by bilalsaeedkhan

My case is the third figure in your link, which is a hyperbola. Is the cross-section that I have taken in my original attachment a parabola or a hyperbola?

Of course you are right and I apologize for the confusion.

Obviously a hyperbolic cross-section is hard to handle as fobos3 has demonstrated. So I suggest to use parabolic cross-sections.

According to Archimedes a segment of a parabola has an area

$A = \frac43 \cdot \underbrace{\frac12 \cdot 2 b \cdot h}_{area\ of\ triangle} = \frac43 b h$

Use the proportion $\dfrac hs = \dfrac{2R-x}{2R}~\implies~h=\dfrac{s(2R-x)}{2R}$

Use Pthagorian theorem to calculate the length of $b = \sqrt{R^2-(R-x)^2}$

Caution: This step probably contains a mistake which causes a wrong final result! I wasn't able to spot my error.

Now the colume of the cone should be:

$V=\displaystyle{\int}_0^{2R}\left(\frac43 \cdot \dfrac{s(2R-x)}{2R} \cdot \sqrt{R^2-(R-x)^2}\right) dx$

My computer tells me that $V = \dfrac \pi3 \cdot R^2 \cdot s$ which is definitely too large.

(... but it looks like a very near miss)

**bilalsaeedkhan** · July 23rd 2010, 11:20 PM

Thanks, I'll try it again using your setup. hopefully i'll get something.

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