Volume of a Cone

Apr 2010
31
2
USA
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?

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earboth

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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

View attachment 18295
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
 
Jun 2008
125
14
Plymouth
It is a lot easier to do it the other way but here it goes.

The equation of a cone is \(\displaystyle x^2+y^2=R^2z^2\).

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

Now we let \(\displaystyle z\in\left[0,h\right]\)

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

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

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

\(\displaystyle R^2z^2-y^2=x_0^2\) or \(\displaystyle y=\pm\sqrt{R^2z^2-x_0^2}\)

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

Next we find the area \(\displaystyle A(x_0)=2hr-\int_{-\sqrt{r^2-x_0^2}}^{\sqrt{r^2-x_0^2}}\dfrac{\sqrt{y^2+x_0^2}}{R}\,dy=\) 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.
 
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Apr 2010
31
2
USA
Thank you! It is complicated beyond my scope but thanks nonetheless.
 

earboth

MHF Hall of Honor
Jan 2006
5,854
2,553
Germany
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

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

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

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

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:

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

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

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

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Apr 2010
31
2
USA
Thanks, I'll try it again using your setup. hopefully i'll get something.